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there is an exercise in a book on complex analysis:

"Let $f$ is a holomorphic function on $\mathbb{C}$ such that exists $a>0$ with property $f(z) = f(z+a) = f(z+ja)$ for each $z \in \mathbb{C}$. Show, that f must be necessarily constant."

The hint is to use Liouville theorem: "Every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number $M$ such that $|f(z)| \le M$ for all $z \in \mathbb{C}$ is constant."

So to finish the exercise it suffices to show $f(z$) is bounded for all $z \in \mathbb{C}$. Could someone help me? Or miss i something trivial?

Thank you.

Bernard
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Ondra
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2 Answers2

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Your function is doubly periodic, it has a real and a purely imaginary period. So its range is just the range of its restriction to any $a\times a$ square on the complex plane, like $\{z: 0\le Re\, z\le a; 0\le Im\, z\le a\}$. It is therefore bounded and you can apply Liouville's Theorem.

GReyes
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Hint: Let $z=x+jy$ with $x,y \in \mathbb R$. There exist integers $n$ and $m$ such that $|x-na| \leq a$ and $|y-ma| \leq a$. Since $f(z)=f(x-na+j(y-ma))$ we see that $|f(z)|$ is bounded by the supremum of $|f|$ on $\{x+iy: |x|\leq a, |y| \leq a\}$. This proves that $f$ is bounded.

Existence of $n$ and $m$: The interval $(\frac x a -1,\frac x a +1)$ has at least one integer since its length exceeds $1$. If $n$ is in this interval then $|x-na| \leq a$. Similarly $m$ exists.

Kavi Rama Murthy
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