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.

  • $\begingroup$ 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. $\endgroup$ Apr 3, 2021 at 20:13
  • 1
    $\begingroup$ It seems that the condition $\lim_{z\to\infty}f(z)=\alpha$ is missing. For example, if $f(z)=z$,what you will get? $\endgroup$
    – Riemann
    Apr 4, 2021 at 2:14

1 Answer 1


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 $|z|<R$ and $f$ is holomorphic on the exterior of $|\zeta| > r >0, r<R$

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!

  • $\begingroup$ Sorry what does primitive mean? $\endgroup$
    – athos
    Apr 3, 2021 at 22:32
  • 1
    $\begingroup$ Antiderivative, $\endgroup$
    – Mittens
    Apr 3, 2021 at 22:50
  • $\begingroup$ @Oliver - corrected as I agree that antiderivative is the better term in this context $\endgroup$
    – Conrad
    Apr 3, 2021 at 23:02

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