# Complex Analysis: Open Mapping Theorem, Argument Principle

1. Apr 16, 2015

### nateHI

1. The problem statement, all variables and given/known data
In each case, state whether the assertion is true or false, and justify your answer with a proof or counterexample.

(a) Let $f$ be holomorphic on an open connected set $O\subseteq \mathcal{C}$. Let $a\in O$. Let $\{z_k\}$ and $\{\zeta_k\}$ be two sequences contained in $O\setminus \{a\}$ and converging to $a$, such that for every $k=1,2,\dots$, $f(z_k)=f(\zeta_k)$ and $z_k\neq \zeta_k$. Then $f^{\prime}(a)=0$.
ANS
This is true. By hypothesis, there exists $D^{\prime}(a,\frac{1}{n})$ such that $f(z_n)=f(\zeta_n)=a$. Let $g=f(z)-a$. By the Identity Theorem, $g=0\implies f(z)=a$, which is a constant function and indeed, $f^{\prime}(a)=0$.

(b) Let g be an entire function such that $\frac{1}{2\pi i}\int_{|z|=R}\frac{g^{\prime}}{g}=0$ for all $R>1000$. Then $g$ is constant.
ANS
This statement is false since $g$ can also be an exponential. Why? Since $g$ is entire it has no poles, and $\frac{1}{2\pi i}\int_{|z|=R}\frac{g^{\prime}}{g}=0$ implies $g$ has no zeroes either. An exponential would be an example of a non-constant entire function that has no zeros.
(c) Let $u$ be a real-valued harmonic function on $D(0,1)$, and let $\gamma$ be a closed curve in that disk. Then $\int_{\gamma}u=0$.
ANS
I'm not sure how to do this. By harmonic does the problem mean analytic or that the 2nd derivative is 0? The book uses it both ways throughout the text hence my confusion.

2. Relevant equations

3. The attempt at a solution
I included my attempt in the problem statement. Although for part (a), I didn't use the hypothesis that $f$ isn't one to one. I believe my answer works though.

2. Apr 16, 2015

### RUber

I didn't see any issues in (a) or (b) that would lead me to question their validity. In part (C), harmonic normally implies that they satisfy Laplace's equation.
$\Delta f - \frac{\partial^2}{\partial t^2} = 0.$ Where $\Delta$ is the second spatial derivative. For a function that doesn't vary in time, this equates to the second derivative being zero.
I think it is implicit in the definition of a harmonic function that it be twice differentiable.

3. Apr 16, 2015

### nateHI

OK thanks, just to be clear, are you saying I did parts (a) and (b) correct?
EDIT:
Actually, I clean my answer part (a) up a little bit below. It's not a significant change but it should make my reasoning more clear.
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This is true. By hypothesis, there exists $z_n\in D^{\prime}(a,\frac{1}{n})$ such that $f(z_n)=f(\zeta_n)=a$. Let $g=f(z)-a$. By the Identity Theorem, $g=0\implies f(z)=a$, which is a constant function and indeed, $f^{\prime}(a)=0$.

Last edited: Apr 16, 2015