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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an entire function such that the multiplicity of each of its zeros is even. Must there exist an entire $g$ such that $f(z) = g(z)^{2}$?

Progress

I thought about using the Weierstrass Factorization Theorem, but I can't seem to get that to work. In the Weierstrass factorization theorem of $f$ we have an infinite product $\prod_{k = 1}^{\infty}(1 - z/b_{k})^{p_{k}}e^{R_{k}(z)}$ where $p_{k}$ is even, $\{b_{k}\}$ are the zeros of $f$ listed without multiplicity and $R_{k}$ is a polynomial. Then how do I know that $\prod_{k = 1}^{\infty}(1 - z/b_{k})^{p_{k}/2}e^{R_{k}(z)/2}$ is entire?

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  • $\begingroup$ What are your thoughts? $\endgroup$ – Ahaan S. Rungta Aug 8 '14 at 22:20
  • $\begingroup$ ...Well let's suppose $f$ could be written as an infinite product of linear functions vanishing at each zero (I know this can't actually happen, nor does it make sense). Then $f(z) = \prod_{i = 1}^{\infty}(z - a_{i})^{2k}$, then we take $g(z) = \prod_{i = 1}^{\infty}(z - a_{i})^{k}$. This reasoning is incorrect, but I suppose it sort of justifies why it should be true. I thought about maybe using the Weierstrass Factorization Theorem, but I can't seem to get that to work. $\endgroup$ – user168041 Aug 8 '14 at 22:23
  • $\begingroup$ What problems have you with the Weierstraß factorization theorem? It should work fairly straightforwardly. $\endgroup$ – Daniel Fischer Aug 8 '14 at 22:26
  • $\begingroup$ @DanielFischer: In the Weierstrass factorization theorem of $f$ we have an infinite product $\prod_{k = 1}^{\infty}(1 - z/b_{k})^{p_{k}}e^{R_{k}(z)}$ where $p_{k}$ is even, $\{b_{k}\}$ are the zeros of $f$ listed without multiplicity and $R_{k}$ is a polynomial. Then how do I know that $\prod_{k = 1}^{\infty}(1 - z/b_{k})^{p_{k}/2}e^{R_{k}(z)/2}$ is entire? $\endgroup$ – user168041 Aug 8 '14 at 23:20
  • $\begingroup$ The latter product is entire because it converges uniformly on bounded subsets, and that is because the original product does. $\endgroup$ – user147263 Aug 8 '14 at 23:58
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On the one hand, from the product representation (we disregard $f \equiv 0$, where the existence of a square root is trivial)

$$f(z) = e^{a(z)} z^{2\nu_0} \prod \left(1-\tfrac{z}{b_k}\right)^{2\nu_k}e^{p_k(z)}\tag{1}$$

of $f$, we directly obtain what should be a product representation of a square root of $f$:

$$g(z) = e^{a(z)/2} z^{\nu_0} \prod\left(1-\tfrac{z}{b_k}\right)^{\nu_k} e^{p_k(z)/2},\tag{2}$$

and then it remains to see that the product in $(2)$ converges locally uniformly on all of $\mathbb{C}$. That is maybe a little tedious if done rigorously.

We can avoid that tedium if we use the Weierstraß product theorem in a different way:
There is an entire function $g_1$ that has zeros of order $\nu_k$ in the $b_k$ (with $b_0 = 0$ and possibly $\nu_0 = 0$), and no other zeros. Then the function

$$q(z) = \frac{f(z)}{g_1(z)^2}$$

is entire and has no zeros. Hence $q$ has a logarithm, $q(z) = e^{g_2(z)}$. Now it is clear that $g(z) = g_1(z) e^{g_2(z)/2}$ is an entire square root of $f$.

An alternative way to establish the existence of a square root of $f$ considers the logarithmic derivative

$$h(z) = \frac{f'(z)}{f(z)}.$$

$h$ is an entire meromorphic function, with simple poles in the zeros of $f$, where the residue in $b_k$ is $2\nu_k$, and no other poles. Therefore the residue of $\tilde{h} = \frac{1}{2}\cdot h$ in all poles is an integer, and hence

$$g(z) = \exp \left(c_0 + \int_{z_0}^z \tilde{h}(\zeta)\,d\zeta\right),$$

where

  • $z_0$ is an arbitrary point such that $f(z_0) \neq 0$,
  • $c_0$ is chosen so that $e^{2c_0} = f(z_0)$,
  • and the integral is over an arbitrary piecewise differentiable path from $z_0$ to $z$ that does not pass through any zero of $f$,

is a well-defined holomorphic function on $\mathbb{C}\setminus f^{-1}(\{0\})$ with $g(z)^2 \equiv f(z)$ there. From that, it follows that the zeros of $f$ are removable singularities of $g$, and the identity $g(z)^2 \equiv f(z)$ holds on all of $\mathbb{C}$ after removing the removable singularities.

Note: the same arguments work on any simply connected domain, but on a domain that is not simply connected, not all holomorphic functions with all zeros of even order have a square root. Also, the arguments, mutatis mutandis, also work for any $m$-th root of $f$ if the order of all zeros of $f$ is a multiple of $m$.

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