Describe all the entire functions,$f(z)$ with $Re(f(z))$ tending to $0$ as $z\to \infty$.
My approach:
$lim_{z\to \infty}Re(f(z))=0$ implies $lim_{z\to \infty}e^{Re(f(z))}=1$. Thus for some $N\in \mathbb{N}$, if $\mid z\mid\geq N$ we will have $\mid e^{Re(f(z))} -1\mid <1$. Therefore $\mid e^{Re(f(z))}\mid <2$ and hence $\mid e^{f(z)} \mid <2$ for $\mid z\mid \geq N$. Also in the compact set $\mid z\mid \leq N$ the entire function $e^{f(z)}$ is bounded. Consequently $e^{f(z)}$ is a bounded entire function and hence constant. This concludes that $f$ is constant and $f\equiv 0$.