Question

If the co-ordinates r=cos∆ and y=rsin∆ then with the help of Cauchy-
Riemann condition ur
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Answer 1

Answer:

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I don't know

Answer 2

Answer:

I happen to have some notes on this question. What follows here is the usual approach, it's just multivariate calculus paired with the Cauchy Riemann equations. I have an idea for an easier way, I'll post it as a second answer in a bit if it works.

If we use polar coordinates to rewrite ff as follows:

f(x(r,θ),y(r,θ))=u(x(r,θ),y(r,θ))+iv(x(r,θ),y(r,θ))

f(x(r,θ),y(r,θ))=u(x(r,θ),y(r,θ))+iv(x(r,θ),y(r,θ))

we use shorthands F(r,θ)=f(x(r,θ),y(r,θ))F(r,θ)=f(x(r,θ),y(r,θ)) and U(r,θ)=u(x(r,θ),y(r,θ))U(r,θ)=u(x(r,θ),y(r,θ)) and V(r,θ)=v(x(r,θ),y(r,θ))V(r,θ)=v(x(r,θ),y(r,θ)). We derive the CR-equations in polar coordinates via the chain rule from multivariate calculus,

Ur=xrux+yruy=cos(θ)ux+sin(θ)uy and Uθ=xθux+yθuy=−rsin(θ)ux+rcos(θ)uy

Ur=xrux+yruy=cos⁡(θ)ux+sin⁡(θ)uy and Uθ=xθux+yθuy=−rsin⁡(θ)ux+rcos⁡(θ)uy

Likewise,

Vr=xrvx+yrvy=cos(θ)vx+sin(θ)vy and Vθ=xθvx+yθvy=−rsin(θ)vx+rcos(θ)vy

Vr=xrvx+yrvy=cos⁡(θ)vx+sin⁡(θ)vy and Vθ=xθvx+yθvy=−rsin⁡(θ)vx+rcos⁡(θ)vy

We can write these in matrix notation as follows:

[UrUθ]=[cos(θ)−rsin(θ)sin(θ)rcos(θ)][uxuy] and [VrVθ]=[cos(θ)−rsin(θ)sin(θ)rcos(θ)][vxvy]

[UrUθ]=[cos⁡(θ)sin⁡(θ)−rsin⁡(θ)rcos⁡(θ)][uxuy] and [VrVθ]=[cos⁡(θ)sin⁡(θ)−rsin⁡(θ)rcos⁡(θ)][vxvy]

Multiply these by the inverse matrix: [cos(θ)−rsin(θ)sin(θ)rcos(θ)]−1=1r[rcos(θ)rsin(θ)−sin(θ)cos(θ)][cos⁡(θ)sin⁡(θ)−rsin⁡(θ)rcos⁡(θ)]−1=1r[rcos⁡(θ)−sin⁡(θ)rsin⁡(θ)cos⁡(θ)] to find

[uxuy]=1r[rcos(θ)rsin(θ)−sin(θ)cos(θ)][UrUθ]=[cos(θ)Ur−1rsin(θ)Uθsin(θ)Ur+1rcos(θ)Uθ]

[uxuy]=1r[rcos⁡(θ)−sin⁡(θ)rsin⁡(θ)cos⁡(θ)][UrUθ]=[cos⁡(θ)Ur−1rsin⁡(θ)Uθsin⁡(θ)Ur+1rcos⁡(θ)Uθ]

A similar calculation holds for VV. To summarize:

ux=cos(θ)Ur−1rsin(θ)Uθ vx=cos(θ)Vr−1rsin(θ)Vθ

ux=cos⁡(θ)Ur−1rsin⁡(θ)Uθ vx=cos⁡(θ)Vr−1rsin⁡(θ)Vθ

uy=sin(θ)Ur+1rcos(θ)Uθ vy=sin(θ)Vr+1rcos(θ)Vθ

uy=sin⁡(θ)Ur+1rcos⁡(θ)Uθ vy=sin⁡(θ)Vr+1rcos⁡(θ)Vθ

The CR-equation ux=vyux=vy yields:

(A.) cos(θ)Ur−1rsin(θ)Uθ=sin(θ)Vr+1rcos(θ)Vθ

(A.) cos⁡(θ)Ur−1rsin⁡(θ)Uθ=sin⁡(θ)Vr+1rcos⁡(θ)Vθ

Likewise the CR-equation uy=−vxuy=−vx yields:

(B.) sin(θ)Ur+1rcos(θ)Uθ=−cos(θ)Vr+1rsin(θ)Vθ

(B.) sin⁡(θ)Ur+1rcos⁡(θ)Uθ=−cos⁡(θ)Vr+1rsin⁡(θ)Vθ

Multiply (A.) by rsin(θ)rsin⁡(θ) and (B.)(B.) by rcos(θ)rcos⁡(θ) and subtract (A.) from (B.):

Uθ=−rVr

Uθ=−rVr

Likewise multiply (A.) by rcos(θ)rcos⁡(θ) and (B.)(B.) by rsin(θ)rsin⁡(θ) and add (A.) and (B.):

rUr=Vθ

rUr=Vθ

Finally, recall that z=reiθ=r(cos(θ)+isin(θ))z=reiθ=r(cos⁡(θ)+isin⁡(θ)) hence

f′(z)=ux+ivx=(cos(θ)Ur−1rsin(θ)Uθ)+i(cos(θ)Vr−1rsin(θ)Vθ)=(cos(θ)Ur+sin(θ)Vr)+i(cos(θ)Vr−sin(θ)Ur)=(cos(θ)−isin(θ))Ur+i(cos(θ)−isin(θ))Vr=e−iθ(Ur+iVr)