Let V be an inner product space over F. Prove the polar identities: for al $x,y$ $\in V, <x,y>=\frac{1}{4} \sum_{k=1}^4 I^k ||x+i^k y||^2$ if $F=C$,where $i^2=-1$.
I am trying to break the right hand side individually, but have some trouble with operations.
For the first one and the second $$1). i||x+iy||^2=i <x+iy,x+iy>=i <x,x>+i<iy,x>+i<x,iy>+i<iy,iy>=i||x^2||+-<y,x>+...$$ $$2). -||x-y||^2=||x||^2-||y||^2+...$$
question: I am not sure how to break open the brackets, need some help with operations. I will then try to do the rest by my self. I am not so used to the conjugate symmetry thing so have hard time dealing with imaginary i.
I will use the convention that $\langle ax,y \rangle = a\langle x,y\rangle$, so that $\langle \cdot,\cdot \rangle$ is linear in the first slot (and conjugate-linear in the second).
Note that $$ \|x + i^k y\|^2 = \langle x + i^ky, x + i^ky \rangle \\ = \langle x, x \rangle + \langle x,i^ky \rangle + \langle i^k y, x \rangle + \langle i^ky, i^k y \rangle\\ = \langle x,x \rangle + i^{-k}\langle x,y \rangle + i^k\langle y,x \rangle + \langle y,y\rangle. $$ Multiplying by $i^{k}$ gives us $$ i^{k}\|x + i^k y\|^2 = i^k\langle x,x \rangle + \langle x,y \rangle + i^{2k}\langle y,x \rangle + i^k\langle y,y\rangle. $$