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Unformatted text preview: MATH 135 Fall 2005 Assignment 9 Solutions HandIn Problems Exercise 864: Express the following complex numbers in standard form. (1 √ 3 i ) 8 Solution:  1 √ 3 i  = 2 and the argument of (1 √ 3 i ) is θ = 5 π 3 , so (1 √ 3 i ) = 2cis ( 5 π 3 ) . By De Moivre’s Theorem, we have (1 √ 3 i ) 8 = ( 2cis ( 5 π 3 )) 8 = 2 8 cis ( 40 π 3 ) = 2 8 cis ( 4 π 3 ) = 2 8 (cos ( 4 π 3 ) + i sin ( 4 π 3 ) ) = 128 128 √ 3 i Exercise 874: Solve each of the following equations for z ∈ C and plot your solutions on the complex plane. z 3 = 9 i Solution: In polar form 9 i = 9cis ( π 2 ) . Hence, to solve for z we take the cube roots of 9 i . By Theorem 8.72, these are z = 3 √ 9 cis π 6 + 2 kπ 3 for k = 0 , 1 and 2. That is, z = 3 √ 9(cos π 6 + i sin π 6 ) = 3 √ 9 √ 3 2 + i 2 = 3 6 √ 3 2 + 3 √ 9 i 2 z 1 = 3 √ 9(cos 5 π 6 + i sin 5 π 6 ) = 3 √ 9 √ 3 2 + i 2 = 3 6 √ 3 2 + 3 √ 9 i 2 z 2 = 3 √ 9(cos 3 π 2 + i sin 3 π 2 ) = 3 √ 9(0 + i ( 1)) = 0 3 √ 9 i Therefore the three solutions for z are ± 3 6 √ 3 2 + 3 √ 9 i 2 and 0 3 √ 9 i. 6 ? 1 i s z s z 1 s z 1 Exercise 882: Solve each of the following equations for z ∈ C and plot your solutions on the complex plane. z 2 (2 + 2 i ) z 1 + 2 i = 0 Solution: Applying the quadratic formula, z = 2 + 2 i ± p (2 + 2 i ) 2 4( 1 + 2 i ) 2 z = 2 + 2 i ± √ 4 2 . Hence the solutions are z = i and z = 2 + i . 6 1 i s s Exercise 8108: Write the following numbers in the exponential form e x + iy . ( √ 3 i ) 100 Solution: In polar form, √ 3 i = 2cis ( 11 π 6 ) . So by DeMoivre’s Theorem, ( √ 3 i ) 100 = 2 100 cis ( 1100 π 6 ) = 2 100 cis ( 550 π 3 ) = 2 100 cis ( 4 π 3 ) . Because e iθ = cis θ and 2 100 = e 100 ln 2 , then ( √ 3 i ) 100 = e 100 ln 2 e i 4 π 3 = e 100(ln 2)+ i 4 π 3 . (Note that ( √ 3 i ) 100 = e 100(ln 2)+ i ( 4 π 3 +2 πk ) for every k ∈ Z as well.) Exercise 8110: If u is a nonreal sixth root of unity, but not a cube root of unity, prove that u 3 = 1, and that u 2 u + 1 = 0. Solution: Note that u satisfies the equation u 6 1 = 0. Factoring the left side, we obtain ( u 3 1)( u 3 + 1) = 0. Thus either u 3 1 = 0 or u 3 + 1 = 0. But u is not a cube root of unity so u 3 1 6 = 0 and hence u 3 + 1 = 0. Since u 3 + 1 = ( u + 1)( u 2 u + 1) = 0, and u is nonreal then u cannot be 1, so u + 1 6 = 0, so u 2 u + 1 = 0. Problem 8118: (a) Prove that the sum of the fifth roots of unity is zero. (b) Generalize this result and prove your generalization. Solution: (a) The fifth roots of unity are z k = cis 2 kπ 5 , k = 0 , 1 , 2 , 3 , 4 giving z = cis 0 = 1 , z 1 = cis 2 π 5 , z 2 = cis 4 π 5 , z 3 = cis 6 π 5 , z 4 = cis 8 π 5 ....
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This homework help was uploaded on 04/18/2008 for the course LINEAR ALG MATH135 taught by Professor Vanderburgh during the Spring '08 term at Waterloo.
 Spring '08
 Vanderburgh

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