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Introduction β’ This chapter extends on what you have learnt in FP1 β’ You will learn how to find the complex roots of numbers β’ You will learn how to use De Moivreβs theorem in solving equations β’ You will see how to plot the loci of points following a rule on an Argand diagram β’ You will see how to solve problems involving transforming a set of values in one plane into another plane π§ = π(πππ π + ππ πππ) Further complex numbers y You can express a complex number in the form z = r(cosΞΈ + isinΞΈ) z (x,y) r You should hopefully remember the modulus-argument form of a complex number z = x + iy from FP1 The value r is the modulus of the complex number, its distance from the origin (0,0) The argument is the angle the complex number makes with the positive x-axis, where: -Ο < ΞΈ β€ Ο To show this visuallyβ¦ y ΞΈ x x r is the modulus of z, its absolute value ο This can be calculated using Pythagorasβ Theorem π= π§ = By GCSE trigonometry, length x = rcosΞΈ and length y = rsinΞΈ π₯ = ππππ π π¦ = ππ πππ π₯2 + π¦2 π§ = π₯ + ππ¦ π§ = ππππ π + πππ πππ π§ = π(πππ π + ππ πππ) Replace x and y using the values above Factorise by taking out r 3A π§ = π(πππ π + ππ πππ) Further complex numbers y You can express a complex number in the form z = r(cosΞΈ + isinΞΈ) Express the following complex number in the modulus-argument form: To do this you need to find both the argument and the modulus of the complex number ο Start by sketching it on an Argand diagram π§ = 2 πππ 5π 5π + ππ ππ 6 6 This is the argument 1 r ΞΈ x β3 ο The βyβ part is positive so will go upwards z = -β3 + i π§ = π(πππ π + ππ πππ) Pay attention to the directions ο The βxβ part is negative so will go in the negative direction horizontally Replace r and ΞΈ ο Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry π= 3 2 + (1)2 ππππ = Calculate π=2 π= Remember that the argument is measured from the positive x-axis! arg π§ = 1 3 π 6 5π 6 Inverse Tan Subtract from Ο 3A π§ = π(πππ π + ππ πππ) Further complex numbers You can express a complex number in the form z = r(cosΞΈ + isinΞΈ) Express the following complex number in the modulus-argument form: z = -β3 + i π§ = 2 πππ 5π 5π + ππ ππ 6 6 Remember that the argument is not unique ο We could add 2Ο to them and the result would be the same, because 2Ο radians is a complete turn To do this you need to find both the argument and the modulus of the complex number ο Start by sketching it on an Argand diagram 3A π§ = π(πππ π + ππ πππ) Further complex numbers y You can express a complex number in the form z = r(cosΞΈ + isinΞΈ) Express the following complex number in the modulus-argument form: To do this you need to find both the argument and the modulus of the complex number ο Start by sketching it on an Argand diagram π§ = π(πππ π + ππ πππ) π π + ππ ππ β 4 4 1 ΞΈ r ο The βyβ part is negative so will go downwards z=1-i π§ = 2 πππ β Pay attention to the directions ο The βxβ part is positive so will go in the positive direction horizontally Replace r and ΞΈ x 1 ο Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry π= 1 π= 2 2 + (1)2 ππππ = Calculate 1 1 π= Inverse Tan π 4 arg π§ = β π 4 Negative as below the x-axis 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ Further complex numbers You can express a complex number in the form z = reiΞΈ In chapter 6 you will meet series expansions of cosΞΈ and sinΞΈ This can be used to prove the following result (which we will do when we come to chapter 6) If z = x + iy then the complex number can also be written in this way: z = reiΞΈ As before, r is the modulus of the complex number and ΞΈ is the argument ο This form is known as the βexponential formβ 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ Further complex numbers y You can express a complex number in the form z = reiΞΈ Express the following complex number in the form reiΞΈ, where -Ο < ΞΈ β€ Ο z = 2 β 3i As with the modulus-argument form, you should start by sketching an Argand diagram and use it to find r and ΞΈ Pay attention to the directions ο The βxβ part is positive so will go in the positive direction horizontally ο The βyβ part is negative so will go downwards π= 2 π = 13 π§ = 13π β0.98π Replace r and ΞΈ r x 3 ο Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry 2 + (3)2 ππππ = Calculate π§ = ππ ππ 2 ΞΈ 3 2 π = 0.98 arg π§ = β0.98 Inverse Tan Negative as below the x-axis 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ πππ βπ = πππ π π ππ βπ = βπ πππ Further complex numbers y You can express a complex number in the form z = reiΞΈ In Core 2, you will have seen the following: cos(-ΞΈ) = cosΞΈ y = cosΞΈ 1 -360º -270º -ΞΈ -180º 0 -90º -ΞΈ ΞΈ ΞΈ 90º 180º 270º ΞΈ -1 You can see that cos(-ΞΈ) = cosΞΈ anywhere on the graph y sin(-ΞΈ) = -sinΞΈ y = sinΞΈ 1 -ΞΈ -360º -270º -180º -90º 0 ΞΈ 90º 180º 270º ΞΈ -1 You can see that sin(-ΞΈ) = -sinΞΈ anywhere on the graph 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ πππ βπ = πππ π π ππ βπ = βπ πππ Further complex numbers You can express a complex number in the form z = reiΞΈ Express the following in the form z = reiΞΈ where βΟ < ΞΈ β€ Ο π§ = 2 πππ π π + ππ ππ 10 10 π§ = 2 πππ You can see from the form that r = β2 π π + ππ ππ 10 10 You can see from the form that ΞΈ = Ο/10 π= π= 2 π 10 π§ = ππ ππ π§= π 2π 10π Replace r and ΞΈ 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ πππ βπ = πππ π π ππ βπ = βπ πππ Further complex numbers You can express a complex number in the form z = reiΞΈ π§ = 5 πππ π π β ππ ππ 8 8 Express the following in the form z = reiΞΈ where βΟ < ΞΈ β€ Ο π§ = 5 πππ π π β ππ ππ 8 8 We need to adjust this first ο The sign in the centre is negative, we need it to be positive for the βrulesβ to work ο We also need both angles to be identical. In this case we can apply the rules we saw a moment agoβ¦ 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ πππ βπ = πππ π π ππ βπ = βπ πππ Further complex numbers You can express a complex number in the form z = reiΞΈ Express the following in the form z = reiΞΈ where βΟ < ΞΈ β€ Ο π§ = 5 πππ π§ = 5 πππ π π β ππ ππ 8 8 π§ = 5 πππ β Apply cosΞΈ = cos(-ΞΈ) Apply sin(-ΞΈ) = -sin(ΞΈ) π π + ππ ππ β 8 8 π π β ππ ππ 8 8 You can see You can see from the form from the form that r = 5 that ΞΈ = -Ο/8 π=5 π=β π 8 π§ = ππ ππ π§= π β π 5π 8 Replace r and ΞΈ 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ πππ βπ = πππ π π ππ βπ = βπ πππ Further complex numbers You can express a complex number in the form z = reiΞΈ Express the following in the form z = x + iy where π₯ β β and π¦ β β π§= 3π 2π 4 π π§= 3π 2π 4 π You can see from the form that r = β2 You can see from the form that ΞΈ = 3Ο/4 π= 2 π= 3π 4 π§ = π(πππ π + ππ πππ) This means that x and y have to be real numbers (ie not complex) Replace r and ΞΈ π§ = 2 πππ 3π 3π + ππ ππ 4 4 You can calculate all of this! Leave the second part in terms of i π§ = β1 + π 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ πππ βπ = πππ π π ππ βπ = βπ πππ Further complex numbers You can express a complex number in the form z = reiΞΈ Express the following in the form r(cosΞΈ + isinΞΈ), where βΟ < ΞΈ β€ Ο π§= 23π 2π 5 π π§= 23π 2π 5 π You can see from the form that r = 2 π=2 You can see from the form that ΞΈ = 23Ο/5 π= 23π 5 π= 13π 5 π= 3π 5 The value of ΞΈ is not in the range we want. We can keep subtracting 2Ο until it is! Subtract 2Ο Subtract 2Ο π§ = π(πππ π + ππ πππ) Replace r and ΞΈ π§ = 2 πππ 3π 3π + ππ ππ 5 5 3A π§ = π(πππ π + ππ πππ) π§ = ππ ππ π ππ βπ = βπ πππ πππ βπ = πππ π Further complex numbers You can express a complex number in the form z = reiΞΈ π ππ = πππ π + ππ πππ π Use: π ππ πππ π = 1 ππ π + π βππ 2 = πππ βπ + ππ ππ βπ π βππ = πππ π β ππ πππ = πππ π + ππ πππ To show that: π(βπ) 1) π ππ = πππ π + ππ πππ 2) π βππ = πππ π β ππ πππ Let ΞΈ = -ΞΈ Use the relationships above to rewrite Add 1 and 2 π βππ + π ππ = 2πππ π 1 βππ π + π ππ = πππ π 2 Divide by 2 3A π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted β you will have seen these in Core 3 π ππ π1 ± π2 = π πππ1 πππ π2 ± πππ π1 π πππ2 πππ π1 ± π2 = πππ π1 πππ π2 β π πππ1 π πππ2 πππ 2 π + π ππ2 π = 1 Multiplying a complex number z1 by another complex number z2, both in the modulus-argument form π§1 = π1 πππ π1 + ππ πππ1 π§2 = π2 πππ π2 + ππ πππ2 π§1 π§2 = π1 πππ π1 + ππ πππ1 × π2 πππ π2 + ππ πππ2 Rewrite π§1 π§2 = π1 π2 πππ π1 + ππ πππ1 πππ π2 + ππ πππ2 Now you can expand the double bracket as you would with a quadratic π§1 π§2 = π1 π2 πππ π1 πππ π2 + ππππ π1 π πππ2 + ππ πππ1 πππ π2 + π 2 π πππ1 π πππ2 π§1 π§2 = π1 π2 πππ π1 πππ π2 + ππππ π1 π πππ2 + ππ πππ1 πππ π2 β π πππ1 π πππ2 Group terms using the identities to the left ο You can also factorise the βiβ out π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 So when multiplying two complex numbers in the modulusargument form: ο Multiply the moduli ο Add the arguments together ο The form of the answer is the same 3B π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 π§1 π§2 = π1 π2 π π(π1 +π2 ) Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Multiplying a complex number z1 by another complex number z2, both in the exponential form To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted β you will have seen these in Core 3 π§1 π§2 = π§1 = π1 π ππ1 π1 π ππ1 π2 π ππ2 π§1 π§2 = π1 π2 π ππ1 +ππ2 π§2 = π2 π ππ2 Rewrite ο Remember you add the powers in this situation You can factorise the power π ππ π1 ± π2 = π πππ1 πππ π2 ± πππ π1 π πππ2 π§1 π§2 = π1 π2 π π(π1 +π2) πππ π1 ± π2 = πππ π1 πππ π2 β π πππ1 π πππ2 You can see that in this form the process is essentially the same as for the modulus-argument form: πππ 2 π + π ππ2 π = 1 ο Multiply the moduli together ο Add the arguments together ο The answer is in the same form 3B π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 π§1 π§2 = π1 π2 π π(π1 +π2 ) π§1 π1 = πππ π1 β π2 + ππ ππ π1 β π2 π§2 π2 Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted β you will have seen these in Core 3 π ππ π1 ± π2 = π πππ1 πππ π2 ± πππ π1 π πππ2 πππ π1 ± π2 = πππ π1 πππ π2 β π πππ1 π πππ2 2 2 πππ π + π ππ π = 1 So when dividing two complex numbers in the modulus-argument form: ο Divide the moduli ο Subtract the arguments ο The form of the answer is the same Dividing a complex number z1 by another complex number z2, both in the modulus-argument form π§1 = π1 πππ π1 + ππ πππ1 π§2 = π2 πππ π2 + ππ πππ2 π1 πππ π1 + ππ πππ1 π§1 = π2 πππ π2 + ππ πππ2 π§2 πππ π2 β ππ πππ2 π1 πππ π1 + ππ πππ1 π§1 × = πππ π2 β ππ πππ2 π2 πππ π2 + ππ πππ2 π§2 Multiply to cancel terms on the denominator π§1 π1 πππ π1 πππ π2 β ππππ π1 π πππ2 + ππ πππ1 πππ π2 β π 2 π πππ1 π πππ2 = π§2 π2 πππ π2 πππ π2 β ππππ π2 π πππ2 + ππ πππ2 πππ π2 β π 2 π πππ2 π πππ2 π§1 π1 πππ π1 πππ π2 β ππππ π1 π πππ2 + ππ πππ1 πππ π2 + π πππ1 π πππ2 = π§2 π2 πππ π2 πππ π2 β ππππ π2 π πππ2 + ππ πππ2 πππ π2 + π πππ2 π πππ2 π1 πππ π1 πππ π2 + π πππ1 π πππ2 + π π πππ1 πππ π2 β πππ π1 π πππ2 π§1 = π2 πππ 2 π2 + π ππ2 π2 π§2 π1 πππ π1 β π2 + ππ ππ π1 β π2 π§1 = π2 π§2 π§1 π1 = πππ π1 β π2 + ππ ππ π1 β π2 π§2 π2 Multiply out Remove i2 Group real and complex Rewrite terms Rewrite (again!) 3B π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 π§1 π§2 = π1 π2 π π(π1 +π2 ) π§1 π1 = πππ π1 β π2 + ππ ππ π1 β π2 π§2 π2 π§1 π1 = π π(π1 βπ2 ) π§2 π2 Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted β you will have seen these in Core 3 π ππ π1 ± π2 = π πππ1 πππ π2 ± πππ π1 π πππ2 πππ π1 ± π2 = πππ π1 πππ π2 β π πππ1 π πππ2 2 2 πππ π + π ππ π = 1 Dividing a complex number z1 by another complex number z2, both in the exponential form π§1 = π1 π ππ1 π1 π ππ1 π§1 = π§2 π2 π ππ2 π1 π§1 = π ππ1 π βππ2 π§2 π2 π1 π§1 = π ππ1 βππ2 π§2 π2 π1 π§1 = π π(π1βπ2) π§2 π2 π§2 = π2 π ππ2 Rewrite terms ο The denominator can be written with a negative power Multiplying so add the powers Factorise the power You can see that in this form the process is essentially the same as for the modulus-argument form: ο Divide the moduli ο Subtract the arguments ο The answer is in the same form 3B π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 π§1 π1 = πππ π1 β π2 + ππ ππ π1 β π2 π§2 π2 π§1 π§2 = π1 π2 π π(π1 +π2 ) π§1 π1 = π π(π1 βπ2 ) π§2 π2 Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number 3 πππ 5π 5π π π + ππ ππ × 4 πππ + ππ ππ 12 12 12 12 3(4) πππ Express the following calculation in the form x + iy: 3 πππ 5π 5π π π + ππ ππ × 4 πππ + ππ ππ 12 12 12 12 12 πππ 5π π 5π π + + ππ ππ + 12 12 12 12 Combine using one of the rules above ο Multiply the moduli ο Add the arguments Simplify terms π π + ππ ππ 2 2 Calculate the cos and sin parts (in terms of i where needed) 12 0 + π(1) Multiply out = 12π 3B π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 π§1 π§2 = π1 π2 π π(π1 +π2 ) π§1 π1 = πππ π1 β π2 + ππ ππ π1 β π2 π§2 π2 π§1 π1 = π π(π1 βπ2 ) π§2 π2 Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Express the following calculation in the form x + iy: 2 πππ 2 πππ π π 2π 2π + ππ ππ × 3 πππ β ππ ππ 15 15 5 5 2 πππ π π 2π 2π + ππ ππ × 3 πππ β + ππ ππ β 15 15 5 5 2(3) πππ π π 2π 2π + ππ ππ × 3 πππ β ππ ππ 15 15 5 5 Combine using a rule from above π 2π π 2π β + ππ ππ β 15 5 15 5 Simplify 6 πππ β π π + ππ ππ β 3 3 cos(-ΞΈ) = cosΞΈ sin(-ΞΈ) = -sinΞΈ The cos and sin terms must be added for this to work! ο Rewrite using the rules you saw in 3A 6 1 3 +π β 2 2 Calculate the cos and sin parts Multiply out = 3 β 3 3π 3B π§1 π§2 = π1 π2 πππ π1 + π2 + ππ ππ π1 + π2 π§1 π§2 = π1 π2 π π(π1 +π2 ) π§1 π1 = πππ π1 β π2 + ππ ππ π1 β π2 π§2 π2 π§1 π1 = π π(π1 βπ2 ) π§2 π2 Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Express the following calculation in the form x + iy: π π + ππ ππ 12 12 5π 5π 2 πππ + ππ ππ 6 6 2 πππ π π + ππ ππ 12 12 5π 5π 2 πππ + ππ ππ 6 6 2 πππ Combine using one of the rules above ο Divide the moduli ο Subtract the arguments π 5π π 5π 2 πππ β + ππ ππ β 12 6 12 6 2 Simplify 3π 3π 2 πππ β + ππ ππ β 4 4 2 You can work out the sin and cos parts 1 1 2 β +π β 2 2 2 Multiply out 1 1 =β β π 2 2 3B Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n Let: z = r(cosΞΈ + isinΞΈ) π§ = π(πππ π + ππ πππ) π§ 2 = π(πππ π + ππ πππ) 1 2 = π(πππ π + ππ πππ) = π 2 (πππ 2π + ππ ππ2π) π§ = π(πππ π + ππ πππ) π§2 = π§ × π§ π§ 2 = π(πππ π + ππ πππ) × π(πππ π + ππ πππ) 2 π§ = π (πππ 2π + ππ ππ2π) 3 = π 3 (πππ 3π + ππ ππ3π) π§ 4 = π(πππ π + ππ πππ) 4 = π 4 (πππ 4π + ππ ππ4π) π π§ = π(πππ π + ππ πππ) π π = π (πππ ππ + ππ ππππ) π§ 3 = π 2 (πππ 2π + ππ ππ2π) × π(πππ π + ππ πππ) 3 π§ = π (πππ 3π + ππ ππ3π) π§4 = π§3 × π§ π§ 4 = π 3 (πππ 3π + ππ ππ3π) × π(πππ π + ππ πππ) π§ 4 = π 4 (πππ 4π + ππ ππ4π) This is De Moivreβs Theorem ο You need to be able to prove this Multiply the moduli, add the arguments π§3 = π§2 × π§ 3 π§ 3 = π(πππ π + ππ πππ) 2 Use the modulusargument form Use the modulusargument form Multiply the moduli, add the arguments Use the modulusargument form Multiply the moduli, add the arguments De Moivre = βDe Mwavreβ (pronunciation) 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n De Moivreβs theorem can be proved using the method of proof by induction from FP1 Basis β show the statement is true for n=1 Assumption β assume the statement is true for n = k Inductive β show that if true for n = k, then the statement is also true for n = k+1 Conclusion β because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n Proving that: π πππ π + ππ πππ π = π π πππ ππ + ππ ππππ is true for all positive integers BASIS ο Show that the statement is true for n = 1 π πππ π + ππ πππ π = π π πππ ππ + ππ ππππ π πππ π + ππ πππ 1 = π1 πππ 1π + ππ ππ1π π πππ π + ππ πππ = π πππ π + ππ πππ Sub in n=1 Simplify each side ASSUMPTION ο Assume that the statement is true for n = k π πππ π + ππ πππ π = π π πππ ππ + ππ ππππ π πππ π + ππ πππ π = π π πππ ππ + ππ ππππ Replace n with k 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n De Moivreβs theorem can be proved using the method of proof by induction from FP1 Basis β show the statement is true for n=1 Assumption β assume the statement is true for n = k Inductive β show that if true for n = k, then the statement is also true for n = k+1 Conclusion β because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n Proving that: π πππ π + ππ πππ π = π π πππ ππ + ππ ππππ is true for all positive integers INDUCTIVE ο Show that if true for n = k, the statement is also true for n = k + 1 π πππ π + ππ πππ π Sub in n = k + 1 π πππ π + ππ πππ π+1 π πππ π + ππ πππ π × π πππ π + ππ πππ You can write this as two separate parts as the powers are added together 1 = π π πππ ππ + ππ ππππ × π(πππ π + ππ πππ) = π π+1 cos(π + 1)π + ππ ππ(π + 1)π We can rewrite the first part based on the assumption step, and the second based on the basis step Using the multiplication rules from 3B ο Multiply the moduli, add the arguments 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n De Moivreβs theorem can be proved using the method of proof by induction from FP1 Basis β show the statement is true for n=1 Proving that: π πππ π + ππ πππ π = π π πππ ππ + ππ ππππ is true for all positive integers CONCLUSION ο Explain why this shows it is trueβ¦ ο We showed the statement is true for n = 1 We then assumed the following: π πππ π + ππ πππ π = π π πππ ππ + ππ ππππ Assumption β assume the statement is true for n = k Using the assumption, we showed that: Inductive β show that if true for n = k, then the statement is also true for n = k+1 As all the βkβ terms have become βk + 1β terms, if the statement is true for one term, it must be true for the next, and so onβ¦ Conclusion β because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n π πππ π + ππ πππ π+1 = π π+1 cos π + 1 π + ππ ππ(π + 1)π ο The statement was true for 1, so must be true for 2, and therefore 3, and so onβ¦ ο We have therefore proven the statement for all positive integers! 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n π πππ π + ππ πππ = We have just proved the theorem for n = k where k is a positive integer = ο Now we need to show it is also true for any negative integerβ¦ = 1 π(πππ π + ππ πππ) βπ Write using a positive power instead π ππ 1 πππ ππ + ππ ππππ ππ 1 πππ ππ β ππ ππππ × πππ ππ + ππ ππππ πππ ππ β ππ ππππ πππ ππ β ππ ππππ ο If n is a negative integer, it can be = π π πππ 2 ππ β π 2 π ππ2 ππ written as β-mβ, where m is a positive integer πππ ππ β ππ ππππ = π You can see that π πππ 2 ππ + π ππ2 ππ the answer has followed the same pattern as De Moivreβs theorem! = Use De Moivreβs theorem for a positive number (which we have proved) 1 ππ Multiply to change some terms in the fraction Multiply out like quadratics β the bottom is the difference of two squares i2 = -1 You can cancel the denominator as it is equal to 1 πππ ππ β ππ ππππ = π βπ cos βππ + ππ ππ(βππ) Use cos(-ΞΈ) = cos(ΞΈ) and sin(-ΞΈ) = -sinΞΈ 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n Having now proved that De Moivreβs theorem works for both positive and negative integers, there is only one left ο We need to prove it is true for 0! ο This is straightforward. As it is just a single value, we can substitute it in to see what happensβ¦ π πππ π + ππ πππ π π πππ π + ππ πππ 0 = π π πππ ππ + ππ ππππ Sub in n = 0 = π 0 πππ 0 + ππ ππ0 1 = 1(1 + 0) Left side = 1 as anything to the power 0 is 1 ο You can find cos0 and sin 0 as well βCalculateβ 1= 1 So we have shown that De Moivreβs Theorem is true for all positive integers, all negative integers and 0β ο It is therefore true for all integers! 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) π§ π = ππ ππ π = π π π πππ Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n It is important to note that De Moivreβs theorem can also be used in exponential form. ππ ππ π = π π π πππ Both parts will be raised to the power βnβ You can remove the bracket! = π π π πππ This is De Moivreβs theorem in exponential form! 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) π§ π = ππ ππ π = π π π πππ Further complex numbers 9π 9π πππ + ππ ππ 17 17 2π 2π πππ β ππ ππ 17 17 You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n Simplify the following: 9π 9π πππ + ππ ππ 17 17 2π 2π πππ β ππ ππ 17 17 5 πππ 9π 9π + ππ ππ 17 17 5 3 3 45π 45π + ππ ππ 17 17 6π 6π πππ β + ππ ππ β 17 17 πππ ο Apply cos(-ΞΈ) = cosΞΈ and sin(-ΞΈ) = -sinΞΈ 5 2π 2π πππ β + ππ ππ β 17 17 πππ The denominator has to have the β+β sign in the middle 3 Apply De Moivreβs theorem (there is no modulus value to worry about here!) ο Just multiply the arguments by the power 45π 6π 45π 6π ββ + ππ ππ ββ 17 17 17 17 πππ 3π + ππ ππ3π = β1 + 0π Apply the rules from 3B for the division of complex numbers ο Divide the moduli and subtract the arguments Simplify the sin and cos terms Calculate the sin and cos terms Simplify = β1 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) π§ π = ππ ππ π = π π π πππ Further complex numbers y You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n Express the following in the form x + iy where x Π R and y Π R 1 + 3π 7 ο You need to write this in one of the forms above, and you can then use De Moivreβs theorem ο This is easier than raising a bracket to the power 7! ο Start with an argand diagram to help find the modulus and argument of the part in the bracket Pay attention to the directions ο The βxβ part is positive so will go in the positive direction horizontally r β3 ΞΈ ο The βyβ part is positive so will go upwards π= 1 2 + 3 x 1 2 Calculate π=2 ππππ = π= 3 1 π 3 Inverse Tan π πππ π + ππ πππ Sub in r and ΞΈ 2 πππ π π + ππ ππ 3 3 3C π§ π = π(πππ π + ππ πππ) π = π π (πππ ππ + ππ ππππ) π§ π = ππ ππ π = π π π πππ Further complex numbers You need to be able to prove that [r(cosΞΈ + isinΞΈ)]n = rn(cos(nΞΈ + isinnΞΈ) for any integer n Express the following in the form x + iy where x Π R and y Π R 1 + 3π 7 ο You need to write this in one of the forms above, and you can then use De Moivreβs theorem ο This is easier than raising a bracket to the power 7! 1 + 3π 7 π π 2 πππ + ππ ππ 3 3 27 πππ 7 7π 7π + ππ ππ 3 3 Rewrite using the different form we worked out before Use De Moivreβs Theorem as above Calculate the cos and sin parts = 128 1 +π 2 3 2 Multiply out and simplify = 64 + 64 3π ο Start with an argand diagram to help find the modulus and argument of the part in the bracket 3C π+π π = ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities This involves changing expressions involving a function of ΞΈ into one without. ο For example changing a cos6ΞΈ into powers of cosΞΈ ο You will need to use the binomial expansion for C2 in this section π+π π ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Remember nCr is a function you can find on your calculator ο The first term has the full power of n ο As you move across you slowly swap the powers over to the second term until it has the full power of n For example: π+π π4 + πΆ1 π3 π + 4 4 πΆ2 π2 π2 + 4 Follow the pattern above πΆ3 ππ 3 + π 4 4 π4 + 4π3 π + 6π2 π2 + 4ππ 3 + π 4 You can work out the nCr parts 3D π+π π = ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities πππ π + ππ πππ 3 Express cos3ΞΈ using powers of cosΞΈ. ο This type of question involves making a comparison between two processes ο One which will give you a βcos3ΞΈβ term β you will use De Moivreβs Theorem for this ο If we apply De Moivreβs theorem to this, we will end up with a βcos3ΞΈβ term ο If we apply the binomial expansion to it, we will end up with some terms with cosΞΈ in ο So this expression is a good starting point! ο One which will give you an expression in terms of cosΞΈ β you will use the binomial expansion for this ο You have to think logically and decide where to start 3D π+π π = ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Express cos3ΞΈ using powers of cosΞΈ. ο This type of question involves making a comparison between two processes πππ π + ππ πππ Apply De Moivreβs theorem πππ π + ππ πππ 3 Follow the rules you know = πππ 3π + ππ ππ3π Apply the Binomial expansion πππ π + ππ πππ ο One which will give you a βcos3ΞΈβ term β you will use De Moivreβs Theorem for this 3 = πππ π 3 3 3 3 + πΆ1 πππ π 2 2 3 ππ πππ + πΆ2 πππ π ππ πππ 2 2 2 3 + ππ πππ 3 = πππ π + 3ππππ ππ πππ + 3π πππ ππ ππ π + π π ππ π ο One which will give you an expression in terms of cosΞΈ β you will use the binomial expansion for this ο You have to think logically and decide where to start = πππ 3 π + 3ππππ 2 ππ πππ β 3πππ ππ ππ2 π β ππ ππ3 π 3 Write out βTidy upβ Replace i2 parts with -1 The two expressions we have made must be equal ο Therefore the real parts in each and the imaginary parts in each must be the same ο Equate the real parts πππ 3π = πππ 3 π β 3πππ ππ ππ2 π 3D π+π π = ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Express cos3ΞΈ using powers of cosΞΈ. ο This type of question involves making a comparison between two processes ο One which will give you a βcos3ΞΈβ term β you will use De Moivreβs Theorem for this πππ 3π = πππ 3 π β 3πππ ππ ππ2 π πππ 3π = πππ 3 π β 3πππ π 1 β πππ 2 π Replace sin2ΞΈ with an expression in cos2ΞΈ Expand the bracket πππ 3π = πππ 3 π β 3πππ π + 3πππ 3 π Simplify 3 πππ 3π = 4πππ π β 3πππ π We have successfully expressed cos3ΞΈ as posers of cosΞΈ! ο One which will give you an expression in terms of cosΞΈ β you will use the binomial expansion for this ο You have to think logically and decide where to start 3D π+π π = ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Express the following as powers of cosΞΈ: π ππ6π π πππ ο So we need something that will give us sin6ΞΈ using De Moivreβs theorem ο It also needs to give us terms of cosΞΈ from the binomial expansion πππ π + ππ πππ 6 ο If we apply De Moivreβs theorem to this, we will end up with a βsin6ΞΈβ term ο If we apply the binomial expansion to it, we will end up with some terms with cosΞΈ in ο So this expression is a good starting point! (and yes you will have to do some expansions larger than powers of 3 or 4!) 3D π+π π = ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities 6 πππ π + ππ πππ Apply De Moivreβs theorem πππ π + ππ πππ Express the following as powers of cosΞΈ: 6 Follow the rules you know = πππ 6π + ππ ππ6π π ππ6π π πππ Apply the Binomial expansion πππ π + ππ πππ = πππ π 6 6 + 6πΆ1 πππ π 5 (ππ πππ) + 6πΆ2 πππ π 4 ππ πππ 2 + 6πΆ3 πππ π 3 ππ πππ 3 + 6πΆ4 πππ π 2 ππ πππ 4 + 6πΆ5 πππ π ππ πππ = πππ 6 π + 6ππππ 5 ππ πππ + 15π 2 πππ 4 ππ ππ2 π + 20π 3 πππ 3 ππ ππ3 π + 15π 4 πππ 2 ππ ππ4 π + 6π 5 πππ ππ ππ5 π + π 6 π ππ6 π = πππ 6 π + 6ππππ 5 ππ πππ β 15πππ 4 ππ ππ2 π β 20ππππ 3 ππ ππ3 π + 15πππ 2 ππ ππ4 π + 6ππππ ππ ππ5 π β π ππ6 π 5 + ππ πππ 6 Replace terms: i2 and i6 = -1 i4 = 1 So the two expressions created from De Moivre and the Binomial Expansion must be equal ο The real parts will be the same, as will the imaginary parts ο This time we have to equate the imaginary parts as this has sin6ΞΈ in ππ ππ6π = 6ππππ 5 ππ πππ β 20ππππ 3 ππ ππ3 π + 6ππππ ππ ππ5 π 5 3 3 5 Divide all by i π ππ6π = 6πππ ππ πππ β 20πππ ππ ππ π + 6πππ ππ ππ π 3D π+π π = ππ + ππΆ1 ππβ1 π + ππΆ2 ππβ2 π 2 + ππΆ3 ππβ3 π3 + β¦ β¦ β¦ β¦ + π π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities πππ π + ππ πππ π ππ6π = 6πππ 5 ππ πππ β 20πππ 3 ππ ππ3 π + 6πππ ππ ππ5 π Express the following as powers of cosΞΈ: π ππ6π π πππ 5 2 3 6 So we have changed the expression we were given into powers of cosΞΈ! 4 3 π ππ6π 6πππ ππ πππ β 20πππ ππ ππ π + 6πππ ππ ππ5 π = π πππ π πππ Divide all terms by sinΞΈ = 6πππ 5 π β 20πππ 3 ππ ππ2 π + 6πππ ππ ππ4 π Replace sin2ΞΈ terms with (1 β cos2ΞΈ) terms = 6πππ 5 π β 20πππ 3 π 1 β πππ 2 π + 6πππ π 1 β πππ 2 π 2 = 6πππ 5 π β 20πππ 3 π + 20πππ 5 π + 6πππ π 1 β 2πππ 2 π + πππ 4 π 5 3 Expand the first bracket, square the second Expand the second bracket 5 = 6πππ π β 20πππ 3 π + 20πππ 5 π + 6πππ π β 12πππ π + 6πππ π = 32πππ 5 π β 32πππ 3 π + 6πππ π Group together the like terms 3D π§+ 1 = 2πππ π π§ Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Let: π§ = πππ π + ππ πππ 1 = πππ π + ππ πππ π§ You also need to be able to work this type of question in a different way: 1 = cos(βπ) + ππ ππ(βπ) π§ For example, you might have a power or cos or sin and need to express it using several linear terms instead 1 = cos π β ππ πππ π§ Eg) Changing sin6ΞΈ to sinaΞΈ + sinbΞΈ where a and b are integers ο To do this we need to know some other patterns first! β1 Write as β1 overβ ο or with a power of -1 Use De Moivreβs theorem Use cos(-ΞΈ) = cosΞΈ and sin(-ΞΈ) = -sinΞΈ We can add our two results together: π§+ 1 = πππ π + ππ πππ + πππ π β ππ πππ π§ 1 π§ + = 2πππ π π§ Simplify 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Let: π§ = πππ π + ππ πππ 1 = πππ π + ππ πππ π§ You also need to be able to work this type of question in a different way: 1 = cos(βπ) + ππ ππ(βπ) π§ For example, you might have a power or cos or sin and need to express it using several linear terms instead 1 = cos π β ππ πππ π§ Eg) Changing sin6ΞΈ to sinaΞΈ + sinbΞΈ where a and b are integers ο To do this we need to know some other patterns first! β1 Write as β1 overβ ο or with a power of -1 Use De Moivreβs theorem Use cos(-ΞΈ) = cosΞΈ and sin(-ΞΈ) = -sinΞΈ We could also subtract our two results: π§β 1 = πππ π + ππ πππ β πππ π β ππ πππ π§ 1 π§ β = 2ππ πππ π§ Simplify 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Let: π§ π = πππ ππ + ππ ππππ 1 = πππ ππ + ππ ππππ π§π You also need to be able to work this type of question in a different way: 1 = cos(βππ) + ππ ππ(βππ) π§π For example, you might have a power or cos or sin and need to express it using several linear terms instead 1 = cos ππ β ππ ππππ π§π Eg) Changing sin6ΞΈ to sinaΞΈ + sinbΞΈ where a and b are integers ο To do this we need to know some other patterns first! ο You can also apply the rules we just saw to powers of z β1 Write as β1 overβ ο or with a power of -1 Use De Moivreβs theorem Use cos(-ΞΈ) = cosΞΈ and sin(-ΞΈ) = -sinΞΈ We could add our two results together: π§π + 1 = πππ ππ + ππ ππππ π§π + πππ ππ β ππ ππππ Simplify 1 π§ π + π = 2πππ ππ π§ 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Let: π§ π = πππ ππ + ππ ππππ 1 = πππ ππ + ππ ππππ π§π You also need to be able to work this type of question in a different way: 1 = cos(βππ) + ππ ππ(βππ) π§π For example, you might have a power or cos or sin and need to express it using several linear terms instead 1 = cos ππ β ππ ππππ π§π Eg) Changing sin6ΞΈ to sinaΞΈ + sinbΞΈ where a and b are integers ο To do this we need to know some other patterns first! ο You can also apply the rules we just saw to powers of z β1 Write as β1 overβ ο or with a power of -1 Use De Moivreβs theorem Use cos(-ΞΈ) = cosΞΈ and sin(-ΞΈ) = -sinΞΈ We could also subtract our two results: π§π β 1 = πππ ππ + ππ ππππ π§π β πππ ππ β ππ ππππ Simplify 1 π§ π β π = 2ππ ππππ π§ 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Letβs now see how we can use these βpatternsβ in solving problems: Express cos5ΞΈ in the form acos5ΞΈ + bcos3ΞΈ + ccosΞΈ ο When working this way round you need to use the identities above to express both cos5ΞΈ and terms with cos5ΞΈ, cos3ΞΈ and cosΞΈ. ο You can then set the expressions equal to each other 5 1 π§+ π§ = 2πππ π 5 = 32πππ 5 π Creating the other cos terms β use the Binomial expansion! 1 π§+ π§ 5 Where a, b and c are constants to be found. Using the Identity above Creating a cos5ΞΈ term 5 = π§ + 5π§ 4 1 1 + 10π§ 3 π§ π§ 2 + 10π§ 2 1 π§ 3 1 + 5π§ π§ 4 1 + π§ 5 Use the B.E. Cancel some z 1 1 1 terms +5 3 + 5 = π§ 5 + 5π§ 3 + 10π§ + 10 π§ π§ π§ Group up terms with the same 1 1 1 power = π§ 5 + 5 + 5 π§ 3 + 3 + 10 π§ + π§ π§ π§ Rewrite using an identity above = 2πππ 5π + 5(2πππ 3π) + 10(2πππ π) Simplify = 2πππ 5π + 10πππ 3π + 20πππ π 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Letβs now see how we can use these βpatternsβ in solving problems: Using the two expressions 1 π§+ π§ 5 1 π§+ π§ 5 = 32πππ 5 π These two expressions must be equal to each other = 2πππ 5π + 10πππ 3π + 20πππ π Express cos5ΞΈ in the form acos5ΞΈ + bcos3ΞΈ + ccosΞΈ Where a, b and c are constants to be found. ο When working this way round you need to use the identities above to express both cos5ΞΈ and terms with cos5ΞΈ, cos3ΞΈ and cosΞΈ. 32πππ 5 π = 2πππ 5π + 10πππ 3π + 20πππ π πππ 5 π = 1 5 5 πππ 5π + πππ 3π + πππ π 16 16 8 Divide both sides by 32 So we have written cos5ΞΈ using cos5ΞΈ, cos3ΞΈ and cosΞΈ ο You can then set the expressions equal to each other 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Show that: 1 3 π ππ3 π = β π ππ3π + π πππ 4 4 ο This is similar to the previous question. You need to use rules above to find a way to create a sin3ΞΈ expression, and an expression containing sin3ΞΈ and sinΞΈ Using the Identity above Creating a sin3ΞΈ term 3 1 π§β π§ = 2ππ πππ 3 = 8π 3 π ππ3 π = β8ππ ππ3 π Creating the other sin terms β use the Binomial expansion! 1 π§β π§ 3 3 = π§ + 3π§ 2 1 1 + 3π§ β β π§ π§ = π§ 3 β 3π§ + 3 = π§3 β 1 π§ β 1 1 β 3 π§ β π§ π§3 = 2ππ ππ3π β 3(2ππ πππ) 1 π§3 2 1 + β π§ 3 Use the B.E. Cancel some z terms Group up terms with the same power Rewrite using an identity above Simplify = 2ππ ππ3π β 6ππ πππ 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities Show that: 1 3 π ππ3 π = β π ππ3π + π πππ 4 4 ο This is similar to the previous question. You need to use rules above to find a way to create a sin3ΞΈ expression, and an expression containing sin3ΞΈ and sinΞΈ Using the two expressions 1 π§β π§ 3 1 π§β π§ 3 = β8ππ ππ3 π These two expressions must be equal to each other = 2ππ ππ3π β 6ππ πππ β8ππ ππ3 π = 2ππ ππ3π β 6ππ πππ β8π ππ3 π = 2π ππ3π β 6π πππ 1 3 π ππ3 π = β π ππ3π + π πππ 4 4 Divide both sides by i Divide both sides by -8 So we have written sin3ΞΈ using sin3ΞΈ and sinΞΈ! 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities a) Express sin4ΞΈ in the form: ππππ 4π + ππππ 2π + π Where d, e and f are constants to be found. 4 1 π§β π§ = 2ππ πππ π 2 4 = 16π 4 π ππ4 π = 16π ππ4 π Creating the cos terms β use the Binomial expansion! 1 π§β π§ 4 4 b) Hence, find the exact value of the following integral: Using the Identity above Creating a sin4ΞΈ term = π§ + 4π§ 3 1 1 + 6π§ 2 β β π§ π§ = π§ 4 β 4π§ 2 + 6 β 4 1 π§2 2 + 1 + 4π§ β π§ 3 1 π§4 π ππ π ππ ο Start exactly as with the previous questions, by finding an expression with sin4ΞΈ and one with cos4ΞΈ, cos2ΞΈ and a number = π§4 + 1 1 β 4 π§2 + 2 + 6 4 π§ π§ 4 Cancel some z terms Group up terms with the same power (use positive values in the brackets so we get cos terms) 4 0 1 + β π§ Use the B.E. Replace using an identity above = 2πππ 4π β 4(2πππ 2π) + 6 Simplify = 2πππ 4π β 8πππ 2π + 6 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers You can apply De Moivreβs theorem to trigonometric identities a) Express sin4ΞΈ in the form: ππππ 4π + ππππ 2π + π Using the two expressions 1 π§β π§ 4 1 π§β π§ 4 = 16π ππ4 π These two expressions must be equal to each other = 2πππ 4π β 8πππ 2π + 6 Where d, e and f are constants to be found. b) Hence, find the exact value of the following integral: π 2 π ππ4 π ππ 16π ππ4 π = 2πππ 4π β 8πππ 2π + 6 1 1 3 π ππ4 π = πππ 4π β πππ 2π + 8 2 8 Divide both sides by 16 0 ο Start exactly as with the previous questions, by finding an expression with sin4ΞΈ and one with cos4ΞΈ, cos2ΞΈ and a number So we have written sin4ΞΈ using cos4ΞΈ and cos2ΞΈ! 3D π§+ 1 = 2πππ π π§ π§β 1 = 2ππ πππ π§ π§π + 1 = 2πππ ππ π§π π§π β 1 = 2ππ ππππ π§π Further complex numbers 1 1 3 π ππ4 π = πππ 4π β πππ 2π + 8 2 8 You can apply De Moivreβs theorem to trigonometric identities π 2 a) Express sin4ΞΈ in the form: 0 ππππ 4π + ππππ 2π + π π 2 Where d, e and f are constants to be found. b) Hence, find the exact value of the following integral: π 2 π ππ4 π ππ 0 ο Start exactly as with the previous questions, by finding an expression with sin4ΞΈ and one with cos4ΞΈ, cos2ΞΈ and a number π ππ4 π ππ 0 1 1 3 πππ 4π β πππ 2π + ππ 8 2 8 1 1 3 = π ππ4π β π ππ2π + π 32 4 8 = = 1 π 1 π 3 π π ππ4 β π ππ2 + 32 2 4 2 8 2 π 2 Cosine Integrals (in C4) πππ 4π = 1 π ππ4π 4 πππ 2π = 1 π ππ2π 2 Replace with an equivalent expression Integrate each term with respect to ΞΈ, using knowledge from C4 0 β 1 1 3 π ππ4 0 β π ππ2 0 + 0 32 4 8 Sub in limits Work out 3 π 16 3D πΌπ: π§ = π πππ π + ππ πππ πβππ: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) Further complex numbers π(πππ π + ππ πππ) π = π π πππ ππ + ππ ππππ You can use De Moivreβs theorem to find the nth roots of a complex number You already know how to find real roots of a number, but now we need to find both real roots and imaginary roots! We need to apply the following results: 1) If: π§ = π πππ π + ππ πππ ο Then: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) where k is an integer This is because we can add multiples of 2Ο to the argument as it will end up in the same place (2Ο = 360º) 2) De Moivreβs theorem π(πππ π + ππ πππ) π = π π πππ ππ + ππ ππππ 3E πΌπ: π§ = π πππ π + ππ πππ πβππ: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) Further complex numbers π(πππ π + ππ πππ) π = π π πππ ππ + ππ ππππ You can use De Moivreβs theorem to find the nth roots of a complex number In this case the modulus and argument are simple to find! Solve the equation z3 = 1 and represent your solutions on an Argand diagram. π=1 ο Now we know r and ΞΈ we can set equal to this expression, when written in the modulus-argument form 1 x π=0 ο First you need to express z in the modulus-argument form. Use an Argand diagram. z3 y π§ 3 = 1 πππ 0 + ππ ππ0 Apply the rule above π§ 3 = cos(0 + 2ππ) + ππ ππ(0 + 2ππ) ο We can then find an expression for z in terms of k π§ = cos 0 + 2ππ + ππ ππ(0 + 2ππ) ο We can then solve this to find the roots of the equation above π§ = cos 0 + 2ππ 0 + 2ππ + ππ ππ 3 3 1 3 Cube root (use a relevant power) Apply De Moivreβs theorem 3E πΌπ: π§ = π πππ π + ππ πππ πβππ: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) Further complex numbers π(πππ π + ππ πππ) π = π π πππ ππ + ππ ππππ You can use De Moivreβs theorem to find the nth roots of a complex number Solve the equation z3 = 1 and represent your solutions on an Argand diagram. π§ = πππ 0 + 2ππ 0 + 2ππ + ππ ππ 3 3 k=0 π§ = πππ 0 + ππ ππ 0 π§=1 ο We now just need to choose different values for k until we have found all the roots ο The values of k you choose should keep the argument within the range: -Ο < ΞΈ β€ Ο So the roots of z3 = 1 are: 1 3 1 3 π§ = 1, β + π and β β π 2 2 2 2 Sub k = 0 in and calculate the cosine and sine parts k=1 π§ = πππ 2π 2π + ππ ππ 3 3 1 3 π§ =β +π 2 2 k = -1 2π 2π π§ = πππ β + ππ ππ β 3 3 1 3 π§ =β βπ 2 2 Sub k = 1 in and calculate the cosine and sine parts Sub k = -1 in and calculate the cosine and sine parts ο (k = 2 would cause the argument to be outside the range) 3E πΌπ: π§ = π πππ π + ππ πππ πβππ: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) Further complex numbers π(πππ π + ππ πππ) π y = π π πππ ππ + ππ ππππ You can use De Moivreβs theorem to find the nth roots of a complex number Solve the equation z3 = 1 and represent your solutions on an Argand diagram. π π β +π π π 2 π 3 ο We now just need to choose different values for k until we have found all the roots ο The values of k you choose should keep the argument within the range: -Ο < ΞΈ β€ Ο So the roots of z3 = 1 are: 1 3 1 3 π§ = 1, β + π and β β π 2 2 2 2 π 2 π 3 x 2 π 3 π π β βπ π π ο The solutions will all the same distance from the origin ο The angles between them will also be the same ο The sum of the roots is always equal to 0 3E πΌπ: π§ = π πππ π + ππ πππ πβππ: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) Further complex numbers π(πππ π + ππ πππ) π = π π πππ ππ + ππ ππππ You can use De Moivreβs theorem to find the nth roots of a complex number Solve the equation z4 - 2β3i = 2 Give your answers in both the modulusargument and exponential forms. By rearrangingβ¦ z4 = 2 + 2β3i ο As before, use an argand diagram to express the equation in the modulusargument form ο Then choose values of k until you have found all the solutions y Find the modulus and argument π= 2 π = π‘ππ 2 + 2 3 β1 2 3 2 2 r π=4 ΞΈ π π= 3 π§ 4 = 4 πππ π π + ππ ππ 3 3 π§ 4 = 4 πππ π π + 2ππ + ππ ππ + 2ππ 3 3 π π π§ = 4 πππ + 2ππ + ππ ππ + 2ππ 3 3 π π + 2ππ + 2ππ 1 π§ = 44 πππ 3 + ππ ππ 3 4 4 π π + 2ππ + 2ππ π§ = 2 πππ 3 + ππ ππ 3 4 4 2β3 2 x Apply the rule above 1 4 Take the 4th root of each side De Moivreβs Theorem Work out the power at the front 3E πΌπ: π§ = π πππ π + ππ πππ πβππ: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) Further complex numbers π(πππ π + ππ πππ) π = π π πππ ππ + ππ ππππ You can use De Moivreβs theorem to find the nth roots of a complex number k=0 π§ = 2 πππ Solve the equation z4 - 2β3i = 2 Give your answers in both the modulusargument and exponential forms. k=1 ο Then choose values of k until you have found all the solutions π π + 2ππ + 2ππ 3 π§ = 2 πππ + ππ ππ 3 4 4 k = -1 π π + ππ ππ 12 12 7π 7π + ππ ππ 12 12 π π β 2π β 2π π§ = 2 πππ 3 + ππ ππ 3 4 4 π§ = 2 πππ β k = -2 Sub k = 0 in and simplify (you can leave in this form) π π + 2π + 2π π§ = 2 πππ 3 + ππ ππ 3 4 4 π§ = 2 πππ By rearrangingβ¦ z4 = 2 + 2β3i ο As before, use an argand diagram to express the equation in the modulusargument form π π π§ = 2 πππ 3 + ππ ππ 3 4 4 5π 5π + ππ ππ β 12 12 Choose values of k that keep the argument between βΟ and Ο π π β 2π β 2π π§ = 2 πππ 3 + ππ ππ 3 4 4 π§ = 2 πππ β 11π 11π + ππ ππ β 12 12 3E πΌπ: π§ = π πππ π + ππ πππ πβππ: π§ = π cos(π + 2ππ) + ππ ππ(π + 2ππ) Further complex numbers π(πππ π + ππ πππ) π = π π πππ ππ + ππ ππππ You can use De Moivreβs theorem to find the nth roots of a complex number Solve the equation z4 - 2β3i = 2 Give your answers in both the modulusargument and exponential forms. By rearrangingβ¦ z4 = 2 + 2β3i ο As before, use an argand diagram to express the equation in the modulusargument form Solutions in the modulus-argument form π§ = 2 πππ π π + ππ ππ 12 12 π§ = 2 πππ 7π 7π + ππ ππ 12 12 π§ = 2 πππ β π§ = 2 πππ β 5π 5π + ππ ππ β 12 12 11π 11π + ππ ππ β 12 12 Solutions in the exponential form π§= π 12 2π π π§= 7π 2π 12 π π§= 5π β π 2π 12 π§= 11π β 12 π 2π ο Then choose values of k until you have found all the solutions π π + 2ππ + 2ππ 3 π§ = 2 πππ + ππ ππ 3 4 4 3E Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram A locus a set of points which obey a rule x The locus of points a given distance from a point O is a circle x O x ο You will need to be able to understand Loci based on Argand diagrams A The locus of points equidistant from two fixed points A and B is the perpendicular bisector of line AB B 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram z = x + iy represents a variable point P(x,y) on an Argand diagram P(x,y) z z1 = x1 + iy1 represents a fixed point A(x1,y1) on an Argand diagram What is represented by: z - z1 A(x1,y1) z1 x π§ β π§1 ο It represents the distance between If we want to get from the fixed point A to the variable point the fixed point A(x1,y1) and the P, we need to travel back along z1 and then out along z variable point P(x,y) (-z1 + z) ο This can be written as a vector, z β z1 ο So |z β z1| represents the distance between the fixed point and the variable point! 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram P(x,y) If: A(5,3) π§ β 5 β 3π = 3 x Sketch the locus of P(x,y) which is represented by z on an Argand diagram π§ β 5 β 3π = 3 π§ β (5 + 3π) = 3 Leave z as it is β this is the variable point Put this part in a bracket - This is the fixed point So we want the locus where the distance between the variable point z and the fixed point (5,3) is equal to 3 This will be a circle of radius 3 units, centre (5,3) 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers πΌπ: π§ = π₯ + ππ¦ You can use complex numbers to represent a locus of points on an Argand diagram πβππ: π§ = π₯ + ππ¦ = π₯2 + π¦2 y P(x,y) If: |z| π§ β 5 β 3π = 3 Use an algebraic method to find a Cartesian equation of the locus of z x y x ο So you have to do this without using the graph you drew ο We will quickly remind ourselves of something that will be useful for this! If: π§ = π₯ + ππ¦ Then: π§ = π₯ + ππ¦ = (By Pythagorasβ Theorem) π₯ 2 + π¦2 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers πΌπ: π§ = π₯ + ππ¦ You can use complex numbers to represent a locus of points on an Argand diagram If: πβππ: π§ β 5 β 3π = 3 ο Now we can find the equation of the locus algebraicallyβ¦ π₯2 + π¦2 Replace z with βx + iyβ π₯ + ππ¦ β 5 β 3π = 3 π§ β 5 β 3π = 3 Use an algebraic method to find a Cartesian equation of the locus of z π§ = π₯ + ππ¦ = π₯ β 5 + π(π¦ β 3) = 3 π₯β5 2 + π¦β3 2 Group the real and imaginary terms Use the rule above to remove the modulus =3 Square both sides 2 (π₯ β 5) + π¦ β 3 2 =9 You (hopefully) recognise that this is the equation of a circle, radius 3 and with centre (5,3)! 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram As a general rule, the locus of: π§ β π§1 = π is a circle of radius r and centre (x1,y1) where z1 = x1 + iy1 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers π§ β π§1 = π ο Circle radius r and centre (x1,y1) You can use complex numbers to represent a locus of points on an Argand diagram Give a geometrical interpretation of each of the following loci of z: a) d) π§ β 3π = 4 ο Circle, centre (0,3) radius 4 2 β 5π β π§ = 3 (β1)(β2 + 5π + π§) = 3 β1 π§ β 2 + 5π = 3 b) π§ β (2 + 3π) = 5 π§ β (2 β 5π) = 3 ο Circle, centre (2,3) radius 5 c) π§ + 3 β 5π = 2 π§ β (β3 + 5π) = 2 Put the βfixedβ part in a bracket ο Circle, centre (-3,5) radius 2 βFactoriseβ the part inside the modulus You can write this as 2 moduli multiplied |-1| = 1, put the βfixedβ part in a bracket ο Circle, centre (2,-5) radius 3 Effectively for d), you just swap the signs of everything in the modulus, its value will not change ο |10 - 8| = |-10 + 8| 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x,y) which is represented by z on an Argand diagram, if: π§ = π§ β 6π π§ π§ β 6π π§ β (6π) This is the distance of P(x,y) from the origin (0,0) This is the distance of P(x,y) from (0,6) y (0,6) ο We therefore need the set of points that are the same distance from (0,0) and (0,6) ο This will be the bisector of the line joining the two co-ordinates y=3 (0,0) x ο You can see that it is the line with equation y = 3 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers πΌπ: π§ = π₯ + ππ¦ You can use complex numbers to represent a locus of points on an Argand diagram πβππ: π§ = π₯ + ππ¦ = π₯2 + π¦2 π§ = π§ β 6π Replace z with x + iy Sketch the locus of P(x,y) which is represented by z on an Argand diagram, if: π₯ + ππ¦ = π₯ + ππ¦ β 6π Factorise the βiβ terms on the right side π₯ + ππ¦ = π₯ + π(π¦ β 6) π§ = π§ β 6π ο Show that the locus is y = 3 using an algebraic method π₯2 + π¦2 = π₯2 + π¦ β 6 2 Use the rule above to remove the moduli Square both sides π₯2 + π¦2 = π₯2 + π¦ β 6 2 Expand the bracket π₯ 2 + π¦ 2 = π₯ 2 + π¦ 2 β 12π¦ + 36 Cancel terms on each side 0 = β12π¦ + 36 Add 12y 12π¦ = 36 Divide by 12 π¦=3 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers πΌπ: π§ = π₯ + ππ¦ You can use complex numbers to represent a locus of points on an Argand diagram πβππ: π§ = π₯ + ππ¦ = π₯2 + π¦2 π§β3 = π§+π Replace z with x + iy a) Use an algebraic method to find the Cartesian equation of the locus of z if: π§β3 = π§+π π = βππ + π b) Represent the locus of z on an Argand diagram π₯ + ππ¦ β 3 = π₯ + ππ¦ + π Group real and imaginary parts π₯ β 3 + ππ¦ = π₯ + π(π¦ + 1) π₯β3 2 π₯β3 2 + π¦2 = π₯2 + π¦ + 1 2 Use the rule above to remove the moduli Square both sides 2 2 +π¦ =π₯ + π¦+1 2 Expand brackets π₯ 2 β 6π₯ + 9 + π¦ 2 = π₯ 2 + π¦ 2 + 2π¦ + 1 Cancel terms β6π₯ + 9 = 2π¦ + 1 Subtract 1 β6π₯ + 8 = 2π¦ Divide by 2 β3π₯ + 4 = π¦ 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram y (0,4) a) Use an algebraic method to find the Cartesian equation of the locus of z if: π§β3 = π§+π x π = βππ + π b) Represent the locus of z on an Argand diagram (3,0) (0,-1) π§β3 = π§+π y = -3x + 4 Distance from (3,0) Distance from (0,-1) 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers πΌπ: π§ = π₯ + ππ¦ You can use complex numbers to represent a locus of points on an Argand diagram πβππ: π§ β 6 = 2 π§ + 6 β 9π Replace z with βx + iyβ π₯ + ππ¦ β 6 = 2 π₯ + ππ¦ + 6 β 9π If: π§ β 6 = 2 π§ + 6 β 9π a) Use algebra to show that the locus of z is a circle, stating its centre and radius π + ππ π + π β ππ π = πππ Circle, centre (-10,12) and radius 10 b) Sketch the locus of z on an Argand diagram π₯2 + π¦2 π§ = π₯ + ππ¦ = Group real and imaginary parts π₯ β 6 + ππ¦ = 2 π₯ + 6 + π(π¦ β 9) π₯β6 2 + π¦2 = 2 π₯+6 2 + π¦β9 2 π₯β6 2 + π¦2 = 4 π₯ + 6 2 + π¦β9 2 Replace the moduli using the rule above Square both sides (remember the β2β) Expand some brackets π₯ 2 β 12π₯ + 36 + π¦ 2 = 4 π₯ 2 + 12π₯ + 36 + π¦ 2 β 18π¦ + 81 Expand another bracket Group all terms on one side π₯ 2 β 12π₯ + 36 + π¦ 2 = 4π₯ 2 + 48π₯ + 144 + 4π¦ 2 β 72π¦ + 324 0 = 3π₯ 2 + 60π₯ + 3π¦ 2 β 72π¦ + 432 0 = π₯ 2 + 20π₯ + π¦ 2 β 24π¦ + 144 0 = π₯ + 10 2 β 10 2 0 = π₯ + 10 2 + π¦ β 12 2 100 = π₯ + 10 2 + π¦ β 12 2 + π¦ β 12 β 100 2 Divide by 3 β 12 2 + 144 Completing the square Simplify Add 100 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram (-10,12) If: The circle shows the set of points that are twice as far from (6,0) as they are from (-6,9)! (-6,9) π§ β 6 = 2 π§ + 6 β 9π a) Use algebra to show that the locus of z is a circle, stating its centre and radius π + ππ π + π β ππ π P(x,y) (6,0) x = πππ Circle, centre (-10,12) and radius 10 b) Sketch the locus of z on an Argand diagram π§ β 6 = 2 π§ + 6 β 9π The distance from (6,0) Is equal Twice the distance to from (-6,9) 3F π§ β π§1 Is the distance between the variable point z and the fixed point z1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Therefore: When π§ β π§1 = π π§ β π§2 ο An Algebraic method will most likely be the best way to find the equation of the locus of z If: π§ β 6 = 2 π§ + 6 β 9π a) Use algebra to show that the locus of z is a circle, stating its centre and radius π + ππ π + π β ππ π ο You will probably have to use completing the square (sometimes with fractions as well!) = πππ Circle, centre (-10,12) and radius 10 b) Sketch the locus of z on an Argand diagram 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram If: ππππ§ = π 4 Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically. The line is not extended back downwards ο It is known as a βhalflineβ π π x ο The locus will be the set of points which start at (0,0) and make an argument of Ο/4 with the positive xaxis 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram y π π If: ππππ§ = x π 4 Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically. ο The locus will be the set of points which start at (0,0) and make an argument of Ο/4 with the positive xaxis ππππ§ = π 4 π πππ(π₯ + ππ¦) = 4 πππβ1 x π¦ π = π₯ 4 Replace z with βx + iyβ The value of the argument is tan-1(opposite/adjacent) π¦ π = πππ π₯ 4 π¦ =1 π₯ π¦=π₯ βNormal tanβ Calculate the tan part Multiply by x (x > 0) 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram If: arg(π§ β 2) = (2,0) π π x π 3 Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically. ο The locus will be the set of values that, when we subtract 2 from them, make an angle of Ο/3 with the origin ο The locus must therefore start at (2,0) rather than (0,0)! 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram y (2,0) If: x-2 π arg(π§ β 2) = 3 Sketch the locus of P(x,y) which is represented by z on an Argand diagram. Then find the Cartesian equation of this locus algebraically. ο The locus will be the set of values that, when we subtract 2 from them, make an angle of Ο/3 with the origin ο The locus must therefore start at (2,0) rather than (0,0)! π π arg(π§ β 2) = π 3 π πππ(π₯ + ππ¦ β 2) = 3 πππβ1 x π¦ π = π₯β2 3 Replace z with βx + iyβ The value of the argument is tan-1(opposite/adjacent) π¦ π = πππ π₯β2 3 π¦ = 3 π₯β2 π¦ = 3π₯ β 2 3 βNormal tanβ Calculate the tan part Multiply by (x β 2) (x > 2) 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram If: πππ(π§ + 3 + 2π) = 3π 4 ππ π x (-3,-2) Sketch the locus of z on an Argand diagram and use an algebraic method to find the equation of the line. ο When we add 3 and 2i to z, the argument from (0,0) and the positive x-axis will be 3Ο/4 ο So the line will have to start at (-3,-2) 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram y+2 If: πππ(π§ + 3 + 2π) = 3π 4 Sketch the locus of z on an Argand diagram and use an algebraic method to find the equation of the line. ο When we add 3 and 2i to z, the argument from (0,0) and the positive x-axis will be 3Ο/4 ο So the line will have to start at (-3,-2) x ππ π x + 3 (-3,-2) πππ(π§ + 3 + 2π) = 3π 4 3π πππ(π₯ + ππ¦ + 3 + 2π) = 4 πππβ1 π¦+2 3π = π₯+3 4 Replace z with βx + iyβ The value of the argument is tan-1(opposite/adjacent) π¦+2 3π = πππ π₯+3 4 π¦+2 = β1 π₯+3 π¦ + 2 = βπ₯ β 3 π¦ = βπ₯ β5 βNormal tanβ Calculate the tan part Multiply by (x + 3) Subtract 2 (x < -3) 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram So therefore: πππ π§ β π§1 = π Is represented by a half line starting at z1 and making an angle of ΞΈ with a line parallel to the x-axis 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Joining the ends of a chord to different points on the circumference will always create the same angle, if the points are in the same sector βAngles in the same sector are equalβ For the next set of Loci, you need to remember some rules relating to circles Major arc β ΞΈ is acute Minor arc β ΞΈ is obtuse ΞΈ ΞΈ A ΞΈ Semi-circle β ΞΈ is 90° ΞΈ B ΞΈ x ΞΈ B A 2x A A B If they are joined to a point on the major arc If they are joined to a point on the minor arc If the chord is the diameter of the circle ο The angle will be acute ο The angle will be obtuse ο The angle will be 90° B ο The angle at the centre is twice the angle at the circumference 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram If: πππ a) π§β6 π = π§β2 4 Sketch the locus of P(x,y), which is represented by z on an Argand diagram ο The argument above can be rewritten using this rule: πππ πππ π§β6 π = π§β2 4 π§1 = πππ π§1 β πππ π§2 π§2 πππ π§ β 6 β πππ π§ β 2 = So what we are doing is drawing the locus of points where the difference between these arguments is Ο/4 π 4 3F Further complex numbers y arg(z β 6) = ΞΈ1 You can use complex numbers to represent a locus of points on an Argand diagram arg(z β 2) = ΞΈ2 If: πππ a) ΞΈ2 π§β6 π = π§β2 4 Sketch the locus of P(x,y), which is represented by z on an Argand diagram πππ π§ β 6 β πππ π§ β 2 = This angle must therefore be ΞΈ1 β ΞΈ2, the difference between the arguments! ΞΈ1 π 4 So what we are doing is drawing the locus of points where the difference between these arguments is Ο/4 However, there are more points that satisfy this rule! ΞΈ1 ΞΈ2 (2,0) (6,0) x ο Imagine drawing both arguments β we will use ΞΈ1 and ΞΈ2 to represent their values ο Using alternate angles, we can show the angle between the arguments is their difference ο We want this difference to be Ο/4 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram If: Ο/ π§β6 π πππ = π§β2 4 a) Ο/ Sketch the locus of P(x,y), which is represented by z on an Argand diagram πππ π§ β 6 β πππ π§ β 2 = π 4 So what we are doing is drawing the locus of points where the difference between these arguments is Ο/4 Geogebra Example 4 4 ΞΈ2 (2,0) ΞΈ1 ΞΈ2 ΞΈ1 (6,0) x ο If we move the point where the lines cross along the major arc of a circle, then the value of Ο/4 will remain the same ο The arguments will change but this doesnβt matter, it is the difference that matters! ο So the locus of a difference between arguments is always given by an arc of a circle 3F Further complex numbers y βThe angle at the centre is twice the angle at the circumferenceβ You can use complex numbers to represent a locus of points on an Argand diagram Ο/ 4 If: πππ a) π§β6 π = π§β2 4 Ο/ Sketch the locus of P(x,y), which is represented by z on an Argand diagram b) Find the Cartesian equation of this locus We need the centre of the βcircleβ and its radius ο We need to use another of the rules we saw: (2,0) 2 (6,0) x We can use this isosceles triangle to find the information we needβ¦ Centre Radius Radius 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Centre Radius Radius If: πππ a) (4,2) π§β6 π = π§β2 4 Sketch the locus of P(x,y), which is represented by z on an Argand diagram b) Find the Cartesian equation of this locus We need the centre of the βcircleβ and its radius Centre (4,2) Radius 2β2 Ο/ 4 2 Ο/ 4 (2,0) 2 (4,0) (6,0) ο Split the triangle in the middle, the smaller angles will both be Ο/ (45α΅) (because the top angle was Ο/ ) 4 2 ο The middle of the base will be (4,0), and you can work out the side lengths from this ο The top will therefore be at (4,2) ο Use Pythagorasβ Theorem to find the diagonal (the radius) 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram Ο/ 4 If: πππ a) π§β6 π = π§β2 4 Sketch the locus of P(x,y), which is represented by z on an Argand diagram b) Find the Cartesian equation of this locus We need the centre of the βcircleβ and its radius Centre (4,2) Radius 2β2 (6,0) (2,0) x The locus is therefore the arc of a circle with the following equation: π₯β4 2 + π¦β2 2 =8 π¦>0 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Generally, for this type of question, you need to follow 3 steps: Step 1: Mark on the Argand diagram the two points where the arguments start Step 2: Decide whether the arc is going to be major, minor, or a semicircle, by considering the angle Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator point ο Anti-clockwise if ΞΈ is positive ο Clockwise if ΞΈ is negative If the value we want is positive, then ΞΈ1 > ΞΈ2 If the value we want is negative, then ΞΈ2 > ΞΈ1 Drawing in the direction indicated in step 3 means you will ensure the arguments are correct to give a positive or negative answer ο As we do some examples we will refer to this! 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x,y) on an Argand diagram if: π§ π πππ = π§ β 4π 2 y Generally, for this type of question, you need to follow 3 steps: Step 1: Mark on the Argand diagram the two points where the arguments start Step 2: Decide whether the arc is going to be major, minor, or a semicircle, by considering the angle Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator point ο Anti-clockwise if ΞΈ is positive ο Clockwise if ΞΈ is negative (0,4) (0,0) and (0,4) The angle to make is Ο/2 ο A semi-circle (0,0) x Ξ is positive, so draw anti-clockwise from (0,0) (numerator point) to (0,4) (denominator point) 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram ο In step 3 we had to choose whether to draw the diagram clockwise or anti-clockwise from the numerator point to the denominator point ο Lets show why this is correct! y y (0,4) (0,4) ΞΈ2 ΞΈ2 ΞΈ2 ΞΈ1 ΞΈ2 ΞΈ1 ΞΈ1 ΞΈ1 x (0,0) ο We drew the angle anticlockwise from (0,0) to (0,4) ο However, as ΞΈ2 is actually negative, the sum is really ΞΈ1 + (-ΞΈ2) ο Using the alternate angles, the angle between the arguments is ΞΈ1 + ΞΈ2 = ΞΈ1 β ΞΈ2 ο This angle is therefore what we were wanting! Basically, always use the rule in step 3! x (0,0) ο If we drew the arc the other way clockwise from (0.0) to (0,4) ο Using the alternate angles, the on the outside is ΞΈ1 + ΞΈ2 ο However, as ΞΈ2 is actually negative, the sum is really ΞΈ1 + (-ΞΈ2) = ΞΈ1 β ΞΈ2 ο But of course it is on the wrong side of the arc so we do not want this part of the circle! 3F Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x,y) on an Argand diagram if: πππ π§ + 3π π = π§β2 3 y Generally, for this type of question, you need to follow 3 steps: Step 1: Mark on the Argand diagram the two points where the arguments start Step 2: Decide whether the arc is going to be major, minor, or a semicircle, by considering the angle Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator point ο Anti-clockwise if ΞΈ is positive ο Clockwise if ΞΈ is negative (0,-3) and (0,2) (0,2) The angle to make is Ο/3 ο A major arc Ξ is positive, so draw anticlockwise from (0,-3) (numerator point) to (0,2) (denominator point) x (0,-3) 3F Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram 3 Given that the complex number z = x + iy satisfies the equation: 3 (12,5) π§ β 12 β 5π = 3 Find the minimum and maximum values of |z| ο Start by drawing this on an Argand diagram ο It is a circle, centre (12,5) radius 3 units 13 x ο The smallest and largest values for |z| will be on the same straight line through the circleβs centre ο You can mark the size of the radius on the diagram ο Find the distance from (0,0) to (12,5), then add/subtract 3 to find the largest and smallest values 52 + 122 = 13 ο So the largest value of |z| will be 16 and the smallest will be 10 3F Further complex numbers y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci (4,2) The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required x Shade on an Argand diagram the region indicated by: π§ β 4 β 2π β€ 2 ο Start with a circle, centre (4,2) and radius 2 units (as 2 is the βlimitβ) The region we want is where the absolute value of z is less than 2 ο This will be the region inside the circle 3G Further complex numbers y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required (4,0) (6,0) x Shade on an Argand diagram the region indicated by: π§β4 < π§β6 ο Start with the perpendicular bisector between (4,0) and (6,0) as this is the βlimitβ π§β4 < π§β6 The distance to |z β 4| must be less than the distance to |z β 6| ο Shade the region closest to (4,0) 3G Further complex numbers y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci π π (2,2) The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required Shade on an Argand diagram the region indicated by: π 0 β€ πππ π§ β 2 β 2π β€ 4 ο Start by drawing the limits of the argument from the point (2,2) x The argument must be between these two values ο Shade the region between the two arguments 3G Further complex numbers y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required x Shade on an Argand diagram the region indicated by: π§ β 4 β 2π β€ 2 Imagine all the regions were on the same diagram π§β4 < π§β6 and 0 β€ πππ π§ β 2 β 2π β€ π 4 ο The region we want will have to satisfy all of these at the same time! 3G Test Your Understanding P6 June 2003 Q4(i)(b) Shade the region for which π π π§ β 1 β€ 1 and β€ arg π§ + 1 β€ 12 2 ? Further complex numbers y You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The z-plane (uses x and y) x Effectively we take a set of points in the complex plane, transform them all and map them on a new complex plane ο You will need to use Algebraic methods a lot for this as visualising the transformations can be very difficult! Transformation from one plane to the next! v The w-plane (uses u and v) u 3H Further complex numbers y You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T1 represents a transformation from the z plane, where z = x + iy, to the w-plane where w = u + iv. v x The z-plane u The w-plane Circle centre (0,0), radius 2 Circle centre (-2,4), radius 2 ο To start with, make z the subject π€ = π§ β 2 + 4π Add 2, subtract 4i Describe the locus of P under the transformation T1, when T1 is given by: π§ = π€ + 2 β 4π π1 : π€ = π§ β 2 + 4π π§ = π€ + 2 β 4π ο We will work out the new set of points algebraicallyβ¦ 2 = π€ + 2 β 4π The modulus of each side must be the same We know |z| from the question Circle, centre (-2,4), radius 2 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T2 represents a transformation from the z plane, where z = x + iy, to the wplane where w = u + iv. Describe the locus of P under the transformation T2, when T2 is given by: π2 : π€ = 3π§ ο We will work out the new set of points algebraicallyβ¦ The z-plane y The w-plane v x Circle centre (0,0), radius 2 u Circle centre (0,0), radius 6 ο To start with, make z the subject π€ = 3π§ π€ =π§ 3 Divide by 3 π€ = π§ 3 π€ =2 3 |z|= 2 Split the modulus up Modulus of both sides π§1 π§1 = π§2 π§2 |3|=3 so multiply by 3 π€ =6 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T2 represents a transformation from the z plane, where z = x + iy, to the wplane where w = u + iv. Describe the locus of P under the transformation T3, when T3 is given by: 1 π3 : π€ = π§ + π 2 ο We will work out the new set of points algebraicallyβ¦ The z-plane y The w-plane v x Circle centre (0,0), radius 2 u Circle centre (0,1), radius 1 ο To start with, make z the subject 1 π§+π 2 1 π€βπ = π§ 2 π€= 1 π€βπ = π§ 2 1 π€βπ = π§ 2 1 π€βπ = 2 2 Subtract i Leaving z like this can make the problem easier! (rather than rearranging completely) Modulus of both sides You can split the modulus on the right π§1 π§2 = π§1 π§2 |z| = 2 Simplify the right side π€βπ =1 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv For the transformation w = z2, where z = x + iy and w = u + iv, find the locus of w when z lies on a circle with equation x2 + y2 = 16 ο It is very important for this topic that you draw information on z or |z| from the question The z-plane y The w-plane x Circle centre (0,0), radius 4 u Circle centre (0,0), radius 16 π€ = π§2 π€ = π§2 ο The equation x2 + y2 = 16 is a circle, centre (0,0) and radius 4 π€ = π§ π§ ο Therefore |z| = 4 π€ =4×4 ο We now proceed as before, by writing the equation linking w and z in such a way that |z| can be replaced v Modulus of both sides Split the modulus up Replace |z| with 4 Calculate π€ = 16 Circle, centre (0,0) and radius 16 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: π€= 5ππ§ + π π§+1 Show that the image, under T, of the circle |z| = 1 in the z-plane, is a line l in the w-plane. Sketch l on an Argand diagram. ο Make z the subject! ο Now eliminate z using what we knowβ¦ π€= 5ππ§ + π π§+1 Multiply by (z + 1) π€(π§ + 1) = 5ππ§ + π Expand the bracket π€π§ + π€ = 5ππ§ + π π€π§ β 5ππ§ = π β π€ Subtract 5iz and subtract w Factorise the left side π§(π€ β 5π) = π β π€ πβπ€ π§= π€ β 5π π§ = 1= πβπ€ π€ β 5π πβπ€ π€ β 5π Divide by (w β 5i) Modulus of both sides |z| = 1 Multiply by |w β 5i| π€ β 5π = π β π€ π€ β 5π = π€ β π |i - w| = |w β i| 3H Further complex numbers Transformation T You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: π€= 5ππ§ + π π§+1 Show that the image, under T, of the circle |z| = 1 in the z-plane, is a line l in the w-plane. Sketch l on an Argand diagram. π€ β 5π = π€ β π π§ =1 Circle centre (0,0), radius 1 The z-plane Perpendicular bisector between (0,1) and (0,5) ο The line v = 3 y The w-plane v x u So a circle can be transformed into a straight line! 3H Test Your Understanding FP2 June 2009 Q6 A transformation π from the π§-plane to the π€-plane is given by π§ π€= , π§ β βπ π§+π The circle with equation π§ = 3 is mapped by π onto the curve πΆ. (a) Show that πΆ is a circle and find its centre and radius. (8) (b) The region π§ < 3 in the π§-plane is mapped by π onto the region π in the π€-plane. Shaded the region π on an Argand diagram. (2) (Hint: be careful and donβt guess! Try a value of π§ where you know that π§ < 3, and then transform it to see where it ends up.) ? ? Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: π€= 3π§ β 2 π§+1 π€ π§ + 1 = 3π§ β 2 Expand the bracket π€π§ + π€ = 3π§ β 2 Subtract wz and add 2 π€ + 2 = 3π§ β π€π§ Factorise the right side π€ + 2 = π§(3 β π€) 3π§ β 2 π€= π§+1 π€+2 =π§ 3βπ€ Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane, is a different circle C in the w-plane. π€+2 = π§ 3βπ€ State the centre and radius of C. Multiply by (z + 1) π€+2 =2 3βπ€ Divide by (3 β w) Modulus of each side Split up the modulus |z| = 2 Multiply by |3 - w| π€+2 =2 3βπ€ |3 - w| = |w - 3| ο Remember that x2 + y2 = 4 is the same as |z| = 2 π€+2 =2 π€β3 We now need to find what the equation of this will be! 3H We will find the equation as we did in the early part of section 3F! Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: π€= 3π§ β 2 π§+1 Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane, is a different circle C in the w-plane. State the centre and radius of C. x2 y2 ο Remember that + = 4 is the same as |z| = 2 π€+2 =2 π€β3 Replace w with βu + ivβ π’ + ππ£ + 2 = 2 π’ + ππ£ β 3 Group real/imaginary terms π’ + 2 + ππ£ = 2 (π’ β 3) + ππ£ Remove the modulus π’+2 2 2 +π£ = 4 π’β3 2 +π£ 2 Expand brackets π’2 + 4π’ + 4 + π£ 2 = 4 π’2 β 6π’ + 9 + π£ 2 π’2 + 4π’ + 4 + π£ 2 = 4π’2 β 24π’ + 36 + 4π£ 2 0 = 3π’2 β 28π’ + 3π£ 2 + 32 0 = π’2 β 2 100 14 = π’β 9 3 2 14 β 3 Move all to one side Divide by 3 28 32 π’ + π£2 + 3 3 14 0= π’β 3 Expand more brackets! 2 + π£2 + + π£2 32 3 Use completing the square Move the number terms across Circle, centre (14/3,0), radius 10/3 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: π€= 3π§ β 2 π§+1 Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane, is a different circle C in the w-plane. π§ =2 2 Transformation T 14 π’β 3 2 π₯ +π¦ =4 Circle centre (0,0), radius 2 The z-plane π€+2 =2 π€β3 2 + π£2 = 100 9 Circle, centre (14/3,0), radius 10/3 y The w-plane x v u State the centre and radius of C. 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv A transformation T of the z-plane to the w-plane is given by: π€= ππ§ β 2 1βπ§ Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram. ο Start by rearranging to make z the subject βas usualβ π€= ππ§ β 2 1βπ§ π€ 1 β π§ = ππ§ β 2 π€ β π€π§ = ππ§ β 2 π€ + 2 = ππ§ + π€π§ Multiply by (1 β z) Expand the bracket Add 2, Add wz Factorise the right side π€ + 2 = π§(π + π€) π€+2 =π§ π+π€ π§= π€+2 π+π€ π€+2 π§= π€+π Divide by (i + w) Write the other way round (if you feel it is easier!) i+w=w+i At this point we have a problem, as we do not know anything about |z| ο However, as z lies on the βrealβ axis, we know that y = 0 ο Replace z with βx + iyβ π₯ + ππ¦ = π€+2 π€+π 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv A transformation T of the z-plane to the w-plane is given by: π€= ππ§ β 2 1βπ§ Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram. ο Now, you need to rewrite the right side so you can separate all the real and imaginary terms ο You must be extremely careful with positives and negatives here! π₯ + ππ¦ = π₯ + ππ¦ = π€+2 π€+π π’ + ππ£ + 2 π’ + ππ£ + π Replace w with u + iv Group real and imaginary terms π₯ + ππ¦ = π’ + 2 + ππ£ π’ + π(π£ + 1) π₯ + ππ¦ = π’ + 2 + ππ£ π’ β π(π£ + 1) × π’ + π(π£ + 1) π’ β π(π£ + 1) π₯ + ππ¦ = π’(π’ + 2) β π(π’ + 2)(π£ + 1) + ππ’π£ β π 2 π£(π£ + 1) π’2 β ππ’(π£ + 1) + ππ’(π£ + 1) β π 2 (π£ + 1)2 π₯ + ππ¦ = π₯ + ππ¦ = Multiply by the denominator but with the opposite sign (this will cancel βiβ terms on the bottom π’(π’ + 2) + π£(π£ + 1) + ππ’π£ β π(π’ + 2)(π£ + 1) π’2 + (π£ + 1)2 π’π£ β (π’ + 2)(π£ + 1) π’ π’ + 2 + π£(π£ + 1) π + π’2 + π£ + 1 2 π’2 + π£ + 1 2 Simplify i2 = -1 Separate real and βiβ terms As z lies on the x-axis, we know y = 0 ο Therefore, the imaginary part on the right side must also equal 0 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv A transformation T of the z-plane to the w-plane is given by: π€= ππ§ β 2 1βπ§ Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram. π₯ + ππ¦ = π’π£ β (π’ + 2)(π£ + 1) π’ π’ + 2 + π£(π£ + 1) π + π’2 + π£ + 1 2 π’2 + π£ + 1 2 ο Set the imaginary part equal to 0 π’π£ β (π’ + 2)(π£ + 1) =0 π’2 + π£ + 1 2 π’π£ β (π’ + 2)(π£ + 1) = 0 Multiply by (u2 + (v + 1)2) (you will be left with the numerator) Multiply out the double bracket π’π£ β (π’π£ + 2π£ + π’ + 2) = 0 Subtract all these terms π’π£ β π’π£ β 2π£ β π’ β 2 = 0 The βuvβ terms cancel out β2π£ β π’ β 2 = 0 Make v the subject 2π£ = βπ’ β 2 ο You can now find an equation for the line in the w-plane 1 π£ =β π’β1 2 Divide by 2 So the transformation has created this line in the w-plane (remember v is essentially βyβ and u is essentially βxβ) 3H Further complex numbers You can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + iv Transformation T π¦=0 (z lies on the real axis) 1 π£ =β π’β1 2 Straight line, gradient is 1/ and y-intercept at 2 (0,-1) A transformation T of the z-plane to the w-plane is given by: ππ§ β 2 π€= 1βπ§ The z-plane y v The w-plane Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an Argand diagram. x π=π -2 u -1 π π=β πβπ π 3H Summary β’ You have learnt a lot in this chapter!! β’ You have seen proofs of and uses of De Moivreβs theorem β’ You have found real and complex roots of powers β’ You have see how to plot Loci and perform transformations of complex functions
