Let $f$ be an entire function on $\mathbb{C}$. Let $g(z)=\overline{f(\bar z)}$. Then which of the following is/are true:
$(1)$ If $f(z) \in \mathbb{R} $ $\forall z\in \mathbb {R}$ then $f(z)=g(z)$
$(2)$ If $f(z)\in \mathbb {R} $ $\forall z\in \{z:im(z)=0\} \cup \{z:im(z)=a\}$ for some $a\gt$0 then $f(z-ia)=f(z+ia)$ $\forall z\in\mathbb {C}$
$(3)$ If $f(z)\in \mathbb {R} $ $\forall z\in \{z:im(z)=0\} \cup \{z:im(z)=a\}$ for some $a\gt$0 then $f(z+2ia)=f(z)$ $\forall z\in\mathbb {C}$
$(4)$ If $f(z)\in \mathbb {R} $ $\forall z\in \{z:im(z)=0\} \cup \{z:im(z)=a\}$ for some $a\gt$0 then $f(z+ia)=f(z)$ $\forall z\in\mathbb {C}$
What I have tried so far: For $(1)$ it is easy to see that the set $\{z \in \mathbb{C} : f(z)=g(z)\}$ contains $\mathbb{R}$ and therefore by Identity Theorem, $f(z)=g(z)$.
And for part $(4)$ Take a function $f(z)=e^z$ and then we see that $(4)$ is NOT true.
Am I correct?
But I have no idea how to proceed for $(2)$ & $(3)$?
Any help would be appreciated.
As $f(z)$ is entire, write $f(z) = c_0 + \sum_{k=1}^{\infty} {c_{k} (z-ia)^k} $
Clearly, $f(ia) = c_0 \in R$
From part (a), we get, $f(z)= \overline {f (\bar {z})} = c_0 + \sum_{k=1}^{\infty} \overline{{c_{k} (\bar z-ia)^k}} = c_0+ \sum_{k=1}^{\infty} {\bar c_{k} (z+ia)^k} $
In the proof of first part, it can be shown that $\overline{c_k} =c_k$ for if
$\overline{f(z)} =f(z)$ whenever $\bar z = z \implies \sum_{k=0}^{\infty} {\overline {b_{k}} z^k = \sum_{k=0}^{\infty} {b_{k} z^k}} \implies \overline {b_k}=b_k$
Thus, $f(z+ia)= c_0 + \sum_{k=1}^{\infty} {c_{k} z^k} = f(z-ia) $