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Let $f$ be an entire non-constant complex function and let $A$ and $B$ be given positive real constants. Is it possible that $|f(z)| \le A + B\log{| z|}$ for all complex $z$ such that $| z| \ge 1$ ?

I've been trying to solve using the fact that since $f$ is continuous in $\{z; |z|\le 1\}$ there exists $M=\sup\{ |f(z); |z|\le 1\}$, and then somehow use the Cauchy's integral formula for derivatives. Any ideas?

Thanks

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Cauchy's integral formula for derivatives is the way to go, but not for a fixed domain. Rather, for every real number $R\geqslant1$ and every complex number $z$ such that $|z|\lt R$, write $ f'(z)=\frac1{2\pi\mathrm i}\oint_{|w|=R}\frac{f(w)}{w^2}\mathrm dw, $ hence $|f'(z)|\leqslant\frac1{2\pi}\cdot\sup\limits_{|w|=R}|f(w)|\cdot\frac1{R^2}\cdot2\pi R\leqslant\frac{A+B\log R}R. $ Fix $z$. When $R\to\infty$, the RHS goes to zero. QED.