A problem of open set, closed set in complex plane Let $f$ be a non-constant analytic function on $\mathbb{C}$ and let $\Omega$ be a open bounded subset of $\mathbb{C}$. If $S=\{\Re(f(z))+\Im(f(z))|z \in \Omega\}$ then which of the following is\are true?


*

*S is open in $\mathbb{R}$.

*S is closed in $\mathbb{R}$.

*S is open in $\mathbb{C}$.

*S is discrete in $\mathbb{R}$.


Please give me some idea about it. Thanks in advance.
 A: Since you asked for some idea, here are some starts for all the statements.


*

*You know that $f$ is analytic and non constant, so by the open mapping theorem (in complex analysis), you know that $f$ is open. You will deduce from that that $S$ is open in $\mathbb R$, because $f(\Omega)$ is open so $\mathfrak I(f(\Omega))$ and $\mathfrak R(f(\Omega))$ are open in $\mathbb R$ so $\mathfrak r(f(\Omega))+\mathfrak I(f(\Omega))$is also open.

*Since $S$ is open, you will only have $S$ closed in $\mathbb R$ is $S=\mathbb R$ or $S=\emptyset$. The second case is impossible, and so is the first one because $\Omega$ is bounded.

*$S$ can not be open in $\mathbb C$ since $\Omega\ne \emptyset$ and $S\subset \mathbb R$ which as an interior empty in $\mathbb C$.

*$S$ can not be discret since $f$ is analytic so $f$ is continuous and so are $\mathfrak R(f)$ and $\mathfrak I(f)$. So if $S$ is discrete $S$, would be a singleton. I think you could use the analyticity of $f$ through Cauchy-Riemann equations to prove that this case will not happen.
