If $H$ is a Hilbert space then integral version of polarisation identity. If $H$ is a Hilbert space, then $\langle x, y\rangle = \frac{1}{2\pi}\int_0^{2\pi}e^{\imath\theta}\lVert x + e^{\imath\theta}y\rVert^2 d\theta$.
I have no idea where to start. So far I had only little troubles solving problems in functional analysis, but now I'm stuck, so I guess I'm just missing something, because it should be quite easy problem.
 A: I'll just write out Daniel Fischer's comment:
\begin{align}
\frac{1}{2\pi}\int_{0}^{2\pi}e^{i\theta}\| x+e^{i\theta}y\|^{2}d\theta &=  \frac{1}{2\pi}\int_{0}^{2\pi}e^{i\theta}\langle x+e^{i\theta}y,x+e^{i\theta}y\rangle d\theta \\
&= \frac{1}{2\pi}\int_{0}^{2\pi}e^{i\theta}\left(\|x\|^{2}+2 \text{Re}\langle x,e^{i\theta}y\rangle+\|y\|^{2}  \right)d\theta \\
&=\frac{1}{2\pi}\int_{0}^{2\pi}2\left(\text{Re}\langle x,y \rangle e^{i\theta}\cos \theta + \text{Im}\langle x,y \rangle e^{i\theta}\sin \theta \right)d\theta \\
&=\frac{1}{2\pi}\left(\text{Re}\langle x,y \rangle\int_{0}^{2\pi}2e^{i\theta}\cos\theta d\theta+\text{Im}\langle x,y \rangle \int_{0}^{2\pi}2e^{i\theta}\sin\theta d\theta \right) \\
&=\frac{1}{2\pi}\left(2\pi\text{Re}\langle x,y \rangle +2i\pi \text{Im}\langle x,y \rangle\right) \\
&=\langle x,y \rangle.
\end{align}
A: Here's another method:
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}e^{i\theta}\|x+e^{i\theta}y\|^2d\theta
&=\frac{1}{2\pi}\int_0^{2\pi}e^{i\theta}\left(\|x\|^2+e^{-i\theta}\langle x,y\rangle+e^{i\theta}\langle y,x\rangle+\|y\|^2\right)d\theta\\
&=\langle x,y\rangle+\frac{1}{2\pi}\left(\|x\|^2+\|y\|^2\right)\int_0^{2\pi}e^{i\theta}d\theta+\frac{1}{2\pi}\langle y,x\rangle\int_0^{2\pi}e^{2i\theta} d\theta\\
&=\langle x,y\rangle.
\end{align*}
