proof to Cauchy-Riemann equations (polar coordinates)

If f(z) is differentialble at z0 then the following limit

f(z0) = limξ0f(z0+ξ)-f(z0)ξ

will remain the same approaching from any direction. First we fix θ as θ0 then we take the limit along the ray where the argument is equal to θ0. Then

f(z0) = limh0f(r0eiθ0+heiθ0)-f(r0eiθ0)heiθ0
= limh0f((r0+h)eiθ0)-f(r0eiθ0)heiθ0
= limh0u(r0+h,θ0)+iv(r0+h,θ0)-u(r0,θ0)-iv(r0,θ0)heiθ0
= 1eiθ0[limh0u(r0+h,θ0)-u(r0,θ0)h+ilimh0v(r0+h,θ0)-v(r0,θ0)h]
= 1eiθ0[ur(r0,θ0)+ivr(r0,θ0)]

Similarly, if we take the limit along the circle with fixed r equals r0. Then

f(z0) = limh0f(r0eiθ0+r0ei(θ0+h))-f(r0eiθ0)r0eiθ0(eih-1)
= limh0f(r0ei(θ0+h))-f(r0eiθ0)heiθ0
= limh0u(r0,θ0+h)+iv(r0,θ0+h)-u(r0,θ0)-iv(r0,θ0)heiθ0
= 1r0eiθ0[limh0u(r0+h,θ0)-u(r0,θ0)hheih-1+ilimh0v(r0+h,θ0)-v(r0,θ0)hheih-1]
= 1r0eiθ0[limh0u(r0+h,θ0)-u(r0,θ0)hlimh0heih-1+ilimh0v(r0+h,θ0)-v(r0,θ0)hlimh0heih-1]
= 1r0eiθ0[uθ(r0,θ0)1i+vθ(r0,θ0)]
= 1r0eiθ0[vθ(r0,θ0)-iuθ(r0,θ0)]

Note: We use l’Hôpital’s rule to obtain the following result used above limh0heih-1=1i.

Now, since the limit is the same along the circle and the ray then they are equal:

1eiθ0[ur(r0,θ0)+ivr(r0,θ0)] = 1r0eiθ0[vθ(r0,θ0)-iuθ(r0,θ0)]
[ur(r0,θ0)+ivr(r0,θ0)] = 1r0[vθ(r0,θ0)-iuθ(r0,θ0)]

which implies that

ur = 1rvθ
vr = -1ruθ


Title proof to Cauchy-Riemann equationsMathworldPlanetmath (polar coordinatesMathworldPlanetmath)
Canonical name ProofToCauchyRiemannEquationspolarCoordinates
Date of creation 2013-03-22 14:06:13
Last modified on 2013-03-22 14:06:13
Owner Daume (40)
Last modified by Daume (40)
Numerical id 5
Author Daume (40)
Entry type Proof
Classification msc 30E99