math 375 solution - Solutions to Exam 1 Review Problems 1....

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Unformatted text preview: Solutions to Exam 1 Review Problems 1. Let z =- 2 1- 3 i . (a) Calculate the rectangular forms of z , z , and z- 1 . (b) Calculate | z | , Arg( z ) and the polar form of z . (c) Calculate the rectangular form of z 42 . ( Hint: Use your answer from part (b).) Solution. (a) z =- 1 2- i 3 2 z =- 1 2 + i 3 2 z- 1 = z =- 1 2 + i 3 2 ( z- 1 6 = z always, its only true whenever | z | = 1) (b) Calculate | z | , Arg( z ) and the polar form of z . | z | = 1 Arg( z ) =- 2 3 z = e- i 2 3 (c) z 42 = e- i 2 3 42 = e- i 2 3 42 = e- i 28 = e i (- 14)2 = 1 2. Solve the equation | e i- 1 | = 2 for real. Solution. | e i- 1 | = 2 | e i- 1 | 2 = 4 ( e i- 1 ) ( e i- 1) = 4 ( e i- 1 ) e i- 1 = 4 ( e i- 1 )( e- i- 1 ) = 4 1- e i- e- i + 1 = 4 e i + e- i =- 2 e i + e- i 2 =- 1 cos( ) =- 1 = (2 k + 1) 3. Prove that 1 e i = e- i , where is a real number. Solution. 1 e i = 1 cos( ) + i sin( ) = 1 cos( ) + i sin( ) cos( )- i sin( ) cos( )- i sin( ) = cos( )- i sin( ) cos 2 ( ) + sin 2 ( ) = cos( )- i sin( ) = cos(- ) + i sin(- ) = e i (- ) 4. Find all the cube roots of w = 3 3 + 3 i . Solution. In polar form, 3 3 + 3 i = 6 e i/ 6 , so, writing z = re i we want z 3 = r 3 e i 3 = 6 e i/ 6 For this we need that r 3 = 6 and 3 - 6 = 2 k or r = 3 6 and = (12 k + 1) 18 So the cube roots of 3 3 + 3 i all look like 3 3 e i (12 k +1) 18 . However, among the infinitely many values for k there are only three distinct roots n e i/ 18 ,e i 13 / 18 ,e i 25 / 18 o 5. Prove that 1 z 2- 1 1 3 for all z on the circle of radius 2 centered at the origin. What can you say about 1 z 2- 1 on the circle of radius R centered at the origin? Solution. By the reverse triangle inequality | z 2- 1 | | z 2 | - | 1 | = | z | 2- 1 for any z . Thus, for z on the circle | z | = 2 we have, | z 2- 1 | 2 2- 1 = 3, so 1 z 2- 1 1 3 Similarly, for any R > 1 and all z on the circle | z | = R , 1 z 2- 1 1 R 2- 1 6. Solve each of the following equations or explain why no solution exists. (a) z 3 =- 27 (b) e z = 2- 2 i (c) Log z = 2 e i/ 6 (d) z 1 / 3 = i Solution. (a) This is solved exactly like number 4 using the polar form- 27 = 37 e i . The solutions are 3 e i/ 3 , 3 e i , 3 e i 5 / 3 (b) In polar form 2- 2 i = 8 e- i/ 4 so, writing e z = e x e iy , we want e x e iy = 8 e- i/ 4 For this we need that e x = 8 and y- - 4 = 2 k or x = ln 8 and y =- 4 + 2 k = (8 k- 1) 4 So the set of solutions of e z = 2- 2 i is ln 8 + i (8 k- 1) 4 | k Z (c) Using the fact that exp(Log( z )) = z for all z we can solve the equation Log( z ) = 2 e i/ 6 as follows z = exp(Log( z )) = exp 2 e i/ 6 = exp(2 (cos( / 6) + i sin( / 6)) = exp 2 3 2 + i 1 2 !!...
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This note was uploaded on 03/21/2010 for the course MATH 375 taught by Professor Marchesi during the Spring '10 term at Binghamton University.

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math 375 solution - Solutions to Exam 1 Review Problems 1....

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