Does there exist non-constant holomorphic $f: \mathbb{C} \to \mathbb{C}$ such that $e^{Re(f(z))}+ (\text{Im}f(z))^2 =1, \forall z \in \mathbb{C} \ ?$ Could anyone advise me how to prove/disprove that there exists non-constant holomorphic function $f: \mathbb{C} \to \mathbb{C}$ such that $e^{Re(f(z))}+ (\text{Im}f(z))^2 =1, \forall z \in \mathbb{C} \ ?$ Do I need to use Open Mapping Theorem?  
Thank you. 
 A: The given equality implies that $\operatorname{Im} f(z)$ is bounded. Apply Liouville's theorem to $\exp(i f(z))$ to see that it is constant. It follows that $f$ is also constant.
A: You are correct that this can also be done using the open mapping theorem. Here is a sketch.


*

*Since $f$ is not constant, its complex derivative is nonzero somewhere. In particular, viewing $f$ as a smooth map $\mathbb{R}^2 \to \mathbb{R}^2$, there is a point $p \in \mathbb{R}^2$ where $Df_p$ has full rank (=2).

*Note that $g : \mathbb{R}^2 \to \mathbb{R}$ given by $g(x,y) = e^x + y^2$ has gradient $(x,y) \mapsto (e^x,2y)$, which never vanishes. In other words, $Dg_p$ has full rank (=1) for all $p \in \mathbb{R}^2$.


Generally, if two smooth functions $f : \mathbb{R}^2 \to \mathbb{R}^2$ and $g : \mathbb{R}^2 \to \mathbb{R}$ are such that $Df_p$ has full rank and $Dg_{f(p)}$ has full rank, then $D(g \circ f)_p$ has full rank by the chain rule. Thus, by the open mapping theorem, $\operatorname{range}(g \circ f)$ must contain an open set, precluding the possibility that $g \circ f \equiv 1$.
