Question: let $f$ be an entire function on $\mathbb{C}$ and let $g(z)=\overline{f(\bar{z})}$ then which of the following is/are correct?
(a) if $f(z) ∈\mathbb{R}$ for all $z∈\mathbb{R}$ then $f=g$
(b) if $f(z)∈\mathbb{R}$ for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a>0$ then $f(z+ ia)=f(z-ia)$ for all $z∈\mathbb{C}$
(c) if $f(z)∈\mathbb{R}$ for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a>0$ then $f(z+ 2ia)=f(z)$ for all $z∈\mathbb{C}$
(d) if $f(z)∈\mathbb{R}$ for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a>0$ then $f(z+ ia)=f(z)$ for all $z∈\mathbb{C}$
My attempt: clearly, if $z∈\mathbb{R}$ then $\bar{z}=z∈\mathbb{R}$. Hence if $f(z)∈\mathbb{R}$ then $f(\bar{z})=f(z)∈\mathbb{R}$ so that $g(z)=\overline{f(\bar{z})}=f(\bar{z})=f(z)$ for $z$ in $\mathbb{R}$ hence (a) is correct.
But as beginner i am unable to prove or discard (b), (c), (d). I am stuck on this from hours, Please help me..
I will show that $(b)$ and $(c)$ are true.
From solving option $(a)$ you found that $f(z)=g(z)=\overline{f(\bar z)} \; \; \forall z \in \Bbb C$
Now in option $(b)$ note that $\{z : \text{Im} (z)=0\}=\Bbb R.$ Let $x \in \Bbb R$, then $f(x+ia)=\overline{f(\overline{x+ia})}=\overline{f(x-ia)}$. But $f(z) \in \Bbb R$ on $\{z : \text {Im}(z)=a\}$ as well. $\therefore f(x+ia)=\overline{f(x-ia)}\ \in \Bbb R \Rightarrow f(x+ia)=f(x-ia) \in \Bbb R.$ By identity theorem, $f(z+ia)=f(z-ia) \; \forall \; z \in \Bbb C.$
For option $(c)$, put $z=w+ia$ then you get $f(w+2ia)=f(w) \; \forall \; w \in \Bbb C$.