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## Homework Statement

a. u(x,y)=(x+2)

^{2}-y

^{2}, v(x,y)=2(x+2)y

b. u(x,y)=cosxcoshy, v(x,y)=-sinxsinhy

c. u(x,y)=(x

^{2}+y

^{2}+y)/(x

^{2}+y

^{2}), v(x,y)=x/(x

^{2}+y

^{2})

d. u(x,y)=e

^{y}cosx, v(x,y)=e

^{y}sinx

## The Attempt at a Solution

a. f(x+iy)=(x+2)

^{2}-y

^{2}+i2(x+2)y

=(x+2+iy)(x+2+iy)

f(z)=(z+2)

^{2}

b. f(x-iy)=cosxcoshy-isinxsinhy

Am I allowed to answer this question as f([itex]\bar{z}[/itex])=cos[itex]\bar{z}[/itex]?

c.f(x-iy)=((x

^{2}+y

^{2}+y)/(x

^{2}+y

^{2}))-i(x/(x

^{2}+y

^{2}))

=((x-iy-i)(x+iy))/((x-iy)(x+iy))

=(x-iy-i)/(x-iy)

Same as b... f([itex]\bar{z}[/itex])=([itex]\bar{z}[/itex]-i)/[itex]\bar{z}[/itex]?

d. du/dx != dv/dy, fails Cauchy-Reimann?

Edit: How do I go from x-iy to x+iy? This escapes me. :(

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