Differentiability of $G_k(w)$

Suppose that $$f$$ is holomorphic in a neighborhood of $$z_0$$, and that all complex derivatives of $$f$$ up to order $$m-1$$ at $$z_0$$ vanish, namely, $$f^{(j)}(z_0)=0$$ for all $$j=0,...,m-1$$, but that $$f^{(m)}(z_0)\ne 0$$.

(a)Prove that there exist $$\epsilon>0$$ and $$\delta>0$$ such that, for every $$k\in\mathbb N$$, the equation

$$G_k(w)=\frac{1}{2\pi i}\int_{|\zeta-z_0|=\epsilon}\frac{\zeta^k f'(\zeta)}{f(\zeta)-w}d\zeta$$ defines a holomorphic function of $$w$$ in the set $$D_\delta(f(z_0))=\{ w\in\mathbb C:|w-f(z_0)|<\delta \}.$$

(b) Prove that, in the context of (a), if $$w\in D_\delta(f(z_0))$$ then the equation $$f(z)-w=0$$ has $$m$$ roots (counted with multiplicity), $$z_1,...,z_m,$$ inside $$|z-z_0|<\epsilon$$, and that $$G_k(w)=\sum_{j=1}^m z_j^k.$$

My attempt:

(a) Suppose $$w\in D_\delta(f(z_0))$$, then \begin{align} \frac{G_k(w+\Delta w)-G_k(w)}{\Delta w}&=\frac{1}{2\pi i}\int_{|\zeta-z_0|=\epsilon}\frac{\zeta^k f'(\zeta)}{(f(\zeta)-w-\Delta w)(f(\zeta)-w)}d\zeta \end{align} I was about to show that the modulus of integrand is bounded by an integrable function so that I can apply the dominated convergence theorem. However, I cannot find such a function...

Edit:

We know that if $$\gamma$$ is a Jordan curve, $$\varphi(\zeta)$$ is continuous on $$\gamma$$, then the function $$F(z)=\frac{1}{2\pi i}\int_\gamma\frac{\varphi (\zeta)}{\zeta-z}d\zeta$$ is analytic on each region of $$\overline{\mathbb C}\setminus\gamma$$. The proof of differentiability of $$F(z)$$ depends on the non-vanishment of $$\zeta-z$$ on $$\overline{\mathbb C}\setminus\gamma$$ which clearly is not the case in this problem. So we have to use different techniques.

Since $$f$$ is analytic $$f(\zeta) = f^{(m)}(z_0) (\zeta-z_0)^m +O((\zeta-z_0)^{m+1})$$ thus for $$\epsilon,\delta/\epsilon$$ small enough $$G_k(w)=\frac{1}{2\pi i}\int_{|\zeta-z_0|=\epsilon}\frac{\zeta^k f'(\zeta)}{f(\zeta)-w}d\zeta$$ is analytic
By the residue theorem $$G_k(w)=\sum_{j=1}^m z_j^k$$ since for $$w\ne 0$$, $$\zeta \mapsto \frac{f'(\zeta)}{f(\zeta)-w}$$ has $$m$$-distinct simple poles of residue $$1$$ on $$|\zeta-z_0| < \epsilon$$.
• How do you conclude $G_k(w)$ is analytic from $f(\zeta)=f^{(m)}(z_0)(\zeta-z_0)^m+O((\zeta-z_0)^{m+1})$? – Bach May 20 at 5:48
• Because $w \mapsto \frac{1}{f(\zeta)-w}$ is analytic (try with $f(z) = z^2, \epsilon=1,\delta = 1/10$) – reuns May 21 at 9:39