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A:

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Q:

Entire function with singular value

Let $f: \mathbb C\setminus \lbrace 0\rbrace\rightarrow \mathbb C\setminus\lbrace 0\rbrace$ be an entire function that has a singular value at $z=0$ with value $a_0=1$. That means that $f(z)=e^{g(z)}$ for some entire function $g$.
Prove that $\lim\limits_{n\to\infty}\left|\frac{1}{z^n}\left(a_0^n-1\right)\right|=0$ for every $z\in\mathbb C\setminus \lbrace 0\rbrace$.

A:

By Cauchy’s inequalities, it’s true that:
$$|\frac{z^n}{a_n}-1|\leq\frac{|z|^n}{|z|^n}=1$$
In particular, it follows that:
$$|z^n|(|a_n|-1)\leq|z^n(a_n-1)|\leq|a_n-1||z^n|$$
The LHS tends to zero so $a_n-1\to0$, and $\frac{z^n}{a_n}\to1$ therefore $\frac{z^n}{a_n}\to1-\frac1z=\frac{1-\frac1z}z=\frac{z-1}{z^2}=z^{ -1}-1$, and you’re done.

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