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Trigonometric Form of a Complex Number Complex Numbers Recall that a complex number has a real component and an imaginary component. bi z = 3 – 2i The absolute value of a complex number is its distance from the origin. z a b Imaginary axis a z = a + bi 2 Argand Diagram Real axis z = 3 – 2i The names and letters are changing, but this sure looks familiar. 2 z 32 2 9 4 13 2 The Trig form of a Complex Number x a r r y b sin r r cos The trig form of the complex number z a bi is z ( r cos ri sin ) r cos i sin . r is called the modulus and is the distance from the origin to the point. r a 2 b2 is called the argument and is the angle formed with the x-axis. b tan 1 a r a 2 b2 b r sin a r cos How is it Different? In a rectangular system, you go left or right and up or down. z 2 2i In a trigonometric or polar system, you have a direction to travel and a distance to travel in that direction. z 2 cos45 i sin 45 Polar form (2,45) Converting from Rectangular form to Trig form 1. Find r. r a 2 b2 b 2. Find . tan 1 a 3. Fill in the blanks in z r cos i sin 1. Find r r 42 32 16 9 r 25 5 2. Find tan 1 3 36.9 4 Convert z = 4 + 3i to trig form. 3. Fill in the blanks z 5 cos 36.9 i sin 36.9 Polar form 5,36.9 Converting from Trig Form to Rectangular Form This one’s easy. 1. Evaluate the sin and cos. 2. Distribute in r 1. Evaluate the sin and cos 2. Distribute the 4. Convert 4(cos 30 + i sin 30) to rectangular form. 3 1 4 i 2 2 2 3 2i Multiplying Complex Numbers To multiply complex numbers in rectangular form, you would FOIL and convert i2 into –1. To multiply complex numbers in trig form, you simply multiply the rs and add the thetas. a bi c di r1 cos 1 i sin 1 r2 cos 2 i sin 2 r1r2 cos 1 2 i sin 1 2 ac adi bci dbi 2 ac adi bci db ac db ad bc i The formulas are scarier than it really is. multiply z1 z2 Example Rectangular form z1 z2 2 3 2i 3 2 3 2i Where z1 2 3 2i 4 cos 30 i sin 30 z2 3 2 3 2i 6 cos 45 i sin 45 Trig form z1 z2 6 6 6 6i 6 2i 6 2i 2 4 cos 30 i sin 30 6 cos 45 i sin 45 4 6 cos 30 45 i sin 30 45 6 6 6 6i 6 2i 6 2 24 cos 75 i sin 75 6 r 6 6 2 6 6 6 2 i 6 6 6 2 6 2 6 6 2 2 r 216 72 12 72 216 72 12 72 r 576 24 6 6 6 2 75 6 6 6 2 tan 1 Dividing Complex Numbers In rectangular form, you rationalize using the complex conjugate. a bi c di a bi c di c di c di ac adi bci bdi 2 c 2 d 2i 2 ac adi bci bd c2 d 2 ac bd bc ad 2 i 2 2 2 c d c d In trig form, you just divide the rs and subtract the theta. r1 cos 1 i sin 1 r2 cos 2 i sin 2 r1 cos 1 2 i sin 1 2 r2 divide Where z1 3 2 3 2i 6 cos 45 i sin 45 Example z2 2 3 2i 4 cos 30 i sin 30 Rectangular form Trig form 3 2 3 2i 2 3 2i 6 cos 45 i sin 45 4 cos30 i sin 30 3 2 3 2i 2 3 2i 2 3 2 i 2 3 2 i 6 6 6 2i 6 6i 6 2i 2 12 4i 2 6 6 6 2i 6 6i 6 2 12 4 6 6 6 2 16 6 6 cos 45 30 i sin 45 30 4 3 cos15 i sin15 2 6 6 2 i 16 z1 z2 6 6 6 2 1 16 15 tan 6 6 6 2 216 72 12 72 216 72 12 72 16 r 2 6 6 6 2 6 6 6 2 r 16 16 256 r 576 9 3 256 4 2 2 De Moivre’s Theorem If z r (cos i sin ) is a complex number And n is a positive integer n n Then z r (cos i sin ) r (cos n i sin n ) n Who was De Moivre? A brilliant French mathematician who was persecuted in France because of his religious beliefs. De Moivre moved to England where he tutored mathematics privately and became friends with Sir Issac Newton. De Moivre made a breakthrough in the field of probability (writing the Doctrine of Chance), but more importantly he moved trigonometry into the field of analysis through complex numbers with De Moivre’s theorem. But, can we prove DeMoivre’s Theorem? Let’s look at some Powers of z. z r (cos i sin ) z r (cos i sin ) 2 2 r 2 (cos i sin )2 r 2 (cos i sin )(cos i sin ) r 2 (cos2 2i cos sin i 2 sin 2 ) r 2 (cos2 sin 2 2i cos sin ) r (cos2 i sin 2 ) 2 Let’s look at some more Powers of z. z r (cos i sin ) 3 3 r(cos i sin )r 2 (cos i sin )2 r (cos3 i sin 3 ) 3 z r (cos i sin ) 4 4 r(cos i sin )r 3 (cos i sin )3 r 4 (cos4 i sin 4 ) It appears that: cos i sin cos n i sinn n Proof: n=1, then the statement is true. Assume We can continue in the previous manor up to some arbitrary k cos i sin cos k i sink cos i sin k Let n = k, so that: Now find k1 cos i sin cos k i sinkcos i sin cos i sin cos k cos sink sin icos k sin sink cos k1 k1 cos i sin cos i sin k1 cos(k ) i sin(k ) k1 cos(k 1) i sin(k 1) Euler’s Formula e cos i sin i We can also use Euler’s formula to prove DeMoivre’s Theorem. cos i sin n e =e i n in = cos n i sin n So what is the use? Find an identity for cos5 using Mr. De Moivre’s fantastic theory cos 5 i sin5 cos i sin 5 Remember the binomial expansion: (a b)5 1(a )5 (b)0 5(a )4 (b)1 10(a )3 (b)2 10(a )2 (b)3 5(a )1 (b)4 1(a )0 (b)5 Apply it: cos 5 i sin 5 (1)(cos 5 ) (5)(cos 4 )(i sin ) (10)(cos 3 )(i 2 sin 2 ) (10)(cos 2 )(i 3 sin 3 ) (5)(cos )(i4 sin4 ) (1)(i5 sin5 ) Cancel out the imaginery numbers: cos 5 cos5 10 cos3 sin 2 5 cos sin 4 Now try these: cos3 cos 3cos sin sin3 3cos sin sin 3 2 2 3 sin4 4 cos sin 4 cos sin 3 3 Powers of Complex Numbers This is horrible in rectangular form. a bi a bi a bi a bi ... a bi n The best way to expand one of these is using Pascal’s triangle and binomial expansion. You’d need to use an i-chart to simplify. It’s much nicer in trig form. You just raise the r to the power and multiply theta by the exponent. z r cos i sin z n r n cos n i sin n Example z 5 cos 20 i sin 20 z 3 53 cos 3 20 i sin 3 20 z 3 125 cos 60 i sin 60 Roots of Complex Numbers There will be as many answers as the index of the root you are looking for Square root = 2 answers Cube root = 3 answers, etc. Answers will be spaced symmetrically around the circle You divide a full circle by the number of answers to find out how far apart they are The formula z r cos i sin n Using DeMoivre’s Theorem we get 360k 360k z n r cos i sin or n n k starts at 0 and goes up to n-1 This is easier than it looks. n 2 k 2 k r cos i sin n n General Process 1. 2. 3. 4. Problem must be in trig form Take the nth root of r. All answers have the same value for r. Divide theta by n to find the first angle. Divide a full circle by n to find out how much you add to theta to get to each subsequent answer. Example Find the 4th root of z 81 cos80 i sin80 1. Find the 4th root of 81 r 4 81 3 2. Divide theta by 4 to get the first angle. 3. Divide a full circle (360) by 4 to find out how far apart the answers are. 360 90 between answers 4 4. List the 4 answers. • • 80 20 4 z1 3 cos 20 i sin 20 The only thing that changes z2 3 cos 20 90 i sin 20 90 3 cos110 i sin110 is the angle. z3 3 cos 110 90 i sin 110 90 3 cos 200 i sin 200 The number of answers z 3 cos 200 90 i sin 200 90 3 cos 290 i sin 290 equals the number of roots. 4