- #1

nateHI

- 146

- 4

## Homework Statement

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.

## Homework Equations

## 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.[/B]