# Show $f(z)$ can be analytically continued and $F(z+4)=F(z)$ for resulting entire function

I'm working on some past qualifying exam problems in complex analysis and I'm quite stuck on this one:

Let $f(z)$ be analytic in $\{z\in\mathbb{C}\,:\,|\text{Re }z|<1\}$ and continuous on the closure of that domain. Suppose that $f(z)$ is real on the lines $\text{Re }z=\pm 1$. Prove that then $f(z)$ can be analytically continued to the whole plane and that the resulting entire function satisfies $F(z+4)=F(z)$ for all $z\in\mathbb{C}$.

Here are my thoughts:

• the problem is clearly (I think) calling for the Schwarz Reflection Principle but I don't quite see how the $F(z+4)=F(z)$ comes out. I assume it will have to do since it can be infinitely reflected left and right throughout the plane,
• With my understanding of the Schwarz reflection principle, I believe $f$ would extend to $\{z\in\mathbb{C}\,:\,-1<\text{Re }z<3\}$ with $\overline{F(\overline{1-z}+1)}=F(z)$ (since $z\mapsto \overline{1-z}+1$ is the reflection about the line $\text{Re }z=1$ and $z\mapsto\overline{z}$ is the usual reflection about the real axis)

TL;DR I understand how it can be analytically continued to an entire function, just not that the resulting entire function satisfies $F(z+4)=F(z)$. Any help is greatly appreciated. Thanks in advance.

You're right that the Schwarz reflection principle is applicable here. We start with $f$ analytic on the strip $\{-1<\operatorname{Re} z<1\}$ and then starting with the right side, we "flip" $f$ over about the line $\operatorname{Re}z=1$. That is, we define $f$ analytically in $\{1\leq\operatorname{Re}z\leq3\}$ by $f(z)=\overline{f(\overline{1-z}+1)}$ as you noted. A nice property of this "flipping" is that we end up with the same values on $\operatorname{Re}z=3$ as on $\operatorname{Re}z=-1$: $$f(3+bi)=\overline{f(2-\overline{3+bi})}=\overline{f(-1+bi)}=f(-1+bi)$$ since $f$ is real-valued on $\{\operatorname{Re}z=-1\}$. Now we have an analytic extension of $f$ to $\{-1\leq\operatorname{Re}z\leq 3\}$, and $f$ takes on the same values at the boundary lines. Therefore, we can from here extend $f$ periodically to an entire function by defining $f(z+4)=f(z)$; this extension is well-defined since $f(z+4)=f(z)$ at the boundary lines, analytic outside the lines $\{\operatorname{Re}z=4n-1:n\in\mathbb{Z}\}$ by induction, and continuous at the boundaries, so Morera's theorem shows that $f$ is an entire function. This sort of procedure can also be used to extend Jacobi's elliptic function $\operatorname{sn}(z)$, the inverse of the elliptic integral $$\int_0^z\frac{d\zeta}{\sqrt{({1-\zeta^2})({1-k^2\zeta^2})}},$$ to a doubly periodic meromorphic function in $\mathbb{C}$.