If $x,y \in \mathbb{C}^m$ prove
$$x^*y = \|x\|_2\|y\|_2 \cos \alpha$$
where $x^*y$ denotes the inner product of two vectors, and alpha is the angle between $x$ and $y$.
Attempt
\begin{align} LHS = \left(x^*y\right)^2 &= \left(\bar x_1y_1 + \bar x_2y_2 + \cdots + \bar x_my_m \right)^2 \\ &= (\bar x_1y_1)^2 + \cdots + (\bar x_my_m)^2+ 2\sum_{j = 1}^m\sum_{i \neq j}^m \bar x_iy_i\bar x_j y_j \dots \end{align}
For more clarity: (2.3)
