Cauchy-Schwarz Inequality/Complex Numbers

Theorem

 * $\displaystyle \left({\sum \left|{w_i}\right|^2}\right) \left({\sum \left|{z_i}\right|^2}\right) \ge \left|{\sum w_i z_i}\right|^2$

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:
 * $\displaystyle \left({\sum_{i=1}^n a_i c_i}\right) \left({\sum_{j=1}^n b_j d_j}\right) = \left({\sum_{i=1}^n a_i d_i}\right) \left({\sum_{j=1}^n b_j c_j}\right) + \sum_{1 \le i < j \le n} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)$

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.