Decomposition of even entire function I am trying to find out how to decompose an even entire function $f(z) = h(z)h(-z)$ for $h$ some entire function. Some results I could think of are that $f(\sqrt z)$ is entire, $f(z)$ has a Laurent series with terms with event powers, etc. However, I still can't think of a way to find the product.
EDIT: I can't think of how to do this for $f(z) = z^2$ for instance . . .
 A: Using the Weierstrass factorisation theorem together with the fact that $f$ is even (so that its zeros come in pairs ($a_n$, $-a_n$) one can represent $f$ as
$$
 f(z) = z^{2m} e^{h(z)} \prod_{n=1}^\infty E_{p_n}\left( \frac{z}{a_n}\right)E_{p_n}\left( -\frac{z}{a_n}\right)
$$
where $h$ is an entire function, $(p_n)$ is a suitable sequence of integers, and $E_{p_n}$ are the so-called elementary factors. It also follows that $e^{h(z)} = e^{-h(z)}$ . Then
$$
 g(z) = (iz)^{m} e^{h(z)/2} \prod_{n=1}^\infty E_{p_n}\left( \frac{z}{a_n}\right)
$$
is an entire function satisfying $g(z)g(-z) = f(z)$.
A: $z^{4k}=z^{2k}(-z)^{2k}, z^{4k+2}=(iz)^{2k+1}(-iz)^{2k+1}$ so by factoring the even power of $z$ that divides $f$ we can assume wlog that $f$ has only non-zero roots which appear in pairs $a_n, -a_n$ (ordered by increasing modulus for example), while we fix $a_n$ the root st $\Re a_n >0$ or $\Re a_n =0, \Im a_n >0$
Consider now $u(z)=\Pi_{n\ge 1}E_{k_n}(z/a_n)E_{k_n}(-z/a_n)$ a converging Weierstrass product with roots $a_n, -a_n$ st the Weierstrass factors $E_{k_n}(w)=(1-w)\exp(w/1+w^2/2+..w^{k_n}/n)$ have same order $k_n$ for $a_n, -a_n$ we notice that also $q(z)=\Pi_{n\ge 1}E_{k_n}(z/a_n)$ is a converging Weierstrass product with roots $a_n$ (well distinguished from $-a_n$ by the above) and by definition $u(z)=q(z)q(-z)$
But now $f/u$ has no zeroes so $f/u=e^w$ for some entire function $w$ and clearly $e^{w(z)}=e^{w(-z)}$ so by continuity $w(z)-w(-z)=2k\pi i$ for some fixed $k$ and using $z=0$ we get that $k=0$ or $e^{w(z)}=e^{w(z)/2}e^{w(-z)/2}$ so taking $r(z)=e^{w(z)/2}$, on clearly has $r(z)r(-z)=e^{w(z)}$ and putting all together with $h=qr$:
$f(z)=u(z)e^{w(z)}=h(z)h(-z)$ hence we are done by putting back the even powers of $z$ as in the first line
