Inner product and its conjugate

Question

How do I prove $\langle x,y \rangle$=$\overline{\langle y,x \rangle}$ using polarization identity

Answer 1

Since$$\langle x,\,y\rangle=\frac14(\Vert x+y\Vert^2-\Vert x-y\Vert^2+i\Vert x-iy\Vert^2-i\Vert x+iy\Vert^2),$$we have$$\overline{\langle y,\,x\rangle}=\frac14(\Vert x+y\Vert^2-\Vert y-x\Vert^2-i\Vert y-ix\Vert^2+i\Vert y+ix\Vert^2).$$Since $\Vert iz\Vert^2=\Vert z\Vert^2$,$$\overline{\langle y,\,x\rangle}=\frac14(\Vert x+y\Vert^2-\Vert x-y\Vert^2-i\Vert x+iy\Vert^2+i\Vert x-iy\Vert^2)=\langle x,\,y\rangle.$$