# Does $f(z+2\pi)=f(z)$ for all $z\in \mathbb{C}$?

If $f:\mathbb{C}\rightarrow\mathbb{C}$ is a differentiable function and $f(x+2\pi)=f(x)$ for all $x\in \mathbb{R}$, would $f(z+2\pi)=f(z)$ for all $z\in \mathbb{C}$?

Is there any theorem/lemma concerning this? Are there any examples/counter examples for this?

• If $f$ is holomorphic, then this works. But differentiability is too weak a condition (take for example $f(x+iy) = \sin((1-y)x)$). Commented May 25, 2012 at 21:47
• Do you have a proof? Commented May 25, 2012 at 21:48
• @Potato : If $f$ is holomorphic, then $g(z) = f(z+2\pi)$ also is. But since $g$ and $f$ are equal on $\mathbb{R}$, the identity theorem (en.wikipedia.org/wiki/Identity_theorem) tells us they are equal on $\mathbb{C}$. Commented May 25, 2012 at 21:52
• You should repost that below as an answer. Commented May 25, 2012 at 21:56
• @JoelCohen: This should really be an answer ... Commented May 25, 2012 at 21:56

If $f$ is holomorphic, then $g(z) = f(z+2\pi)$ also is. But since $f$ and $g$ are equal on $\mathbb{R}$, the identity theorem tells us they are equal on $\mathbb{C}$.
Now if you only assume differentiability as a function on $\mathbb{R}^2$, there are counter examples. Take for example $f(x+iy) = \sin(yx)$.
• Arguably, if the OP says that the domain of the function is $\mathbb C$, then differentiable means holomorphic. Commented May 25, 2012 at 22:13
• @Phira : You're right, this is probably what was implied. I added the last part just to avoid confusion with being differentiable as a function on $\mathbb{R}^2$. Commented May 25, 2012 at 22:28