From Greene:
Using the definition of the complex line integral, and using the parametrisation $\gamma\colon[0,2\pi]\to C\colon t\mapsto e^{it}$, I derived $$ f(z)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{1}{e^{it}-z}dt. $$ However, I still don’t see how $f$ vanishes everywhere? Any idea?
EDIT
From the hint given in the comments, we rewrite $$ \frac{1}{e^{it}-z}=\frac{1}{e^{it}}\frac{1}{1-ze^{-it}}=\sum_{k=0}^\infty z^ke^{-it(k+1)}, $$ where the use of the geometric series is justified by $\vert ze^{-it}\vert=\vert z\vert<1$. Now if this series converges uniformly, we can integrate term by term, and then we're done. However, we do need to show uniform convergence. The only theorem I know is about power series, while here we basically have a power series in a power series (due to the exponential). Any ideas on how to show uniform convergence?
$EDIT2$
Right, so I will write out my final answer (in full detail), based on the help I got in the comments. Let $u\in\mathbb C$ such that $\vert u\vert<1$; then we know that $(1-u)^{-1}$ is given by the geometric series $$ \sum_{k=0}^\infty u^k. $$ Now for $\vert z\vert<R<1$, this series converges uniformly. Uniform convergence just means that for each $\epsilon>0$, we can find $N$ such that for all $n>N$ we have $$ \sup_{\vert u\vert\leq R}\left\{\left\vert(1-u)^{-1}-\sum_{k=0}^n u^k\right\vert\right\}<\epsilon. $$ Now obviously, if we have $u=ze^{-it}$, then the condition of uniform convergence is satisfied over $t\in[0,2\pi]$, hence our integrand is uniform over $t$, and we can integrate term by term.