# Polarization identity

If $\lVert \cdot \rVert$ is a norm on a linear space $X$ which satisfies the parallelogram law, then the polarization identity defined as $4 \langle x,y\rangle =\lVert x+y\rVert^2-\lVert x-y\rVert^2+i\lVert x+iy\rVert^2-i\lVert x-iy\rVert^2$ gives an inner product on $X$.

I am having difficulty with prooving $\langle ix,y \rangle =i \langle x,y\rangle$, for $i=\sqrt{-1}$.

Recall $\lVert \lambda x \rVert^2 = \lvert \lambda \rvert^2 \lVert x \rVert^2$. Since $\lvert i \rvert = 1$, you have $$4 \langle ix,y\rangle =\lVert ix+y\rVert^2-\lVert ix-y\rVert^2+i\lVert ix+iy\rVert^2-i\lVert ix-iy\rVert^2$$ $$= \lVert x-iy\rVert^2-\lVert x-iy\rVert^2+i\lVert x+y\rVert^2-i\lVert x-y\rVert^2 =$$ $$= i\left(-i\lVert x-iy\rVert^2+i\lVert x-iy\rVert^2+\lVert x+y\rVert^2-\lVert x-y\rVert^2 \right) = 4i\langle x,y\rangle.$$