Let $f$ be an entire function on $\mathbb{C}$. Let $g(z)=\overline{f(\bar{z})}$. Which of the following statements is/are correct?
1) If $f(z) \in \mathbb{R}$ for all $z \in \mathbb{R}$ then $f=g$.
2) If $f(z) \in \mathbb{R}$ for all $z \in \mathbb{R} \cup \{z | Im z=a\}$ for some $a>0$, then $f(z+ia)=f(z-ia)$ for all $z \in \mathbb{C}$.
3) If $f(z) \in \mathbb{R}$ for all $z \in \mathbb{R} \cup \{z | Im z=a\}$ for some $a>0$, then $f(z+2ia)=f(z)$ for all $z \in \mathbb{C}$.
4) If $f(z) \in \mathbb{R}$ for all $z \in \mathbb{R} \cup \{z | Im z=a\}$ for some $a>0$, then $f(z+ia)=f(z)$ for all $z \in \mathbb{C}$.
option 1) is true as $g(z)=f(z)$ for all $z \in \mathbb{R}$ and therefore by identity theorem $f=g$.
In option 2) I get $g(z)=f(\bar{z})$ but I don't know how to proceed. Someone please help. Thanks
$\textbf{EDIT:}$ In option 2) since $f$ and $g$ agree on $\mathbb{R}$ $f=g \ \forall z \in \mathbb{C}$.
i.e. $f(z)=g(z)=\overline{f(\bar{z})}$
Therefore for $z \in \mathbb{R}$ $f(z+ai)=g(z+ai)=\overline{f(\overline{(z+ai)})}=f(z-ai)$
Therefore by identity theorem $f=g$ for all $z \in \mathbb{C}$
Replacing $z$ by $z+ai$ in option 2) gives option 3)
For 2) consider the function $f_1(z)=f(z+ia)$ and apply 1).
3) follows immediately from 2) by changing $z$ to $z+ia$.
Answer for 4): $f(z)=e^{z}, a=\pi$ gives a counterxample.
[Details for 2): let $f_1(z)=f(z+ia)$ and $g_1(z)=\overline {f_1(\bar{z})}$. If $z \in \mathbb R$ then the imaginary part of $z+ia$ is $a$. Hence $f_1(z) \in \mathbb R$. By 1) this gives $f_1=g_1$ at every point. In particular $f(z+ia)=\overline{ f(z-ia)}$. Now if $z$ is real then the left side of this equation is real so we can take complex conjugate on both sides to get $f(z+ia)=f(z-ia)$ for $z$ real. However $f$ is an entire function and if two entire functions agree on the real line they agree on $\mathbb C$. Hence $f(z+ia)=f(z-ia)$ for all $z \in \mathbb C$.