135-s10-2011

135-s10-2011 - Assignment 10 Solutions 1. Prove that (1 +...

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Assignment 10 – Solutions 1. Prove that (1 + i 2 n )(1 + i n ) is either 0 or 4 for n P . Describe the values of n that yield the various answers. Solution: We have (1+( i 2 ) n )(1+ i n ) = (1+( - 1) n )(1+ i n ). By Proposition 8.23, we only need to check the congruence of n modulo 4 in order to determine the value of i n . n 0 (mod 4) , i n = 1 and ( - 1) n = 1 . So (1 + ( i 2 ) n )(1 + i n ) = 2(2) . n 1 (mod 4) , i n = i and ( - 1) n = - 1 . So (1 + ( i 2 ) n )(1 + i n ) = 0(1 + i ) . n 2 (mod 4) , i n = - 1 and ( - 1) n = 1 . So (1 + ( i 2 ) n )(1 + i n ) = 2(0) . n 3 (mod 4) , i n = - i and ( - 1) n = - 1 . So (1 + ( i 2 ) n )(1 + i n ) = 0(1 - i ) . Therefore (1 + i 2 n )(1 + i n ) is 0 for n 1 , 2 , 3 (mod 4) and is 4 for n 0 (mod 4). 2. Suppose z = x + iy with x,y R . Determine all complex numbers z for which z 2 = 2 z . Solution: Suppose z = x + iy , with x,y R . Then
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135-s10-2011 - Assignment 10 Solutions 1. Prove that (1 +...

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