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The mean-value property for holomorphic functions states that if f is holomorphic in the neighbourhood of the closed disc centered at $z0$ of radius $R$, then

$f(z_{0}) = \frac{1}{2\pi}\int_{0}^{2\pi}f(z_{0} + re^{i\theta})\, d\theta$

if $f(z)=\sum_{n=0}^{+\infty}a_nz^n$, We have by dominated convergence:

$\int_{0}^{2\pi}f(z_{0} + re^{i\theta})d\theta = \int_{0}^{2\pi}\sum_{n=0}^{+\infty}a_n(z_{0} + re^{i\theta})^nd\theta=\sum_{n=0}^{+\infty}a_n\int_{0}^{2\pi}(z_{0} + re^{i\theta})^nd\theta =[\frac{(z_0+re^{i\theta})^{n+1}}{ir(n+1)}]_0^{2\pi}=0$

Which means evry holomorphic function is equal to 0, a non sense. I can't find my mistake, any help much appreciated.

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    $\begingroup$ Your antiderivative is incorrect. $\endgroup$ – Mark Viola Apr 21 '18 at 13:09
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HINT:

Your antiderivative is incorrect. Use the binomial theorem to write

$$(z_0+re^{i\theta})^n =\sum_{k=0}^n \binom{n}{k} z_0^{n-k}e^{ik\theta}$$

then exploit the fact that $\int_0^{2\pi}e^{ik\theta}\,d\theta=0$ for $k\ne 0$.

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  • $\begingroup$ Thank you. using the binomial theorem I found: $\int_{0}^{2\pi}(z_{0} + re^{i\theta})^nd\theta = z_0^n$ so that makes sense! $\endgroup$ – PerelMan Apr 21 '18 at 13:21
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    $\begingroup$ Yes, and hence you recover $f(z_0)=\sum_{n=0}^\infty a_nz_0^n$, which is the Taylor series of $f$ $\endgroup$ – Mark Viola Apr 21 '18 at 13:24

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