MAT135-Assignment10-Solutions

# MAT135-Assignment10-Solutions - MATH 135 Fall 2010 Solution...

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MATH 135, Fall 2010 Solution of Assignment #10 Problem 1 . If z,w C , prove that | z + w | 2 + | z - w | 2 = 2 | z | 2 + 2 | w | 2 . Solution. Solution 1 | z + w | 2 + | z - w | 2 = ( z + w )( z + w ) + ( z - w )( z - w ) = ( z + w )( z + w ) + ( z - w )( z - w ) (by Prop. 8.42.i) = z z + z w + w z + w w + z z - z w - w z + w w = 2 z z + 2 w w = 2 | z | 2 + 2 | w | 2 as required. Solution 2 Suppose z = x + iy and w = u + iv with x,y,u,v R . Then | z + w | 2 + | z - w | 2 = | ( x + u ) + ( y + v ) i | 2 + | ( x - u ) + ( y - v ) i | 2 = ( x + u ) 2 + ( y + v ) 2 + ( x - u ) 2 + ( y - v ) 2 = x 2 + 2 ux + u 2 + y 2 + 2 yv + v 2 + x 2 - 2 ux + u 2 + y 2 - 2 yv + v 2 = 2 x 2 + 2 y 2 + 2 u 2 + 2 v 2 = 2( x 2 + y 2 ) + 2( u 2 + v 2 ) = 2 | z | 2 + 2 | w | 2 as required. Problem 2 . For real numbers x and y , we deﬁne e x + iy = e x cos y + ie x sin y . For a complex number z , we deﬁne cos z = e iz + e - iz 2 and sin z = e iz - e - iz 2 i . (a) Show that for all

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## This note was uploaded on 12/10/2010 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.

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MAT135-Assignment10-Solutions - MATH 135 Fall 2010 Solution...

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