# Examples of complex functions with infinitely many complex zeros

What are some examples of complex functions with infinitely many complex zeros? There are no particular restrictions on the functions I am just curious and having a hard time finding examples. Also what can be said about a complex function with infinitely many complex zeros, must they have any special properties?

There is nothing special about functions with infinitely many zeros. In fact, it is the norm.

If $f(z)$ is any entire function which isn't a polynomial, then $\infty$ is an essential signularity of $f(z)$. By Big Picard's theorem, $f(z)$ takes on all possible values of $\mathbb{C}$, with at most a single exception, infinitely often.

This means if your $f(z)$ is entire, not a polynomial with finitely many zeroes, then $f(z) + \alpha$ for any constant $\alpha \ne 0$ has infinitely many zeros.

• Thank you for the answer! I do not know much about complex functions and did not know about Big Picard's theorem, thank you for the link! – Twiltie Oct 27 '13 at 0:04
• The answer may not have explicit examples, but it helps me understand whats going on with complex functions in a more general way, for this reason I think it is a fitting answer. – Twiltie Oct 27 '13 at 0:07

Remember that the real numbers are a proper subset of the complex numbers and so infinitely many real zeros satisfies the criterion of infinitely many complex zeros. Hence: $$\operatorname{f}(z) := \sin z$$ If you want infinitely many non-real, complex zeros then note that $\cos(\operatorname{i}\!z) \equiv \cosh z$, and so $$\operatorname{g}(z) := \cosh(z)-1$$ has infinitely many non-real complex zeros: $z \in \{2\pi\operatorname{i}n : n \in \mathbb{Z}\}$

• Thank you for the example! I think I was a little of, I really should have asked about functions with infinitely many complex zeroes. I will edit my question. EDIT: Just saw your edit, that's the kind of example I was looking for thank you! – Twiltie Oct 26 '13 at 23:55
• @Twiltie My pleasure. I'm glad that I could help. – Fly by Night Oct 26 '13 at 23:57

$f(z)=0$ for all $z \in \mathbb C$

• Thanks for the answer! But I was hoping for something a little less trivial. – Twiltie Oct 26 '13 at 23:56