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Let $f:\mathbb C \rightarrow \mathbb C$ be holomorphic and $f(z)=f(-z)$ for all $z\in \mathbb C$. Show that there exists a holomorphic function $g$ such that $g(z^2)=f(z)$.

If I take $g(z):=(f(z)+f(-z))/2$ then I can prove that $g(z)=\sum_0^\infty a_{2n}z^n$. Of course $g$ thus defined is holomorphic. But how does this show that $g(^2)=f(z)$ as well?. I mean the coefficients of the two are different when you compare term by term. Can you guys help?. Or should this $g$ be defined differently? Thanks for your help.

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up vote 4 down vote accepted

If $f$ is even, the Maclaurin series of $f$ contains only even powers. (Fill in the details of this!)

Hence, $$ f(z) = a_0 + a_2z^2 + a_4z^4 + a_6z^6 +\cdots = a_0 + a_2(z^2) + a_4(z^2)^2 + a_6(z^2)^3 + \cdots$$

Set $$g(z) = a_0 + a_2z + a_4z^2 + a_6z^3 + \cdots$$

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Thanks I got it – Jack Dawkins Jan 5 '13 at 9:03

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