# Conformal Mapping - Can't Prove Analyticity

## Homework Statement

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

Prove this mapping is indeed conformal.

## Homework 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)

## 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?

Related Calculus and Beyond Homework Help News on Phys.org
Never mind I made a dumb mistake, it works.

pasmith
Homework Helper
The Cauchy-Riemann equations are conditions for two functions $\mathbb{R}^2 \to \mathbb{R}$ to be the real and imaginary parts of a differentiable function $\mathbb{C} \to \mathbb{C}$. But to prove analyticity you didn't need to show that the real and imaginary parts of $f(z) = z + K/z$ satisfy the Cauchy-Riemann equations, because you already knew that $f'(z)$ exists for all $z \neq 0$ and is equal to $1 - K/z^2$.

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