Question

Derive cauchy - riemann equations in polar coordinates

Answer 1

Proof of Polar C.R Let f=u+ivf=u+iv be analytic, then the usual Cauchy-Riemann equations are satisfied
∂u∂x=∂v∂y and ∂u∂y=−∂v∂x (C.R.E)
∂u∂x=∂v∂y and ∂u∂y=−∂v∂x (C.R.E)
Since z=x+iy=r(cosθ+isinθ)z=x+iy=r(cos⁡θ+isin⁡θ), then x(r,θ)=rcosθx(r,θ)=rcos⁡θ and y(r,θ)=rsinθy(r,θ)=rsin⁡θ. By the chain rule:
∂u∂r=∂u∂xcosθ+∂u∂ysinθ=(C.R.E)1r(∂v∂yrcosθ−∂v∂xrsinθ)=1r(∂v∂θ)
∂u∂r=∂u∂xcos⁡θ+∂u∂ysin⁡θ=(C.R.E)1r(∂v∂yrcos⁡θ−∂v∂xrsin⁡θ)=1r(∂v∂θ)
and again, by the chain rule:
∂v∂r=∂v∂xcosθ+∂v∂ysinθ=(C.R.E)−1r(∂u∂yrcosθ−∂u∂xrsinθ)=−1r(∂u∂θ)
∂v∂r=∂v∂xcos⁡θ+∂v∂ysin⁡θ=(C.R.E)−1r(∂u∂yrcos⁡θ−∂u∂xrsin⁡θ)=−1r(∂u∂θ)
So indeed
(∂u∂r)=1r(∂v∂θ) and (∂v∂r)=−1r(∂u∂θ) ■
(∂u∂r)=1r(∂v∂θ) and (∂v∂r)=−1r(∂u∂θ) ◼
Logarithm Example log(z)=ln(r)+iθlog(z)=ln⁡(r)+iθ with z=reiθz=reiθ, r>0r>0 and −π<θ<π−π<θ<π. Then
u(r,θ)=ln(r) and v(r,θ)=θ
u(r,θ)=ln⁡(r) and v(r,θ)=θ
and
(∂u∂r)=1r=1r⋅1=1r⋅(∂v∂θ) and (∂v∂r)=0=−1r⋅0=−1r(∂u∂θ)
(∂u∂r)=1r=1r⋅1=1r⋅(∂v∂θ) and (∂v∂r)=0=−1r⋅0=−1r(∂u∂θ)
So indeed, log(z)log(z) is analytic.

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