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We cannot apply liouville theorem as $f$ is not bounded in $\mathbb{C}$. So instead I made use of the fact that growth of $\log(\mid z \mid)$ is slower than any power of $z$ and hence to dominate $\mid f(z) \mid$ which is a power series, $f$ must be a constant.

So is the proposition and the reasoning right?

P.S Suppose $ f(i) = 2i$, then what is the value of $f(1)$. I think by proving $f$ is constant, I can get the value to be 2i which happens to be the correct answer.

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In a way yes it's right, but it's a bit more technical. An entire function has a Taylor expansion $$f(z)=a_0+\sum\limits_{n=1}a_nz^n \tag{1}$$ where $$a_n=\frac{f^{(n)}(0)}{n!}=\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{f(z)}{z^{n+1}}dz$$ As a result $$|a_n|=\left|\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{f(z)}{z^{n+1}}dz\right|\leq \frac{1}{2\pi}\int\limits_{|z|=r} \left|\frac{f(z)}{z^{n+1}}\right||dz|\leq \\ \frac{1}{2\pi}\int\limits_{|z|=r} \frac{100\log|z|}{|z|^{n+1}}|dz|= \frac{1}{2\pi}\int\limits_{|z|=r} \frac{100 \log{r}}{r^{n+1}}|dz|=\\ \frac{1}{2\pi}\frac{100 \log{r}}{r^{n+1}}\cdot 2\pi r= \frac{100 \log{r}}{r^{n}}$$ Taking the limit $\lim\limits_{r\rightarrow\infty}$ we have that $|a_n|=0$ for all $n\geq1$, thus from $(1)$ $$f(z)=a_0$$

Remark: It applies to $|z|\geq 1$ too.

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