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Conformal Mapping - Can't Prove Analyticity

by chill_factor
Tags: analyticity, conformal, mapping, prove
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chill_factor
#1
Dec11-12, 12:59 AM
P: 900
1. The problem statement, all variables and given/known data

We have the conformal map w = f(z) = z + K/z.

Prove this mapping is indeed conformal.

2. Relevant equations

z = x + iy

A map w = f(z) is conformal if it is analytic and df/dz is nonzero.

f(z) = u(x,y) + iv(x,y)

3. The attempt at a solution

df/dz = 1 - Kz^-2 =/= 0 for finite z, nonzero derivative condition met.

Attempting to prove its analytic:

f(z) = z + K/z = x+iy + K/(x+iy) = x+iy + K(x-iy)/(x+iy)(x-iy) = x+iy + K(x-iy)/(x^2+y^2)

= [x + Kx/(x^2+y^2)] + i [y - Ky/(x^2+y^2)] = u(x,y) + iv(x,y)

u(x,y) = x + Kx(x^2+y^2)^-1
v(x,y) = y - Ky(x^2+y^2)^-1

du/dx = 1 - Kx(x^2+y^2)^-2 * 2x
dv/dy = 1 + Ky(x^2+y^2)^-2 * 2y

they aren't equal therefore they do not satisfy Cauchy-Riemann relations, and the function is not analytic. However I'm directly told that it MUST be analytic because it is a conformal map. Did I make a mistake or is the problem mistyped?
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chill_factor
#2
Dec11-12, 01:49 AM
P: 900
Never mind I made a dumb mistake, it works.
pasmith
#3
Dec11-12, 12:22 PM
HW Helper
Thanks
P: 1,025
The Cauchy-Riemann equations are conditions for two functions [itex]\mathbb{R}^2 \to \mathbb{R}[/itex] to be the real and imaginary parts of a differentiable function [itex]\mathbb{C} \to \mathbb{C}[/itex]. But to prove analyticity you didn't need to show that the real and imaginary parts of [itex]f(z) = z + K/z[/itex] satisfy the Cauchy-Riemann equations, because you already knew that [itex]f'(z)[/itex] exists for all [itex]z \neq 0[/itex] and is equal to [itex]1 - K/z^2[/itex].

Also, doesn't [itex]f'(\sqrt K) = f'(-\sqrt K) = 0[/itex]? Your function is then conformal on the open set [itex]\mathbb{C} \setminus \{\sqrt K, 0, -\sqrt K\}[/itex].


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