If $f$ is an entire function then what about the set $S=\{Ref(z)+Imf(z) :z\in D\}.$ Let, $f$ be an entire function on $\mathbb C$ and let $D$ be a bounded open subset of $\mathbb C$. Let, $$S=\{Ref(z)+Imf(z) :z\in D\}.$$

Which of the following(s) is(/are) necessarily true ?

(a) $S$ is open set in $\mathbb R$.

(b) $S$ is closed set in $\mathbb R$.

(c) $S$ is open set in $\mathbb C$.

(d) $S$ is discrete set in $\mathbb R$.

I tried through an example : Let, $f(z)=5$ , a constant function.

Then $S=\{5\}$ , is closed set in $\mathbb R$. So, it can not be open set in $\mathbb R$.

So, clearly option (a) is false. But we can not say that option (b) is true.

Am I right or wrong ? If I am wrong then where my fallacy ?

What about the other option ?
Please anyone help me....
If $f$ is constant then $S$ consists of one point.

If $f:\&gt;z\mapsto w:=f(z)$ is not constant then $\Omega:=f(D)$ is open. The map
$$w=u+iv \quad  \mapsto\quad  {\rm Re}(w)+{\rm Im}(w)=u+v$$
producing $S$ from $\Omega$ is an orthogonal projection followed by a rotation and a scaling, whence maps the region $\Omega$  to an open interval.

(It is a key feature of the  topology on ${\mathbb R}^n$ that a projection maps open sets to open sets.)
consider
$$f(z)=z$$
on $Re[x]\in(-1,+1)$, $Im[x]\in(-1,+1)$, then $S=(-2,+2)$ which is open and b is wrong too.

similarly c is wrong.

d is wrong for obvious reason as well.