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I need to prove the following:

If $a\gt 0$, $n\in \mathbb{N}-\{0\}$, and if $f$ analytic on $(0, 2a)$ with $$|f^{(n)}(0)| = \frac{n!}{a^n}\cdot \sup\left\{|f(z)|\mid : |z|=a\right\},$$ then $f(z) = rz^n$ for some $r\in \mathbb{C}$.

I have tried to borrow ideas from the proof of Liouville's Theorem and go backwards. However, the problem doesn't quite seem to fit the hypothesis of Liouville's Theorem. Any suggestion would be appreciated. Thanks!

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Put for $z\neq 0$: $g(z):=\frac{f(z)}{z^n}$. We can show thanks to Cauchy's integral formula that for $|z|<a$ $$\left|g(z)-\frac{f^{(n)}(0)}{n!}\right|\leq \frac{|z|}{(a-|z|)a^{n+1}}\sup_{|\xi|=a}|f(\xi)|,$$ so we can put $g(0):=\frac{f^{(n)}(0)}{n!}$. $g$ is still analytic and by hypothesis we have $$|g(0)|=\left|\frac 1{2\pi i}\int_{C(0,a)}\frac{f(\xi)}{\xi^{n+1}}d\xi\right|=\sup_{|\xi|=a}|g(\xi)|,$$ and by maximum principle we get that $g$ is constant on $B(0,a)$, hence on $B(0,2a)$.

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