Suppose you have a real quadratic form on a complex vector space $\mathbb{C}^n$. As far as I understand, this is just a map $q:\mathbb{C}^n \to \mathbb{R}$ such that $q(\lambda v)=|\lambda|^2q(v)$ for all $\lambda \in \mathbb{C}$. How does one find the associated Hermitian matrix?
All I can find in searches online is that one should take the matrix to be $A \in \mathbb{C}^{n \times n}$ by $A_{ij}=q(e_i+e_j)-q(e_i)-q(e_j)$, but it seems this equation only works when the base field is $\mathbb{R}$ because this would automatically make $A$ a real matrix. Of course these are Hermitian matrices with complex entries! There must be some analogous equation for the complex setting.
Thanks to @user10354138 for the tipoff. The Wikipedia article on the polarization identity shows what to do. I'll just reproduce it here for wider distribution.
The associated Hermitian form $H:\mathbb{C}^n \times \mathbb{C}^n \to \mathbb{C}$ is given by $$H(x,y) = \frac{1}{4}\left( q(x+y)-q(x-y)+i\cdot q(x+iy)-i\cdot q(x-iy) \right)$$ if you want a form that it $\mathbb{C}$-linear in the first argument and conjugate-linear in the second.