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...


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 $|\zeta-z_0| < \epsilon$.

  • $\begingroup$ How do you conclude $G_k(w)$ is analytic from $f(\zeta)=f^{(m)}(z_0)(\zeta-z_0)^m+O((\zeta-z_0)^{m+1})$? $\endgroup$ – Bach May 20 at 5:48
  • $\begingroup$ Because $w \mapsto \frac{1}{f(\zeta)-w}$ is analytic (try with $f(z) = z^2, \epsilon=1,\delta = 1/10$) $\endgroup$ – reuns May 21 at 9:39

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