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<math>e^{\pm f(z)}</math>
 
<math>e^{\pm f(z)}</math>
  
I came up with a proof by looking at <math>\frac{1}{1+f(z)}</math> and use Louisville's Theorem.--[[User:Zhug|Zhug]] 14:20, 11 February 2011 (UTC)
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I came up with a proof by looking at <math>\frac{1}{1+f(z)}</math> and use Liouville's Theorem.--[[User:Zhug|Zhug]] 14:20, 11 February 2011 (UTC)
  
 
Problem 10 hint:
 
Problem 10 hint:

Revision as of 10:23, 11 February 2011

Homework 4 discussion area

Problem 7 hint:

$ e^{\pm f(z)} $

I came up with a proof by looking at $ \frac{1}{1+f(z)} $ and use Liouville's Theorem.--Zhug 14:20, 11 February 2011 (UTC)

Problem 10 hint:

Parametrize the circular part of the boundary via

$ C_R:\quad z(t)=Re^{it}, 0<t<\pi/4. $

You need to show that

$ I_R := \int_{C_R}e^{-z^2}\ dz\to 0 $

as R goes to infinity. You won't be able to use the standard estimate to do this. Write out the definition of the integral to find that

$ |I_R|\le\int_0^{\pi/4} Re^{-R^2\cos(2t)}\,dt $

and use freshman calculus ideas to show that this integral tends to zero. (Don't hit it with the big stick, the Lebesgue Dominated Convergence Theorem.) Hint: Draw the graph of cos_2t on the interval and realize that the line connecting the endpoints is under the graph. Compare the integral with what you would get by replacing cos_2t by the simple linear function underneath it.


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