If $f$ is an entire function with $|f(z)|\le 100\log|z|$ and $f(i)=2i$, what is $f(1)$? Let $f$ be an entire function with $|f(z)|\le 100\log|z|,\forall |z|\ge 2,f(i)=2i, \text{ Then} f(1)=?$
I have no idea how to solve this one!
$g(z)={f(z)\over \log|z|}$ Then Can I say $g$ is constant by Liouville Theorem?

You can't directly use Liouville's theorem, since dividing $f$ by $\log \lvert z\rvert$ or $\log z$ doesn't produce an entire function.
But you can use the Cauchy estimates to show that that bound on $\lvert f(z)\rvert$ actually implies that $f$ is constant, for $R > 2$ and $\lvert z\rvert \leqslant \frac{R}{2}$, you can compute
$$\left\lvert f'(z) \right\rvert = \left\lvert \frac{1}{2\pi i}
\int\limits_{\lvert \zeta \rvert = R} \frac{f(\zeta)}{(\zeta-z)^2}\, d\zeta\right\rvert \leqslant \frac{1}{2\pi} \int\limits_{\lvert\zeta\rvert = R} \frac{\lvert f(\zeta)\rvert}{\lvert \zeta - z\rvert^2}\, \lvert d\zeta\rvert \leqslant \frac{4R}{R^2}100\log R$$
and, since the right hand side tends to $0$ for $R \to \infty$, conclude that $f' \equiv 0$.
