# periodic holomorphic function other than the exponent function

Is there a function f, in the complex plane, such that f(z) is both analytic and periodic and doesn't involve e^z in it? I've tried:

g(Real(z)) ( cos(Im(z))+i sin(Im(z)) )

For function of real number g, but haven't been able to find an appropriate g.

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Any constant function is both holomorphic and periodic... do you mean for your question to exclude constant functions? –  AWertheim Jun 9 '13 at 20:18
@AWertheim : Maybe those have $e^z$ in them: $c(e^z)^0$. –  Michael Hardy Jun 9 '13 at 20:26
"Does not involve" is a bit vague. For example, every periodic entire function $f$ with period $2\pi i$ can be written as $f(z) = g(e^z)$ where $g$ is holomorphic on $\mathbb{C} \setminus \{0\}$. Does this involve $e^z$? –  WimC Jun 9 '13 at 20:29
@MichaelHardy, fair enough, though I figured that was a trivial "inclusion"... if there is a $g(z)$ satisfying his query, then it could be immediately be disqualified by noting that $g(z) = g(z)(e^{z})^{0}$. –  AWertheim Jun 9 '13 at 20:30
Role reversal: Last time someone asked this question, I'm the one who wrote "In one sense, the answer is 'no'." and pointed out Fourier series, and someone else mentioned $\sum_n\exp(-(x-n)^2)$. math.stackexchange.com/questions/81411/… –  Michael Hardy Jun 9 '13 at 21:33

In a certain sense there is none. Assume that $f$ is analytic and $2\pi$-periodic (the period doesn't really matter, it's just to simplify the notation), defined on some horizontal strip $|\operatorname{Im} z| < a$. Then $f$ can be written as a Fourier series $$f(z) = \sum_{n=-\infty}^\infty c_n e^{inz}$$ with uniform convergence on $|\operatorname{Im} z| \le b$ for every $b < a$.
Thanks, When we consider an expression such as $e^z$ the problem is that $|e^z|$ may grow large. As a result it may be hard to estimate the integral of a function that involves $e^z$ on ,say, a circular domain with sufficiently large radius. –  Yonatan Jun 12 '13 at 19:03
How about this: $$\sum_{n\in\mathbb Z} \exp\left(-(z-n)^2\right).$$ That has $\exp((z-n)^2)$ rather than $\exp((z-\text{whatever})^1)$.