How to generalise this complex equation? 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.
 A: I don't think your generalized statement is correct.  In a very simplified example, let a=b=c=1 and d=-1.  We would have $$|{ab+ac+ad+bc+bd+cd\over a+b+c+d}|={0\over2}\ne r$$
I believe if you write the statement as $$|{abc+abd+acd+bcd\over a+b+c+d}|$$ that would be true (although this still needs proof for all values of n).
In other words, instead of the numerator being all possible combintions of two different factors, it should be all possible combinations of n-1 different factors
A: Consider to rewrite your starting point as follows:$$
\eqalign{
  & N_{\,2,\,3} (a_{\,1} ,a_{\,2} ,a_{\,3} ) = a_{\,1} \,a_{\,2}  + a_{\,1} \,a_{\,3}  + a_{\,2} \,a_{\,3}  =   \cr 
  &  = \left( {a_{\,1} \,a_{\,2} \,a_{\,3} } \right)\left( {{1 \over {a_{\,1} }} + {1 \over {a_{\,2} }} + {1 \over {a_{\,3} }}} \right) =   \cr 
  &  = r^{\,2} e^{\,i\,\alpha _{\,1} } e^{\,i\,\alpha _{\,2} }  + r^{\,2} e^{\,i\,\alpha _{\,1} } e^{\,i\,\alpha _{\,3} }  + r^{\,2} e^{\,i\,\alpha _{\,2} } e^{\,i\,\alpha _{\,3} }  =   \cr 
  &  = r^{\,2} \left( {e^{\,i\,\left( {\alpha _{\,1}  + \alpha _{\,2} } \right)}  + e^{\,i\,\left( {\alpha _{\,1}  + \alpha _{\,3} } \right)}  + e^{\,i\,\left( {\alpha _{\,2}  + \alpha _{\,3} } \right)} } \right) =   \cr 
  &  = r^{\,2} e^{\,i\,\left( {\alpha _{\,1}  + \alpha _{\,2}  + \alpha _{\,3} } \right)} \left( {e^{\, - i\,\alpha _{\,1} }  + e^{\, - i\,\alpha _{\,2} }  + e^{\, - i\,\alpha _{\,2} } } \right) =   \cr 
  &  = r^{\,2} e^{\,i\,\left( {\alpha _{\,1}  + \alpha _{\,2}  + \alpha _{\,3} } \right)} e^{\,i\,\pi } \left( {e^{\,i\,\alpha _{\,1} }  + e^{\,i\,\alpha _{\,2} }  + e^{\,i\,\alpha _{\,2} } } \right) \cr} 
$$
so that:
$$
\left| {N_{2,\,3} (a_{\,1} ,a_{\,2} ,a_{\,3} )} \right| = r^{\,2} \left| {\left( {e^{\,i\,\alpha _{\,1} }  + e^{\,i\,\alpha _{\,2} }  + e^{\,i\,\alpha _{\,2} } } \right)} \right| = r\left| {\left( {a_{\,1}  + a_{\,2}  + a_{\,3} } \right)} \right|
$$
That makes clear how to proceed and generalize further.
