If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$
I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that $$f(z) = \sum a_n z^n = \sum a_n (-z)^n$$
which implies that a_n does not depend on z and we can write
$h(z^2) = \sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.
It follows from your equality$$\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty a_n(-z)^n$$than $(\forall n\in\mathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=\sum_{n=0}^\infty a_{2n}z^{2n}=\sum_{n=0}^\infty a_{2n}(z^2)^n.$$So, define$$h(z)=\sum_{n=0}^\infty a_{2n}z^n.$$