If $f(z) = f(\frac{1}{z})$ then $f(z) = g(z + \frac{1}{z})$ 
Let $f$ be a holomorphic function on $\mathbb{C} \backslash \{0\}$ such that $f(z) = f(\frac{1}{z})$ for all $z \ne 0.$
Show that there exists a holomorphic function $g$ on $\mathbb{C}$ such that
$$f(z) = g(z+ \frac{1}{z})$$

Attempt: Consider the Laurent expansion of $f$
$$f(z) = \sum_{n \in \mathbb{Z}} a_n z^n$$
From hypothesis we get $a_n = a_{-n}$ for all $n$. Now we group the terms and get $$f(z) = a_0 + \sum_{n>0} a_n( z^n + \frac{1}{z^n}).$$
We can use the fact that $z^n + \frac{1}{z^n}$ is a polynomial of degree $n$ in $z+ \frac{1}{z}$. If the expansion of $f$ has finitely many terms we get $g$ as a polynomial. But in general, each term $(z+\frac{1}{z})^k$ can appear in the expansion of infinitely many $z^n + \frac{1}{z^n},$ with $n\ge k,$ so there are some issues of convergence.
Moreover, it's not obvious that the formal series thus obtained for $g$ is convergent.
 A: The function $z\mapsto z+\frac1z$ maps the exterior of $S^1$ to the complement of $[-2,2]$. There is a holomorphic inverse $h\colon\Bbb C\setminus[-2,2]\to \Bbb C\setminus \overline D$ (you can even write it down by solving a quadratic). If we let $g(z)=f(h(z))$, then $g(z+\frac1z)=f(z)$ at least for $|z|>1$. By the condition on $f$, this also holds for $0<|z|<1$. Can you see why it also holds for $|z|=1?
A: Let $\Omega_1$ be obtained from removing the imaginary axis and the segment $[-2,2]$ from the plane. Let $\Omega_2$ be obtained by removing the segments $(-\infty,2]$ and $[2,\infty)$ from the plane. Using obvious analytic branches $A_1$ and $A_2$ of square roots define $g_j(z)=f((z+A_j(z^2-4))/2$, $j=1,2$. The union of $\Omega_1$ abd $\Omega_2$ is the entire plane. $g_1(z)$ and $g_2(z)$ are both analytic on the intersection of the two domains and they coincide on ${1+1/n+i/n}$, n=1,2.... Hence they coincide on the common domain. This gives a  consistently defined analytic function $g$ on the entire plane and the required formula connecting this with $f$ holds.
