Proposition 12.6.2 (Polarisation Identity).label Let $E$ be a vector space over $\complex$, $\lambda: E \times E \to \complex$ be a sesquilinear form, and
\[q: E \to K \quad x \mapsto \lambda(z, z)\]
be its quadratic form, then for any $x, y \in E$,
\begin{align*}\lambda(x, y)&= \frac{1}{4}\braks{q(x + y) - q(x - y) + iq(x + iy) - iq(x - iy)}\\&= \frac{1}{4}\sum_{k = 0}^{3} i^{k} q(x + i^{k}y)\end{align*}
Proof. On the real side,
\begin{align*}q(x + y)&= \lambda(x, x) + \lambda(x, y) + \lambda(y, x) + \lambda(y, y) \\ q(x - y)&= \lambda(x, x) - \lambda(x, y) - \lambda(y, x) + \lambda(y, y) \\ q(x + y) - q(x - y)&= 2[\lambda(x, y) + \lambda(y, x)]\end{align*}
On the imaginary side,
\begin{align*}q(x + iy)&= \lambda(x, x) - i\lambda(x, y) + i\lambda(y, x) + \lambda(y, y) \\ q(x - iy)&= \lambda(x, x) + i\lambda(x, y) - i\lambda(y, x) + \lambda(y, y) \\ q(x + iy) - q(x - iy)&= -2i[\lambda(x, y) - \lambda(y, x)] \\ iq(x + iy) - iq(x - iy)&= 2[\lambda(x, y) + \lambda(y, x)]\end{align*}
Therefore
\[\sum_{k = 0}^{3} i^{k} q(x + i^{k}y) = 4\lambda(x, y)\]
$\square$
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