This particular question was asked in masters exam for which I am preparing .
I have proved other 3 options , so I will not write them .
Let f:$\mathbb{C} -> \mathbb{C}$be a holomorphic function . Choose the correct statements from below :
1.Suppose f($\mathbb{C}$) is a proper subset of $\mathbb{R}$ . Then f(A) is closed in $\mathbb{C}$ for every
closed subset A of $\mathbb{C} $ .
Attempt: An entire function assumes all values with 1 possible exception , so f($\mathbb{C}$) is a proper subset of $\mathbb{R}$ implies f($\mathbb{C}$) is constant and hence closed set is mapped to closed set .
Is my reasoning right ?
Best Answer
An holomorphic function $f:\Omega\subset\mathbb{C}\to\mathbb{C}$ seen as a differentiable map $f:U\subset\mathbb{R}^2\to\mathbb{R}^2$ is either of rank $2$ or of rank $0$: indeed, the $\mathbb{C}$-differentiability implies that its differential is of the form
$$Df_{(x,y)}=\begin{pmatrix}a(x,y)&-b(x,y)\\b(x,y)&a(x,y)\end{pmatrix},$$
that is multiplication by the complex number $a(x,y)+ib(x,y)$. Thus, if the image is lying in a subspace of dimension $1$, the rank of your map is bounded by $1$, and you have no choice to be a constant function. Since singletons are closed in $\mathbb{C}$, constant functions are closed, and then the claim $1.$ is correct.