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soln12 - MATH 135 Algebra Solutions to Assignment 12 1...

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MATH 135 Algebra, Solutions to Assignment 12 1: Express each of the following complex numbers in cartesian form. (a) 4 e i 5 π/ 3 Solution: We have 4 e i 5 π/ 3 = 4 ( cos 5 π 3 + i sin 5 π 3 ) = 4 1 2 - 3 2 i = 2 - 2 3 i (b) (1 + i 3) 10 Solution: We have (1 + i 3) 10 = 2 e i π/ 3 10 = 2 10 e i 10 π/ 3 = 2 10 ( cos 10 π 3 + i sin 10 π 3 ) = 1024 - 1 2 - 3 2 i = - 512 - 512 3 i . (c) 5 e i θ , where θ = tan - 1 2 Solution: We have 5 e i θ = 5 (cos θ + i sin θ ) = 5 1 5 + 2 5 i = 5 + 2 5 i . 2: Express each of the following complex numbers in the polar form re i θ . (a) - 2 + 2 i Solution: We have - 2 + 2 i = 2 2 e i 3 π/ 4 . (b) (1 - i ) 2 (1 + i 3) Solution: We have (1 - i ) 2 (1 + i 3) = ( 2 e - i π/ 4 ) 2 2 e iπ/ 3 = 2 e - i π/ 2 2 e i π/ 2 = e - i ( π 2 + π 3 ) = e - i 5 π/ 6 . (c) - 3 - i Solution: We have - 3 - i = 10 e i θ , where θ = π + tan - 1 1 3 .
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3: Solve each of the following for z C . Express your answers in the polar form re i θ . (a) z 3 + 8 i = 0 Solution: Write z = r e i θ with r > 0 and θ [0 , 2 π ). Then z 3 + 8 i = 0 ⇐⇒ ( r e i θ ) 3 = - 8 i ⇐⇒ r 3 e i 3 θ = 8 e i 3 π/ 2 ⇐⇒ r 3 = 8 and 3 θ = 3 π 2 + 2 π k for some k Z ⇐⇒ r = 2 and θ = π 2 + 2 π 3 k where k = 0 , 1 or 2 . Thus the solutions are z = 2 e i π/ 2 , z = 2 e i 7 π/ 6 and z = 2 e i 11 π/ 6 .
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