Cauchy - Schwarz for complex numbers 
Let $z_1, . . . , z_n$ and $w_1, . . . , w_n$ be complex numbers. Show
  that $$|z_1w_1 + ··· + z_n w_n|^2 ≤ \sum ^n _{j=1} |z_j|^2 \sum ^n
_{j=1}|w_j|^2$$

I basically tried to use the proof given for real numbers but I feel that something must be wrong. Could somebody look over the proof? 
Let $z=(z_1,...,z_n)$ and $w=(w_1,...,w_n)$. Then:
$$0 \leq \sum_j^n\bigg(||z|||w_j|-||w|||z_j|\bigg)^2=||z||^2||w||^2-2||z||||w||\sum_j^n|z_jw_j|+||z||^2||w||^2=2||z||||w||\bigg(||z||||w||-\sum_j^n|z_jw_j|\bigg)$$
But then
$$|\sum_j^nz_jw_j|\leq \sum_j^n|z_jw_j|\leq ||z||||w||$$
And just squaring is left.
Thanks!
 A: You have a canonical inner product in $\Bbb C^n$ given by $x\cdot y=\displaystyle \sum_{i=1}^n x_i\overline{y_i}$. Note $x\cdot x=\displaystyle \sum_{i=1}^n x_i\overline{x_i}=\displaystyle \sum_{i=1}^n |x_i|^2$
Claim Let $V$ be a $\Bbb C$-vector space, and $\langle \cdot,\cdot\rangle$ an inner product. If we define a norm $\lVert x\rVert :=\langle x,x\rangle^{1/2}$, then we always have $$|\langle x,y\rangle|^2\leqslant \lVert x\rVert ^2\lVert y\rVert^2$$
Proof If $x$ or $y$ are zero, the inequality is true, since $\langle x,0\rangle=0$ and $\lVert 0\rVert =0$. We may assume then that both norms are nonzero. Since $\langle z,z\rangle\geqslant 0$ for any choice of $z$, we may take $$z=x-\frac{\langle x,y\rangle y}{\lVert y\rVert ^2} $$
Using the inner product is (sesqui)linear we get 
$$\begin{align}
  \langle z,z\rangle  
   &= \langle {x,x} \rangle  - \frac{1}{\lVert y\rVert^2}\langle {x,\langle x,y \rangle y} \rangle  - \frac{1}{\lVert y\rVert^2}\langle {\langle x,y \rangle y,x} \rangle  + \frac{{\langle x,y \rangle \overline {\langle x,y \rangle}  }}{{\lVert y\rVert^2\lVert y\rVert^2}}\langle {y,y} \rangle  \\ 
   &= {\| x \|^2} - \frac{{| \langle x,y \rangle  |^2}}{\lVert y\rVert^2} - \frac{{| \langle x,y \rangle  |^2}}{\lVert y\rVert^2} + \frac{{| \langle x,y \rangle  |^2}}{\lVert y\rVert^2} \\
   &= {\| x \|^2} - \frac{{| \langle x,y \rangle  |^2}}{\| y \|^2} \geqslant 0\end{align}  $$
as we wanted. We made repeated use of $$\langle a+b,c\rangle=\langle a,c\rangle+\langle b,c\rangle\\  \langle a,b+c\rangle=\langle a,b\rangle+\langle a,c\rangle\\\langle \alpha a,b\rangle=\alpha\langle a,b\rangle\\\langle a,\beta b\rangle=\overline \beta\langle a,b\rangle \\z\overline z=|z|^2$$
A: Let $\omega = z_1w_1 + z_2w_2 + \cdots + z_nw_n$. In general, $|\omega|^2 = \omega \overline{\omega}$.
$$\omega \overline{\omega} = (z_1w_1+z_2w_2 + \cdots + z_nw_n)(\overline{z_1}\overline{w_1} + \cdots + \overline{z_n}\overline{w_n}) \\
= z_1\overline{z_1}w_1\overline{w_1} + z_1\overline{z_2}w_1\overline{w_2} \cdots + z_n\overline{z_n}w_n\overline{w_n} \\
 = |z_1|^2|w_1|^2 + \cdots + |z_n|^2 |w_1|^2 + \textrm{mixed terms}$$
now It should be clear from here.
