I can't answer my own question, as Igor Khavkine points out, it may be too hard. For the paper I'm writing, I was able to find a suitable "square root" to my analytic function. This is what I'm providing here, in case anyone else needs it. Recall that a unit is a nonzero holomorphic function. Also recall that a Weierstrass polynomial of degree $m$ is a function of the form
$$p(x,y) = x^m + \sum_{j=0}^{m-1} a_j(y)x^j,$$
where $x \in \mathbb{C}$ and each $a_j(y)$ is holomorphic in $y$.
Theorem. Let $U \subset \mathbb{C}^n$ be a domain, and let $f(z)$ be holomorphic on $U$. Let $K \subset U \cap \mathbb{R}^n$ be compact in $U$. Write $z = (x,y)$ with $x \in \mathbb{C}$. Assume that, if $z \in \mathbb{R}^n \cap U$ then $f(z) \geq 0$. For every $z_0 \in K$ there is a neighborhood $B$ of $z_0$ and a Weierstrass polynomial $p(z)$ and a unit $u(z)$ on $U$ such that $f(x,y) = p(x,y)\;\overline{p(\bar{x},y}\,) \; u^2(x,y)$ on $B$.
Proof.
By the Weierstrass preparation theorem, $f = qv$ on a neighborhood $B$ of $z_0$, where $q$ is a Weierstrass polynomial, and $v$ is a unit. Since $v$ is a unit, shrinking $B$ if necessary, the range of $v$ lies in some half-space that excludes the origin, and then $u = \sqrt{v}$ is holomorphic on $B$. Furthermore, one looks up in a book the form of the Weierstrass polynomial, to find that it is given by
$$
q(x,y) = \prod_{j=1}^m (x-\alpha_j(y)),
$$
where the $\alpha_j$ are roots of $x \to f(x,y)$ on $B$. Because $f \geq 0$, the coefficients of the Taylor series are all real, and the complex roots must occur in conjugate pairs, and also the real roots must have even multiplicities. Thus we may rewrite
$$
q(x,y) =
\prod_{j=1}^{m/2} (x-\beta_j(y))(x-\bar{\beta}_j(y))
,
$$
e.g with $\Im \beta_j \geq 0$. Finally, put
$$
p(x,y) = \prod_{j=1}^{m/2} (x-\beta_j(y)).
$$
