Let $K=\mathbb C((z))$ be the field of Laurent series in the variable $z$, and consider the involution on $K$ that sends $f(z)$ to $f(-z)$. A complex symmetric matrix of size $r$ over $K$ is a matrix $A(z)\in M\_r(K)$ such that $$^tA(z)=A(-z),$$ where $^tA(z)$ is the transpose (not conjugate transpose) matrix of $A(z)$.



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> Is it true that any complex symmetric matrix can be written in the form $^tB(z)\cdot B(-z)$? 
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Complex symmetric Matrices over the the field of Laurent series
The answer is yes -- this follows from the general theory of reduction of hermitian matrices.


The matrix $A$ such that $^tA(z) = A(-z)$ is an hermitian matrix in the terminology of Bourbaki, Algèbre, Chap. 9, $\S$ 3, n°1. The reduction theory ($\S$ 6, n°1, Cor. 2 in loc.cit.) tells you that there exists an invertible matrix $P \in \mathrm{GL}\_r(K)$ such that $A(z)= {}^tP(z) D(z) P(-z)$ with $D(z)=\operatorname{diag}(f\_1(z), \ldots , f\_m(z),0, \ldots, 0)$, where $f\_i$ are elements of $K$ such that $f\_i(z)=f\_i(-z)$ and $m$ is the rank of $A$.


It remains to check that any $f \in K$ fixed under the involution can be written $f(z)=g(z) g(-z)$ for some $g \in K$. Note that $f \in \mathbb{C}((z^2))$ and we may assume without loss of generality that $f(z)=1+\sum\_{n\geq 1} a\_n z^{2n}$. Using the Taylor expansion of the square root, we get $f(z) =g(z^2)^2$ for some $g \in K$, and thus $f(z)=g(z^2)g((-z)^2)$ as desired.


The proof of the reduction result is entirely similar to the proof for classical hermitian matrices. Letting $\phi$ be the hermitian form associated to $A$, the starting point is to find a vector $x \in K^r$ such that $\phi(x,x) \neq 0$, and then to proceed by induction on $r$. To find $x$, choose vectors $x\_0,y\_0 \in K^r$ such that $\phi(x\_0,y\_0) \neq 0$, then at least one of the vectors $x\_0,y\_0,x\_0+y\_0,x\_0+zy\_0$ will work. This provides an algorithm (but I don't claim it is optimal).


