My question is the following: Let $f$ be a continuous function on $\{z:0<|z|\le 1\}$ and holomorphic on $\{z:0<|z|<1\}$. Assume $f=0$ for $|z|=1$, then $f=0$ for all $z$ such that $0<|z|\le 1$.
My approach is the following: We will study the singularities at $z=0$. If $z=0$ is removable then the claim is obvious following maximum principle.
Now if $z=0$ is a pole say of order $n$, then we have $|z^nf(z)|$ again vanishes on $|z|=1$ so $z^nf(z)$ is $0$ so $f(z)=0$ for all $z$ such that $0<|z|\le 1$.
However, if $z=0$ is an essential singularity, I don't know what a good way to tackle this. Thank you.
Also, I am wondering that if $z=0$ is an essential singularity, what property might the harmonic function $Re f$ have?
Best Answer
Let $\sum_{n=-\infty}^\infty a_nz^n$ be the Laurent series of $f$ at the origin and, for each $r\in(0,1]$, let $\gamma_r\colon[0,2\pi]\longrightarrow\Bbb C$ be the loop defined by $\gamma_r(\theta)=re^{i\theta}$. If $r<1$ and $n\in\Bbb Z$, then$$a_n=\frac1{2\pi i}\int_{\gamma_r}\frac{f(z)}{z^{n+1}}\,\mathrm dz$$and so$$a_n=\lim_{r\to1^-}\frac1{2\pi i}\int_{\gamma_r}\frac{f(z)}{z^{n+1}}\,\mathrm dz=\frac1{2\pi i}\int_{\gamma_1}\frac{f(z)}{z^{n+1}}\,\mathrm dz=0.$$Therefore,$$f(z)=\sum_{n=-\infty}^\infty a_nz^n=0.$$