$f$ entire and $Im(f(z)) \leq Re(f(z))^2$ implies $f$ is constant

Question

My notes say to use Little Picard, with the hint hat $f$ will miss a disk. I don't see how to do that, instead I have:

$f$ entire $\Rightarrow$ $e^{-f}$ entire.

Now, $\mid e^{-f}\mid = e^{-Re(f(z))} \leq e^{Im(f(z))\frac{1}{2}}$ therefore $e^{-f}$ is bounded, hence constant by Liouville. Then, $f$ is also constant.

Is this correct? Also, how would you go about showing that $f$ misses a disk to use Little Picard instead?

Answer 1

Your inequality is wrong. Use Picard's Theorem. $|(a+ib) -i| <\frac 1 2$ implies $1-b <\frac 1 2 $ so $b >\frac 1 2 >a^{2}$. Hence no point of the disk $D(i, \frac 1 2)$ can be in the range of $f$.