$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant
Solution 1:
$|\mathrm{Im}f(z)|\le |\mathrm{Re}f(z)|$ implies that $|f(z)-i|\ge \dfrac{\sqrt{2}}{2}$. It follows that $g(z)=\dfrac{1}{f(z)-i}$ is a bounded entire function, and due to Liouville's theorem, it must be a constant.
Solution 2:
Alternatively, $\textrm{Re} \, f(z)^2 = \left(\textrm{Re} \, f(z) \right)^2 - \left(\textrm{Im} \, f(z) \right)^2 \geq 0$. This implies that $f(z)^2$ is constant.