Cauchy-Schwarz Inequality/Complex Numbers

Theorem

 * $\displaystyle \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$

where all of $w_i, z_i \in \C$.

Proof
Let $w_1, w_2, \ldots, w_n$ and $z_1, z_2, \ldots, z_n$ be arbitrary complex numbers.

Take the Binet-Cauchy Identity:
 * $\displaystyle \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$

and set $a_i = w_i, b_j = \overline {z_j}, c_i = \overline {w_i}, d_j = z_j $.

This gives us:

Hence the result.