Contents
Practice material for Exam 1 collaboration space
You can easily talk about math here, like this:
$ e^{i\theta} = \cos \theta + i \sin \theta. $
Is this the Cauchy Integral Formula?
$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a} \ dz $
This isn't directly related to the practice exam, but is concerning a fact discussed in class.
In one of the first lessons an important fact was provided. Namely, Suppose u is continuously a differentiable function on a connected open set $ \Omega $ and that $ \nabla u \equiv 0 $ Then u must be constant on $ \omega $.
How/Why is
$ 0 = \int\nabla u\ ds $
If we have a domain, $ \Omega $ and $ \gamma $ a curve in $ \Omega $, where $ A, B $ are end points of $ \gamma $, from vector calculus, we have
$ u(B) - u(A) = \int_\gamma \nabla u \cdot ds $.
but the integral is zero since $ \nabla u $ is zero. Hence $ u(A) = u(B) $ for any arbitrary point $ A $ and $ B $ and $ u $ is constant in $ \Omega $.
Practice Problems: Suggested Solutions/Thoughts
Problem 1.
Daniel: I interpret "an analytic function f has constant modulus on a domain" as a function f that maps all domain to some f(z) where |f(z)| = r. Then f is probably not constant, since this is a circle. So I am interpreting the problem wrong?
Hooram: This is my attempt. It may or may not be a correct proof.
The problem is to show that if f is analytic and |f| is constant on $ \Omega $, then f is constant on $ \Omega $. First, write f=u+iv. Then $ |f| = u^2+v^2 = c $, some constant. Now, we take the derivative of the modulus of f:
$ \dfrac{\partial}{\partial x} |f| = 2u u_x + 2v u_x = 0 $
$ \dfrac{\partial}{\partial y} |f| = 2u u_y + 2v u_y = 0. $
We know the derivatives $ u_x, u_y, v_x, v_y $ exist because f is assumed to be analytic.
So we have
$ u u_x + v u_x = 0 $
$ u u_y + v u_y = 0. $
Since f is analytic, by Cauchy-Riemann equations, we have $ u_x = v_y $ and $ u_y = -v_x $.
Before we carry on, let's think about what we're gonna do. We're gonna do two things:
1) eliminate $ u_y $ and conclude $ u_x = 0 $
$ u_y = \dfrac{-vu_x}{u} $
$ uu_x + \dfrac{-v^2u_x}{u} = 0 $
$ u_x (u^2+v^2) = u_x c = 0 $
$ u_x = 0 $
2) Similarly eliminate $ u_x $ and conclude $ u_y = 0 $
Then, we will have $ u_x = u_y = 0 $, which implies that $ f'(z) = u_x + iu_y = 0 $.
Hence f is constant. This completes the proof.
Daniel: Thanks Hooram, although I am still unsure whether |f| = $ u^2 + y^2 $ is valid? Also isn't it
$ \dfrac{\partial}{\partial x} |f| = \dfrac{\partial u^2}{\partial x} + \dfrac{\partial v^2}{\partial x} = 2uu_x + 2vv_x $
Problem 2.
Daniel: Both a and b are simple - I would simply invoke the Fund. Theorem of Calc. Problem 2 could have been tricky if it involved going over the branch cut, but since it did not, I think we are safe. (check out pg. 175)
Problem 3.
Daniel: I saw a similar problem on pg. 164. Shouldn't be too tricky since t is a real variable. Remembering what actually is $ e^{3i\pi} $ may be the hard part.
Problem 4.
Daniel: Will return to it.
Problem 5.
Daniel: hmm
