| Line 11: | Line 11: | ||
<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 Liouville's Theorem.--[[User:Zhug| | + | I came up with a proof by looking at <math>\frac{1}{1+f(z)}</math> and use Liouville's Theorem. --[[User:Zhug|Guangwei Zhu]] 15:21, 11 February 2011 (UTC) |
Problem 10 hint: | Problem 10 hint: | ||
Latest revision as of 10:21, 11 February 2011
Homework 4 discussion area
Problem 3
This is what I come up with. I am not sure if the following is sharp enough:
The successive derivatives of an analytic function at a point $ a $ can never satisfy $ |f^{(n)}(a)|>n!h^n(n) $, where $ h(n) $ is a function of $ n $ such that $ \limsup_{n\to\infty}h(n)=+\infty $. --Guangwei Zhu 15:20, 11 February 2011 (UTC)
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. --Guangwei Zhu 15:21, 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.
