Identity with complex numbers related to the Cauchy-Schwarz inequality I have this equation 
$ a_j,b_j\in \mathbb{C} , j=1,2,...,n$
$$  
\left| \sum\limits_{j=1}^n a_jb_j  \right|^2 = \sum\limits_{j=1}^n |a_j|^2 \sum\limits_{j=1}^n |b_j|^2
-\sum_{1\leq i \leq j \leq n} {| a_i \overline{b_j} -a_j \overline{b_i}|}^2 $$
I just want to know about the last sum which i and j we use ? 
If it is possible tell me this with an example
 A: Following KittyL's example:
Explicit Example Let $n=2$ (so we only have to focus on four terms for simplicity and demonstration)
Let $a_1=i,\quad a_2=2i, \quad$
$b_1=3,\quad b_2=4$
Then we just plug everything into the formula as shown:
First we calculate
$$\left|{\sum_{j=1}^{2} a_j b_j}\right|^2 = \left| i\cdot3+2i\cdot 4 \right|^2=\left| 3i +8i \right|^2=\left| 11i \right|^2=121
$$
So this is the left hand side.
Now the right hand side we will take term by term:
$$\sum_{j=1}^{2}\left| a_j\right|^2=\left| i\right|^2 + \left| 2i\right|^2=1+4=5
$$
$$\sum_{j=1}^{2}\left| b_j\right|^2=\left|3 \right|^2+\left|4 \right|^2=9+16=25
$$
So
$$\sum_{j=1}^{2}\left| a_j\right|^2\sum_{j=1}^{2}\left| b_j\right|^2=125
$$
Now the final term is
$$\sum_{1\leq i \leq j \leq n} \left| a_i \bar{b_j}-a_j \bar{b_i} \right|^2
$$
$$=\left| a_1 \bar{b_1}-a_1 \bar{b_1} \right|^2 +\left| a_1 \bar{b_2}-a_2 \bar{b_1} \right|^2 +\left| a_2 \bar{b_2}-a_2 \bar{b_2} \right|^2
$$
(remember we only consider terms where $i\lneq j$, otherwise there would be another term)
$$=0+\left| 4i-6i \right|^2+0=4
$$
So the RHS reads:
$$\sum_{j=1}^{2}\left| a_j\right|^2 \sum_{j=1}^{2}\left| b_j\right|^2-\sum_{1\leq i \leq j \leq n} \left| a_i \bar{b_j}-a_j \bar{b_i} \right|^2=125-4=121
$$
As required
A: For example, let $n=4$. The last term is
$$|a_1\bar{b_2}-a_2\bar{b_1}|^2+|a_1\bar{b_3}-a_3\bar{b_1}|^2+|a_1\bar{b_4}-a_4\bar{b_1}|^2\\+|a_2\bar{b_3}-a_3\bar{b_2}|^2+|a_2\bar{b_4}-a_4\bar{b_2}|^2+|a_3\bar{b_4}-a_4\bar{b_3}|^2$$
I didn't include the terms where $i=j$ since they would be zeros. You can see these include all the terms where $i\ne j$ and $i<j$ ensures no repetition. 
