I am trying to generalise the statement for $n$ complex numbers: For any complex numbers $a,b,c$ with property $|a|=|b|=|c|=r\neq 0$. Prove

$|\frac{ab+bc+ca}{a+b+c}|=r$ 

I proved this by showing $|\frac{\alpha}{\beta}|^2$ = $\frac{\alpha\alpha^\*}{\beta\beta^\*}=r^2$  

where $\alpha= ab+bc+ca$,  

$\beta=a+b+c$  

and $\alpha^\*$ and $\beta^\*$ are the conjugates.

How can I generalise this for $n$ complex numbers instead of three?

My attempt: For $n$ complex number, we can write the statement as: 

$|\frac{\frac{1}{2}\sum\_{i\neq j}^n a\_ia\_j}{\sum\_{i=1}^n a\_i}| =r $ 

which is the same as proving: 

$\frac{(\frac{1}{2}\sum\_{i\neq j}^n a\_ia\_j)\*(\frac{1}{2}\sum\_{i\neq j}^n a\_i^\*a\_j^\*)}{(\sum\_{i=1}^n a\_i)(\sum\_{i=1}^n a\_i^\*)}$

I'm stuck on how to now proceed.