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Let $f(z)=u(x,y)+iv(x,y)$ be a holomorphic function. Then how to show $$\frac{1}{2\pi i}\oint_\gamma \frac{u(\zeta)-iv(\zeta)}{\zeta-z}d\zeta = u(0)-iv(0)$$ where $\gamma$ is a circle centered at $0$ and $z$ is in the interior of $\gamma$.

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    $\begingroup$ It seems that this follows immediately from the Cauchy integral formula $\endgroup$
    – Matt
    Commented Nov 16, 2016 at 5:25
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    $\begingroup$ @Matt I'm not sure that it does, because $u-iv$ is not holomorphic. But the question is clearly wrong because it would give not $(u-iv)(0)$ but $(u-iv)(z)$ instead. $\endgroup$
    – Bob Jones
    Commented Nov 16, 2016 at 6:00

2 Answers 2

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We assume for simplicity that $\gamma $ is the unit circle and $$f(z)=\sum_{k=0} ^\infty a_kz^k$$ is analytic on $|z|\le 1$, that is, $$\sum_{k=0} ^\infty a_kz^k$$ converges absolutely and uniformly on $\gamma $ .

First we note that \begin{align} \int_\gamma \frac{1}{\zeta -z}\, d\zeta &=2\pi i,\\ \int_\gamma \frac{\overline{\zeta^n}}{\zeta -z}\, d\zeta &=0, \quad (n\ge 1)\tag{1} \end{align} for $z$ inside $\gamma $.
Then we have \begin{align} \int_\gamma \frac{\overline{f(\zeta)}}{\zeta -z}\,d\zeta&=\int_\gamma \left(\sum_{k=0}^\infty \overline{a_k\zeta^k}\right)\frac{1}{\zeta -z}\, d\zeta \\ &=\sum_{k=0}^\infty \overline{a_k}\int _\gamma \frac{\overline{\zeta^k }}{\zeta -z}\,d\zeta \\ &=2\pi i\,\overline{a_0}=2\pi i\,\overline{f(0)}, \end{align} which completes the proof.

Proof of $(1)$.
Since $\overline{\zeta }=\frac{1}{\zeta}$ for $\zeta \in \gamma $, we have $$ \int_\gamma \frac{\overline {\zeta ^n}}{\zeta -z}\, d\zeta =\int_\gamma \frac{d\zeta }{(\zeta -z)\zeta ^n}. $$ Notice that $\frac{1}{(\zeta -z)\zeta ^n}$ is analytic on the annulus $1\le |\zeta |\le R$. Therefore we have by Cauchy theorem that $$ \int_\gamma \frac{d\zeta }{(\zeta -z)\zeta ^n}=\int_{|\zeta |=R} \frac{d\zeta }{(\zeta -z)\zeta ^n} $$ and it is easy to see that $$ \lim_{R\to \infty} \int_{|\zeta |=R} \frac{d\zeta }{(\zeta -z)\zeta ^n} =0, $$ which completes the proof of $(1)$.

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$u$ and $v$ are harmonic, thus the Mean Value Property of harmonic functions gives

$\frac{1}{2\pi i}\oint_\gamma \frac{u(\zeta)-iv(\zeta)}{\zeta-z_0}d\zeta = u(z_0)-iv(z_0)$

where $\gamma(t)=z_0 +re^{it}$ and $r$ so that $\{z: |z-z_0| \le r\}$ is contained in the domain of $f$

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  • $\begingroup$ That is not the Mean Value Property of harmonic functions. The Mean Value Property of harmonic functions is $u(z_0)=\frac{1}{2\pi}\oint_{\gamma}u(\zeta)$. $\endgroup$
    – Bob Jones
    Commented Nov 17, 2016 at 4:00
  • $\begingroup$ No ! The mean Value Property reads as follows: $u(z_0)=\frac{1}{2 \pi}\int_{0}^{2 \pi}u(z_0+re^{it}) dt$. Your integral: $\frac{1}{2\pi}\oint_{\gamma}u(\zeta) d \zeta=\frac{1}{2 \pi}\int_{0}^{2 \pi}u(z_0+re^{it})rie^{it} dt$ $\endgroup$
    – Fred
    Commented Nov 17, 2016 at 5:56
  • $\begingroup$ I guess I should have written $d\theta$ at the end to make it correct. Anyway, neither of these are what you wrote, and neither correspond to what the OP wants. $\endgroup$
    – Bob Jones
    Commented Nov 17, 2016 at 5:59
  • $\begingroup$ What ?? $\frac{1}{2 \pi i}\oint_{\gamma}\frac{u(\zeta)}{\zeta-z_0} d \zeta=\frac{1}{2 \pi i} \int_{0}^{2 \pi}\frac{u(z_0 +r e^{it})}{re^{it}}rie^{it} dt=\frac{1}{2 \pi}\int_{0}^{2 \pi}u(z_0+re^{it}) dt = u(z_0)$ $\endgroup$
    – Fred
    Commented Nov 17, 2016 at 6:07
  • $\begingroup$ Ok, you're right. I guess I don't see things if they're not spelled out for me. $\endgroup$
    – Bob Jones
    Commented Nov 17, 2016 at 6:11

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