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Suppose $f$ is analytic on ${\mathbb C}\setminus \{0\}$ and satisfies $f(z^{-1}) = f(z)$ for all nonzero $z$. Show that there exists an entire analytic function $g$ such that $f(z) = g(z + \frac{1}{z})$ for all nonzero $z$.

I thought maybe using either Cauchy's formula or Laurent series but I don't see why a Laurent series $\sum_{n \in {\mathbb Z}} a_n z^n$ necessarily has $a_n = a_{-n}$ or why it must converge.

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  • $\begingroup$ Are you sure you don't mean $f(z)=g(z)+g(1/z)$? $\endgroup$ – Alex R. Mar 1 '15 at 7:23
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Expand $f$ in a Laurent series with coefficients $a_n$ about $0$. The condition shows that $a_n=a_{-n}$.

Now use the fact that $a_n(z^n+1/z^n)$ can be written as a sum of powers of $(z+1/z)$.

For example, $$(z+1/z)^2=2+(z^2+1/z^2),$$ $$(z+1/z)^3=3(z+1/z)+(z^3+1/z^3),$$ and so on.

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  • $\begingroup$ Why does this imply $f(z)=g(z+1/z)$? $\endgroup$ – Alex R. Mar 1 '15 at 7:18
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    $\begingroup$ Are you sure? This gives $f(z) = g(z)+g(1/z)$, not $g(z+1/z)$. $\endgroup$ – marty cohen Mar 1 '15 at 7:18
  • $\begingroup$ @martycohen You're both right, but luckily it's fixable. $\endgroup$ – Potato Mar 1 '15 at 7:25

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