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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online