Let $\langle x,y\rangle=\displaystyle\frac{1}{4}\sum\limits_{k=1}^4 i^k\lVert x+i^ky\rVert^2 $ where $F=\mathbb C$ and $i^2=-1$.
Proof: We have $$ \begin{align} \lVert x+i^ky\rVert^2 &= \langle x+i^ky,x+i^ky\rangle \\ &= \langle x,x\rangle + \langle i^ky,x\rangle + \langle x,i^ky\rangle + \langle i^ky,i^ky\rangle \\ &= \langle x,x\rangle + \overline{\langle x,i^ky\rangle} + \langle x,i^ky\rangle + \langle i^ky,i^ky\rangle \\ &= \langle x,x\rangle + 2Re \langle x,i^ky\rangle + \langle i^ky,i^ky\rangle \\ &= \lVert x\rVert^2 + 2Re \langle x,i^ky\rangle + \langle i^ky,i^ky\rangle \\ &= \lVert x\rVert^2 + 2Re[\overline{ i^k}\langle x,y\rangle] + \lVert i^ky\rVert^2 \end{align} $$
From here I'm not getting how to move on. Please guide me.
We have \begin{align} \frac{1}{4}\sum_{k=1}^4i^k\Vert x+i^ky\Vert^2 &=\frac{1}{4}\left(i\Vert x+iy\Vert^2-\Vert x-y\Vert^2 -i\Vert x-iy\Vert^2+\Vert x+y\Vert^2\right)\\ &=\frac{i}{4}\left(\Vert x+iy\Vert^2-\Vert x-iy\Vert^2\right) +\frac{1}{4}\left(\Vert x+y\Vert^2-\Vert x-y\Vert^2\right)\\ &=\frac{i}{4}\cdot2\left(\langle x,iy\rangle+\langle iy,x\rangle\right) +\frac{1}{4}\cdot2\left(\langle x,y\rangle+\langle y,x\rangle\right)\\ &=-\frac{1}{4}\cdot2\left(-\langle x,y\rangle+\langle y,x\rangle\right) +\frac{1}{4}\cdot4\cdot{\rm Re}\langle x,y\rangle\\ &=-\frac{1}{4}\cdot2\left(-2\cdot{\rm Im}\langle x,y\rangle\right) +{\rm Re}\langle x,y\rangle\\ &={\rm Im}\langle x,y\rangle+{\rm Re}\langle x,y\rangle\\ &=\langle x,y\rangle. \end{align}