Verifying if a function is analytic

Question

I'm trying to solve the third exercise from page 126 of Conway's complex variables textbook. It says:

Let $f$ be analytic in $\bar{B}(0,R)$, with $f(0)=0$, $f'(0)\neq 0$ and $f(z)\neq 0$ for $0<|z|\leq R$. Put $\rho=min\{|f(z)|:|z|=R\}>0$. Define $g:B(0,\rho )\to \mathbb{C}$ by $$g(w)=\dfrac{1}{2\pi i}\int_{|z|=R}\dfrac{z\,f'(z)}{f(z)-w}dz.$$

I have to show that $g$ is analytic. Any ideas?

Answer 1

We notice that $|w|<\rho \le |f(z)|$ for $|z|=R$ means that $\dfrac{z\,f'(z)}{f(z)-w}=\frac{zf'(z)}{f(z)}\sum{\frac{w^k}{f(z)^k}}$ and the convergence is uniform and absolute for $|z|=R$ and $|w|$ in any compact set in the smaller disc of radius $\rho$, so we can integrate term by term on $|z|=R$ and obtain that $g$ is analytic with Taylor series:

$g(w)=\sum{(\dfrac{1}{2\pi i}\int_{|z|=R}\dfrac{z\,f'(z)}{f(z)^{k+1}}dz)w^k}$