The proof about Cauchy's Integral Formula in Ahlfors' Complex Analysis

Question

On the third edition of Ahlfors' Complex Analysis, page 122 Lemma 3 it states: Now, if we divide the identity by $z-z_0$ and let $z$ tend to $z_0$, the quotient in the first term tends to a derivative which by the induction hypothesis equals $(n-1)F_{n+1}(z_0)$.

I'm confused about it.

We have: \begin{equation*} \lim_{z\rightarrow z_{0}}\frac{1}{z-z_{0}}\left(\int_{\gamma}\frac{\varphi(\zeta)}{(\zeta-z)^{n-1}}d\zeta-\int_{\gamma}\frac{\varphi(\zeta)}{(\zeta-z_{0})^{n-1}}d\zeta\right)=(n-1)\int_{\gamma}\frac{\varphi(\zeta)}{(\zeta-z_{0})^{n}}d\zeta \end{equation*}

But I don't know how to prove: \begin{equation*} \lim_{z\rightarrow z_0}\frac{1}{z-z_0}\left[\int_{\gamma}\frac{\varphi(\zeta)}{(\zeta-z)^{n-1}(\zeta-z_{0})}d\zeta-\int_{\gamma}\frac{\varphi(\zeta)}{(\zeta-z_{0})^{n}}d\zeta\right]=(n-1)\int_{\gamma}\frac{\varphi(\zeta)}{(\zeta-z_{0})^{n+1}}d\zeta \end{equation*}

This a question from Ahlfors' Complex Analysis. I show you the whole context of the question below. I have trouble understanding the statement with red line.

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$Let $\psi:\zeta\mapsto\frac{\varphi(\zeta)}{\zeta-z_0}$, for $\zeta$ in the image of $\gamma$ and $z_0$ fixed not in $\gamma$. This is evidently continuous. Really, it's the same trick that they use to demonstrate differentiability of $F_1$.

$$\begin{align}&\quad\frac{1}{z-z_0}\left[\oint_\gamma\frac{\varphi(\zeta)}{(\zeta-z)^{n-1}(\zeta-z_0)}\d\zeta-\oint_\gamma\frac{\varphi(\zeta)}{(\zeta-z_0)^n}\d\zeta\right]\\&=\frac{1}{z-z_0}\left[\oint_\gamma\frac{\varphi(\zeta)}{(\zeta-z)^{n-1}(\zeta-z_0)}\d\zeta-\oint_\gamma\frac{\varphi(\zeta)}{(\zeta-z_0)^{n-1}(\zeta-z_0)}\d\zeta\right]\\&=\frac{1}{z-z_0}\left[\oint_\gamma\frac{\psi(\zeta)}{(\zeta-z)^{n-1}}\d\zeta-\oint_\gamma\frac{\psi(\zeta)}{(\zeta-z_0)^{n-1}}\d\zeta\right]\\&=\frac{F_{n-1}(z;\psi)-F_{n-1}(z_0;\psi)}{z-z_0}\\&\overset{\text{induction}}{\longrightarrow}(n-1)F_n(z_0;\psi)\\&=(n-1)\oint_\gamma\frac{\psi(\zeta)}{(\zeta-z_0)^n}\d\zeta\\&=(n-1)\oint_\gamma\frac{\varphi(\zeta)}{(\zeta-z_0)^{n+1}}\d\zeta\\&=(n-1)F_{n+1}(z_0;\varphi)\end{align}$$As desired. In sum, we get $F'_n(z_0;\varphi)$ exists and equals $(n-1)F_{n+1}(z_0;\varphi)+F_{n+1}(z_0;\varphi)=nF_{n+1}(z_0;\varphi)$ as desired.

By induction, we obtain the crucial result (for continuous $\varphi$ defined on a continuous curve $\gamma$ and $z$ not in the trace of $\gamma$): $$\frac{\mathrm{d}^n}{\mathrm{d}z^n}\oint_\gamma\frac{\varphi(\zeta)}{\zeta-z}\d\zeta=n!\cdot\oint_\gamma\frac{\varphi(\zeta)}{(\zeta-z)^{n+1}}\d\zeta$$Demonstrating (using estimates similar to those used to demonstrate differentiability of $F_1$) analyticity of $F_0$.