# Prove $\frac{1}{2\pi i}\oint_\gamma \frac{zf(\zeta)}{z\zeta-\zeta^2}d\zeta=0$

Let $$\gamma$$ be a rectifiable simple close curve, $$D_1$$ be the inside of $$\gamma$$ and $$D_2$$ be the outside of $$\gamma$$. Suppose a function $$f(z)$$ is holomorphic in $$D_2$$ and continuous on $$D_2 \cup \gamma$$. If both the origin and $$z$$ are in $$D_1$$, show that

$$\frac{1}{2\pi i}\oint_\gamma \frac{zf(\zeta)}{z\zeta-\zeta^2}d\zeta=0$$

I've done so far as:

$$\frac{1}{2\pi i}\oint_\gamma \frac{zf(\zeta)}{z\zeta-\zeta^2}d\zeta=\frac{1}{2\pi i} \left(\oint_\gamma \frac{f(\zeta)}{\zeta}d\zeta-\oint_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta \right)=\frac{1}{2\pi i}\left(\oint_\Gamma \frac{f(\zeta)}{\zeta}d\zeta-\oint_\Gamma \frac{f(\zeta)}{\zeta-z}d\zeta\right)$$

, where $$\Gamma$$ is a big enough circle around the origin, i.e. it is $$|z|=R$$ where $$R$$ is sufficiently large.

But then I'm stuck here as cannot show it is $$0$$.

Pls kindly enlighten me.

• I'm just as stuck as you are because it seems like by Cauchy integral formula the last line should be $f(0)-f(z)$, which is not always zero as you say. Apr 3, 2021 at 20:13
• It seems that the condition $\lim_{z\to\infty}f(z)=\alpha$ is missing. For example, if $f(z)=z$,what you will get? Apr 4, 2021 at 2:14

As noted in the original post (and by Cauchy) we only need to prove (for some large $$R$$)
$$\oint_{|\zeta|=R} \frac{f(\zeta)}{\zeta}d\zeta-\oint_{|\zeta|=R} \frac{f(\zeta)}{\zeta-z}d\zeta =0$$, where $$|z| and $$f$$ is holomorphic on the exterior of $$|\zeta| > r >0, r
Keeping in mind that $$\zeta^{-k}$$ has an antiderivative in our domain for $$k \ge 2$$ so its circle integral is zero, looking at the Laurent series of $$f(w)=\sum_{k \ge 0}a_kw^{-k}$$ and integrating term by term (by absolute convergence) we notice that in the first integral only the term $$a_0/\zeta$$ has a (potentially) non zero integral.
On the other hand expanding $$\frac{f(\zeta)}{\zeta-z}=(\sum_{k \ge 0}a_k\zeta^{-k})(\sum_{m \ge 0}\frac{z^m}{\zeta^{m+1}})$$ and again multiplying term by term and integrating (allowed by absolute convergence), only the term $$a_0/\zeta$$ has a non zero integral as all the other terms are scalar multiples of $$\zeta^{-k}, k \ge 2$$, so we are done!