Showing that $E\left[\left|\sum_{j=1}^{N} e^{i \phi_j}\right|^{2}\right]=N$ I believe
$$E\left[\left|\sum_{j=1}^{N} e^{i \phi_j}\right|^{2}\right]=N$$
where $E$ denotes the expectation value and where the $\phi_j$'s are mutually independent and uniformly distributed in $[0,2\pi)$. However, I am not sure how to show that.
 A: $$\Bbb{E}\left[\left|\sum_{j=1}^Ne^{i\phi_j}\right|^2\right]=\Bbb{E}\left[\sum_{j=1}^n\sum_{k=1}^n(\cos\phi_j\cos\phi_k+\sin\phi_j\sin\phi_k)\right]=\Bbb{E}\left[N+\sum_{1\le j<k\le n}(\cos\phi_j\cos\phi_k+\sin\phi_j\sin\phi_k)\right]$$
For $j\ne k$,
$$\Bbb{E}(\cos\phi_j\cos\phi_k)=\int_0^{2\pi}\int_0^{2\pi}\cos x\cos y\frac{1}{2\pi}\frac{1}{2\pi} dxdy=0$$
$$\Bbb{E}(\sin\phi_j\sin\phi_k)=\int_0^{2\pi}\int_0^{2\pi}\sin x\sin y\frac{1}{2\pi}\frac{1}{2\pi} dxdy=0$$
A: By linearity of expectation,
$$\mathsf{E}\left|\sum_{j=1}^N e^{i\phi_j}\right|^2=\mathsf{E}\left\langle\sum_{j=1}^N e^{i\phi_j},\sum_{k=1}^N e^{i\phi_k}\right\rangle=\mathsf{E}\sum_{j=1}^N \sum_{k=1}^N\left\langle e^{i\phi_j}, e^{i\phi_k}\right\rangle=\sum_{j=1}^N \sum_{k=1}^N \mathsf{E} \langle e^{i\phi_j},e^{i\phi_k}\rangle.$$
You get an $N$ factor from the terms corresponding to $j=k$. When $j\neq k$, the inner term in the sum is simply the expected inner product between two independent and uniform points on the circle -- which is zero, as once you fixed one vector, the other one has equal probability to be to the left or right.
