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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.
 Spring '10
 MARCHESI
 Math

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