Let $f(z)$ be entire. Suppose $\mathrm{Im} f(x) > 0$ for all $z$. Prove that $f =$ const. I don't know how can I do. A indication is use $L(z)=(iz+1)/(1-iz)$ map the upper half plane into the disk unitary.

Composing $f$ with $L$, we obtain
$$L\circ f=z\mapsto \frac{1+if(z)}{1-if(z)}$$
which is an entire function, and takes all values in the unit disk, hence bounded, so it's constant.
Then $f=L^{-1}\circ L\circ f$ is also constant. 
