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Let $f$ be an entire function such that $|f'(z)|\leq|f(z)|$ for all $z\in \mathbb{C}$. Prove that there exist complex numbers $a$ and $c$ such that $f(z)=ae^{cz}$ for all $z\in\mathbb{C}$.

My try: take $g(z)$ such that $$ f(z)=e^{g(z)}. $$ Then by chain rule $$ f'(z)=e^{g(z)}g'(z). $$ Now we have $$ |g'(z)||f(z)|=|g'(z)||e^{g(z)}|=|f'(z)|\leq |f(z)| $$ which implies $$ |g'(z)|\leq 1. $$ Thus $g(z)=cz$, and then $f(z)=e^{cz}$. However, how to make $g(z)$ is entire function? and where is the constant $a$?

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You have to justify writing $f$ as $e^{g}$. Here is a correct solution: If $f \equiv 0$ we can take $a=0$ and $c=1$. Otherwise the zeros of $f$ form a set $\{z_n\}$ with no limit points. On $\mathbb C \setminus \{z_n\}$ we have $|\frac {f'(z)} {f(z)}| \leq 1$. This implies that $\frac {f'(z)} {f(z)}$ has a removable singularity at each $z_n$. Hence it extends to a bounded entire function. By Louiville's Theorem we have $\frac {f'(z)} {f(z)}=A$ for some constant $A$. Now observe that $(e^{-Az}f(z))'=0$ so $e^{-Az}f(z)=B$ for some constant $B$. Hence $f(z)=Be^{Az}$.

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Note that $f$ has not zeros. In fact, if $a$ is a zero of $f$ of order $n\in\mathbb N$, then $a$ is a zero of $f'$ os order $n-1$. But then, near $a$, the inequality $\bigl\lvert f'(z)\bigr\rvert\leqslant\bigl\lvert f(z)\bigr\rvert$ means$$\bigl\lvert na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+\cdots\bigr\rvert\leqslant\bigl\lvert a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+\cdots\bigr\rvert.$$Now, dividing both sides by $\bigl\lvert(z-a)^n\bigr\rvert$, you get that$$\bigl\lvert na_n(z-a)^{-1}+(n+1)a_{n+1}+\cdots\bigr\rvert\leqslant\bigl\lvert a_n+a_{n+1}(z-a)+\cdots\bigr\rvert,$$which is impossible, because the LHS tends to $+\infty$ when $z\to a$, whereas the RHS tends to $\lvert a_n\rvert$.

So, since $f$ has no zeros and $\mathbb C$ is simply connected, there is indeed an entire function $g$ such that $f=e^g$. You proved that $g'$ is bounded. Therefore, by Liouville's therem, it is constant. So there are constants $k$ and $c$ such that $(\forall z\in\mathbb{C}):g(z)=cz+k$and therefore$$(\forall z\in\mathbb{C}):f(z)=e^{g(z)}=e^{cz+k}=e^ke^{cz}.$$

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