Proof $f(u,v)=\frac{1}{4}(q(u+v)-q(u-v))+\frac{i}{4}(q(u+iv)-q(u-iv))$

Question

Let $q$ be the Hermitian quadratic form associated with a symmetric bilinear form $f$, on a vector space $V$ over the field $F$. Prove that $$f(u,v)=\frac{1}{4}(q(u+v)-q(u-v))+\frac{i}{4}(q(u+iv)-q(u-iv))$$

I have no idea how to start$?$ All I know by definition is

A mapping $q:V\rightarrow F$ is a quadratic form if $q(v)=f(v,v)$ for some symmetric bilinear form $f$ on $V$

Then if $1+1\neq 0$ in $F$, then bilinear form $f$ can be obtained from the quadractic form $q$ by the following polar form of $f$: $$f(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$$ But I don't think this help me to solve that proof. Any kind of help will be appreciated.
Thanks for your time. Thanks in advance .

Answer 1

You basically approach this by starting on the right side with the expressions for $q$. Since we have $q(v) = f(v,v)$ we can then start computing each of these quantities one at a time: \begin{eqnarray*} q(u+v) & = & f(u+v, u+v) \\ & = & f(u, u+v) + f(v,u+v) \\ & = & f(u,u) + f(u,v) + f(v,u) + f(v,v). \end{eqnarray*}

Repeat this with the others and collect like terms.