# solut3 - Solutions to homework 3 II.16(a f z = z 2 sin z...

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Unformatted text preview: Solutions to homework # 3 II.16. (a) f ( z ) = z 2 sin z . Since sin z = e i z- e- i z 2i = e i( x +i y )- e- i( x +i y ) 2i = e- y (cos x + i sin x )- e y (cos x- i sin x ) 2i = ( e- y + e y ) sin x 2 + i cos x ( e y- e- y ) 2 and z 2 = ( x 2- y 2 ) + 2i xy , we get f ( x, y ) = u ( x, y ) + i v ( x, y ) where u ( x, y ) = ( x 2- y 2 )( e y + e- y ) sin x 2- xy ( e y- e- y ) cos x v ( x, y ) = xy ( e y + e- y ) sin x + ( x 2- y 2 )( e y- e- y ) cos x 2 . Then ∂u ∂x = (2 x sin x + ( x 2- y 2 ) cos x ) e y + e- y 2- y ( e y- e- y )(cos x- x sin x ) = ∂v ∂y ∂u ∂y = x sin x ( e y + e- y + y ( e y- e- y )) + cos x (( x 2- y 2 ) e y + e- y 2- y ( e y- e- y )) =- ∂v ∂x . (b) f ( z ) = 1 / (1 + z ). We have 1 1 + z = 1 1 + x + i y = 1 + x- i y (1 + x ) 2 + y 2 , so f ( x, y ) = u ( x, y ) + i v ( x, y ) where u ( x, y ) = 1 + x (1 + x ) 2 + y 2 , v ( x, y ) =- y (1 + x ) 2 + y 2 . Hence ∂u ∂x = y 2- (1 + x ) 2 ((1 + x ) 2 + y 2 ) 2 = ∂v ∂y ∂u ∂y =- 2(1 + x ) y ((1 + x ) 2 + y 2 ) 2 =- ∂v ∂x . II.17. Note that ∂f ∂ ¯ z = ∂f ∂x ∂x ∂ ¯ z + ∂f ∂y ∂y ∂ ¯ z , and x = 1 2 ( z + ¯ z ) , y = 1 2i ( z- ¯ z ) , so ∂x ∂ ¯ z = 1 2 , ∂y ∂ ¯ z =- 1 2i ....
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## This note was uploaded on 12/12/2009 for the course MATHEWS SOC 2000 taught by Professor M during the Spring '09 term at AIB College of Business.

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solut3 - Solutions to homework 3 II.16(a f z = z 2 sin z...

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