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BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS COMPLEX NUMBERS Jerome Cardan is widely credited as being the first person to create the concept of complex numbers, to solve previous unsolvable problems, in 1545. He stated that we can think of ( 1) as having a solution i. i ( 1) Examples: 1. 25 25 (1) 5 1 5i 2. 7 7 (1) 7 1 7i A complex number has two parts to it; a real part and an imaginary part. z a bi We will define complex number as where a R( z) and b I ( z) . For example, z1 2 3i and z2 5 7i are both complex numbers. It is worth noting that i ( 1) , i 2 1, i3 i, i 4 1, i5 i etc. Also, we can show that if z1 a bi and z2 c di and z1 z2 then a c and b d . For z1 a bi and z2 c di where a, b, c, d R we can observe the following results: (i) z1 z2 a bi c di (ii) a c bi di z1z2 a bi c di ac adi bci bdi 2 a c b d i ac bd adi bci ac bd ad bc i which is a complex number. which is a complex number. (iii) z1 2 a bi 2 (iv) where k R ka kbi a 2 2abi b 2i 2 ka kb i which is a complex number. a 2 b 2 2abi kz1 k a bi a 2 b 2 2ab i which is a complex number. Now try: Page 90, Exercise 1 – questions 1, 2(f to i), 4f, 5, 6i, 7b, 8 PAGE 1 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Complex numbers can be used when solving quadratic equations. Example: 3. x2 4x 5 0 discriminant b2 4ac 4 2 4 1 5 4 roots are distinct and not real (i.e. complex). 2 x b b 4ac 2a 4 4 2 4 2i 2 2 i or 2 i 4. Find all three roots that satisfy x3 6 x 2 12 x 7 0 . PAGE 2 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS The Complex Conjugate is used to rationalise denominators etc. Let z a bi . Examples: Result (v): Therefore, the conjugate of z is, z a bi . 5. The conjugate of 3 i is 3 i . 6. z 5 3i z 5 3i . Therefore, for z a bi and z a bi (where a, b R ) we get the following result: zz a bi a bi [a difference of two sqaures] a bi 2 2 a 2 b 2i 2 a 2 b2 [which is a real number] Example: 7. To rationalise the denominator of a complex fraction: 3 2i 3 2i 1 2i 1 2i 1 2i 1 2i 3 2i 1 2i 1 2i 1 2i 2 3 8i 4i2 12 2i 3 8i 4 1 4 1 8i 5 1 8i 5 5 PAGE 3 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS We can use the above results to calculate the square root of a complex number: Example: 8. We know that 3 4i a bi because complex number complex number 2 Hence, 3 4i a bi [see result (iii) above] where a, b R . Square both sides to give 3 4i a bi 2 3 4i a 2 b 2 2abi By equating real and imaginary coefficients: a 2 b2 3 2ab 4 equation 1 b 2 a equation 2 Substitute equation 2 into equation 1 to get, a2 2 a 2 3 a 2 42 3 a 4 a 4 3 a2 a2 a4 4 3 a2 a 4 4 3a 2 a 4 3a 2 4 0 a 2 4 a 2 1 0 a2 4 a 2 or a 2 1 a i invalid as a R When a 2, b 1 and Therefore, a 2, b 1 3 4i 2 i or 2 i . Now try: Page 91, Exercise 2 – questions 3e, 5a, 2(c, e), 3(a, c, f), 4b, 5b, 6b(iv, v), 4c, 5c, 6b(vi), 8 PAGE 4 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Solving Equations with Complex Roots In example 3 above we saw that for x 2 4 x 5 0 we have roots that are distinct and not real (i.e. complex). 2 x b b 4ac 2a 4 4 2 2 i or 2 i Notice that when we use the quadratic formula to solve an equation we will always get two roots: in the above example we obtain the roots z 2 i and z 2 i . In fact, for any equation with a complex root the conjugate of that complex number will also be a root. These roots provide us with factors and the product of the two factors will also be a factor (think of the prime factors of 20. 20 2 2 5 . The product of any two of these factors is also a factor). Example: 9. Given that 4 i is a root of z 4 8z3 13z 2 32z 68 0 , find all other roots that satisfy this equation. [N.B. to check that 4 i is a root we could substitute it into the equation and observe that it works out to be zero.] z (4 i) is a factor z 4 i is a root so z 4 i is a root z (4 i) is also a factor and z (4 i) z (4 i) is also a factor i.e. z (4 i) z (4 i) ( z 4) i ( z 4) i ( z 4)2 i 2 z 2 8 z 16 (1) z 2 8 z 17 is also a factor We can use this quadratic factor to divide the polynomial z2 4 z 2 8 z 17 z 4 8 z 3 13 z 2 32 z 68 z 4 8 z 3 17 z 4 z 2 32 z 68 4 z 2 32 z 68 0 PAGE 5 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS z 4 8 z 3 13z 2 32 z 68 z 2 8 z 17 z 2 4 z 2 8 z 17 z 2 4 0 10. z 2 8 z 17 0 or z2 4 0 z 4 i, z 4 i or z 2, z 2 Find all the roots of 2z 4 3z3 17 z 2 12z 10 0 given that ( z 1 3i ) is a factor of 2z 4 3z3 17 z 2 12z 10 . Now try: Page 108, Exercise 8 – questions 4, 5, 6b, 7b, 8 PAGE 6 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Argand Diagrams A complex number can be represented geometrically by using an Argand diagram (displaying the complex number as we would a vector). The x-axis represents the real part of the complex number and the y-axis represents the imaginary part. This plane is called the complex plane and the complex number x + yi is represented by the point P(x, y). P y (Imaginary) r y O x x (Real) P(x, y) represents the complex number x yi . OP is considered to be a vector that has been rotated off the x-axis, the length of OP is r. The size of the rotation is called the argument of z and is often denoted by Arg z (for the diagram above notice that Arg z is 2n where n Z ). We call the value of Arg z that lies in the range the principal argument of z and denote this as arg z (notice the lower case is used here). r z , the modulus of z and so, r x2 y 2 We can see also that and and therefore tan 1 y x where x r cos y r sin z r cos i sin z x yi but written in polar form z r cos i sin . PAGE 7 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Examples: 11. y O P r 32 22 13 = 3 61 to 3 sig. fig. 2 Arg z tan 1 2 2n 3 x 3 (3, 2) is in the first quadrant so arg z = 0 187 radians (to 3 s.f.) Sketch the Argand diagram for z = -4 – 3i and find modulus and argument of z. 12. y -4 -3 P r (4)2 (3) 2 5 O x Arg z tan 1 3 2n (remember, ). 4 (-4, -3) is in the third quadrant so arg z = (0 205 ) 0 795 radians (to 3 s.f.) or arg z = (36 9 180) 143 (to 3 s.f.) Now try: Page 94, Exercise 3 – questions 3(f, h), 4, 6(a, e), 7(b, c, i, h), 8 PAGE 8 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS The Polar Form of Complex Numbers We have already seen that complex numbers can be written as z1 r cos i sin z2 s cos i sin and Using the polar form of complex numbers we get the following results: (vi) z1 z2 r cos i sin s cos i sin rs cos i sin cos i sin rs cos cos i cos sin i sin cos i 2 sin sin rs cos cos sin sin i cos sin sin cos rs cos( ) i sin( ) From this we can observe that and (vii) z1z2 z1 z2 Arg z1z2 Arg z1 Arg z2 . r (cos θ i sin θ) z1 z2 s(cos α i sin α) (cos θ i sin θ) (cos α i sin α) r s (cos α i sin α) (cos α i sin α) 2 r ( cos cos i sin sin2 i 2cos 2 sin i sin sin ) s cos i sin cos cos sin sin i(sin sin cos cos ) r s 1 r cos( ) i sin( ) s From this we can observe that and z z1 1 z2 z2 z Arg 1 Arg z1 Arg z2 . z 2 PAGE 9 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Examples: 13. Simplify 3 cos i sin 4 cos i sin . 3 3 2 2 14. z z1 9 cos i sin and z2 3 cos i sin . Find 1 . 2 2 6 6 z2 PAGE 10 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS 15. ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Let z r cos i sin . Find expressions for z 2 , z 2 z z 3, z 4, z 5 and z n. N.B. an easier way is to realise that we already know that: z2 z z r2 and Arg z 2 Arg z Arg z 2 z 2 r 2 cos 2 i sin 2 Now try: Page 98, Exercise 5 – questions 1(d, e, f, i) 2(b, d) From example 12 we can deduce that z n r n cos n i sin n . This is not a comprehensive proof: it can be proved by induction. This is called De Moivre’s Theorem. PAGE 11 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS z n r n cos n i sin n De Moivre’s Theorem Examples: 16. Let z 3 i . Write z in polar form and find and expression for z 4 (remember that arg z ). Argand diagram: Polar form: r= Arg z = z= Therefore, z4 PAGE 12 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS 17. ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Treat z 1 i like an algebraic expression (e.g. 1 x ). Use the binomial expansion to find and expression for z 5 (remember i 2 1, i3 i, i 4 1, i5 i ). Use De Moivre’s Theorem to find an expression for z 5 in polar form and show the solution on an Argand diagram. Hence, by equating the real parts of each of the above expressions, show that cos 3 1 . 4 2 PAGE 13 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS 18. ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS (from SQA AH 2005) Let z cos i sin . (a) Use the binomial expansion to express z 4 in the form u iv , where u and v are expressions involving sin and cos . 3 (b) Use De Moivre’s theorem to write down a second expression for z 4 . 1 (c) Using the results of (a) and (b), show that cos 4 p cos 2 q sec2 r, where , 2 2 cos 2 stating the values of p, q and r. 6 Now try: Page 98, Exercise 5 – question 5 Page 101, Exercise 6 – questions 1a, 2(c, d), 3d, 4(h, i, j), 6, 7 PAGE 14 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Sets of Loci on the Complex Plane Complex equations and inequations can be used to define a set of points in an Argand diagram. Examples: 19. | z | 4 (where z= x + i y) represents the shaded area shown below: y z 4 4 x2 y 2 4 x2 y 2 16 Which is a circle with centre (0, 0) and radius 4. x 4 z 4 z 4 is the circumference and enclosed are of the circle with centre (0, 0) and radius 4 20. Sketch the loci of the points given by z 1 2 . 21. On an Argand diagram indicate the loci of the points represented by z 2 z 2i . PAGE 15 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS nth Roots of Unity - Roots of a Complex Number 1 An extension of de Moivre’s Theorem, z1 r n cos 2k i sin 2k n n where k 0, 1, 2, 3...... n 1 . z r cos i sin De Moivre’s Theorem states that for z n r n cos n i sin n we get or to be more precise z n r n cos n Arg z i sin n Arg z where Arg z 2k for k Z . Therefore, 1 1 z n r n cos 1 2k i sin 1 θ 2kπ n n Examples: 22. for k 0, 1, 2, 3...... n 1 . Simplify z = 4 81(cos i sin ) = 81 cos i sin 2 2 2 2 = 1 4 1 81 4 cos 1 2k i sin 1 2k where k = 0, 1, 2, 3 4 2 4 2 = 3 cos 84k i sin 84k remember to adjust for . For k = 0, z 3 cos i sin 8 8 For k =1, z 3 cos 5 i sin 5 8 8 For k = 2, z 3 cos 9 i sin 9 3 cos 7 i sin 7 8 8 8 8 For k = 3, z 3 cos 13 i sin 13 3 cos 3 i sin 3 8 8 8 8 PAGE 16 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS 23. ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Solve for z when z 4 2 2i . z 4 (2) 2 (2) 2 2 2 arg z 4 tan 1 2 3 2 4 4 N.B. it's in the 3 rd quadrant 3 3 1 1 cos 4 4 2k i sin 4 4 2k 3 2 8 cos 1 3 2k i sin 1 3 2k where k 0, 1, 2, 3 4 4 4 4 z 2 2 1 4 k = 0 gives, 3 z 2 8 cos 3 i sin 3 16 16 3 2 8 cos 3 i sin 3 16 16 k = 1 gives, 3 z 2 8 cos 5 i sin 5 16 16 k = 2 gives, 3 z 2 8 cos 13 i sin 13 16 16 k = 3 gives, 3 z 2 8 cos 3 2 8 cos 3 2 8 cos 2116 i sin 2116 1116 i sin 1116 1116 i sin 1116 Now try: Page 106, Exercise 7 – questions 1(b, c, d, g), 2 PAGE 17 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Examples (from SQA AH exams 2001, 2002, 2005 and 2003): 24. Verify that i is a solution of z 4 4 z 3 3z 2 4 z 2 0 . Hence find all other solutions. 5 PAGE 18 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS 25. (a) ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Given that 1 cos i sin , , state the value of . 1 (b) Use de Moivre’s Theorem to find the non-real solutions, z1 and z2 , of the equation z 3 1 0 . 5 Hence show that z12 z2 and z22 z1 . 2 (c) Plot all the solutions of z 3 1 0 on an Argand diagram and state their geoemtrical significance. 3 PAGE 19 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS 26. ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Identify the locus in the complex plane given by z 1 2 . 3 27. Given the equation z 2i z 8 7i , express z in the form a ib . 4 PAGE 20 OF 21 BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS 28. ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS Given that w cos i sin , show that 1 cos i sin . w 1 Use de Moivre’s theorem to prove wk wk 2 cos k , where k is a natural number. 3 Expand w w1 4 by the binomial theorem and hence show that cos4 1 cos 4 1 cos 2 3 . 8 2 8 5 PAGE 21 OF 21