Problem. Find all the entire functions $f:\mathbb{C}\rightarrow\mathbb{C}$ such that
$$|f(z)|\leq-\ln(|z|^2)+|z|^2, f(0)=i$$
Since $-\ln(|z|^2)+|z|^2\leq2|z|^2,$ we have$|f(z)|\leq2|z|^2$. hence $f$ is polynomial degree of 2. Since $f$ is entire and so analytic (power series representation),
$$f(z)=f(0)+f'(0)z+\frac{f''(0)}{2}z^2=i+f'(0)z+\frac{f''(0)}{2}z^2$$
(I don't even know if I wrote it right up to here. Please point out if there is anything wrong.) I need more exact conditions for $f'(0)$ and $f''(0)$, but I don't know what to do. I'd appreciate it if you could give me a hint.
$|f(z)|\leq-\ln(|z|^2)+|z|^2$ does imply that $f$ is a polynomial of degree at most two, because $|f(z)| \le |z|^2$ for $|z| \ge 1$.
But it is actually much simpler: For $|z| = 1$ the given condition gives $|f(z)| \le 1$, so that $|f(z)| \le 1$ in the unit disk by the maximum modulus principle. But then $f(0) = i$ implies that $|f|$ attains its maximum inside the unit disk.
It follows that $f$ is constant in the unit disk, and therefore constant in $\Bbb C$. So the only solution is the constant function $f(z) = i$.