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Suppose that $|f(z)|\leq A+B|z|^M$ and that $f$ is entire. Show that for all coefficients $c_j$ with $M<j$ in its power series extansion are $0$.

Attampt:

$$ |f(z)|=\left|\sum_{k=0}^\infty c_kz^k\right| \leq A+B|z|^M \Rightarrow \\ \left|\sum_{k=0}^\infty c_kz^{k-M}\right|\leq{A\over |z|^M}+B\underset{z \to \infty}{\rightarrow}B $$ $g(z):=\sum_{k=0}^\infty c_kz^{k-M}$ is continuous and thus $|g|$ is bounded by some $Q$.

How can I finish the exercise?

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  • $\begingroup$ Do you already have this theorem for $M = 0$ (bounded entire function is constant)? $\endgroup$ – mihaild Apr 26 at 14:57
  • $\begingroup$ This does not work since in general $g$ will have a pole at $z=0$. $\endgroup$ – Hans Engler Apr 26 at 15:01
  • $\begingroup$ But we can first replace $f$ with $f(z) - c_0 - c_1 z - \ldots c_M z^M$, then $g$ will have no poles (and such replacement doesn't affect boundary or $c_j = 0$ for $j > M$). $\endgroup$ – mihaild Apr 26 at 15:11
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It should be $c_j=0$ for $j>M$. By the way, it would be useful to express $c_j$ by the Cauchy Integral Formula:$$c_j=\frac{f^{(j)}(0)}{j!}=\frac{1}{j!}\frac{1}{2\pi i}\int_{\partial B_R}\frac{f(w)}{w^{j+1}}dw$$ Because then a bound for $c_j$ is easy to obtain: $$|c_j|\le\frac{2\pi R}{j!2\pi}\frac{A+BR^M}{R^{j+1}}\to 0 \text{ as }R\to\infty,\text{ for all }j>M$$

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  • $\begingroup$ I know Cauchy Integral Formula for $j=0$ can you please write the formula for the $j$'s derivative? Thanks $\endgroup$ – J. Doe Apr 26 at 15:23

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