Conformal Mapping  Can't Prove Analyticityby chill_factor Tags: analyticity, conformal, mapping, prove 

#1
Dec1112, 12:59 AM

P: 887

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(xiy)/(x+iy)(xiy) = x+iy + K(xiy)/(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 CauchyRiemann 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? 



#2
Dec1112, 01:49 AM

P: 887

Never mind I made a dumb mistake, it works.




#3
Dec1112, 12:22 PM

HW Helper
P: 775

The CauchyRiemann 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 CauchyRiemann 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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