## 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.