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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:

$g\big(\mbox{Real}(z)\big) \Big( \cos(Im(z))+i \sin(\mbox{Im}(z)) \Big)$

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

Yonatan
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  • Any constant function is both holomorphic and periodic... do you mean for your question to exclude constant functions? – Alex Wertheim Jun 09 '13 at 20:18
  • @AWertheim : Maybe those have $e^z$ in them: $c(e^z)^0$. – Michael Hardy Jun 09 '13 at 20:26
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    "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 09 '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 $g(z) = g(z)(e^{z})^{0}$. – Alex Wertheim Jun 09 '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 $\sum_n\exp(-(x-n)^2)$. http://math.stackexchange.com/questions/81411/are-sin-and-cos-the-only-continuous-and-infinitely-differentiable-periodic-funct – Michael Hardy Jun 09 '13 at 21:33
  • "Does not involve exp" meaning that we can express this function without using the exponent function at all. – Yonatan Jun 10 '13 at 13:40

2 Answers2

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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 $b < a$.

mrf
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  • 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
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My first thought was that Jacobi's and Weierstrass' elliptic functions are doubly periodic, but then I thought you might object to their poles.

How about this: $$ \sum_{n\in\mathbb Z} \exp\left(-(z-n)^2\right). $$ That has $\exp((z-n)^2)$ rather than $\exp((z-\text{whatever})^1)$.

Michael Hardy
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