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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.

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  • $\begingroup$ In the numerator in the first identity, shouldn't it be $ab+ac+bc$ ? $\endgroup$
    – G Cab
    Commented Apr 10, 2016 at 15:56
  • $\begingroup$ @GCab Yes. Edited. $\endgroup$
    – Nadia
    Commented Apr 10, 2016 at 15:59
  • $\begingroup$ Then you should correct $\alpha=..$ as well. $\endgroup$
    – G Cab
    Commented Apr 10, 2016 at 16:53

2 Answers 2

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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

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  • $\begingroup$ I was trying to generalise the original question (three complex numbers) for n complex number. It seems like your example shows it cannot be done. $\endgroup$
    – Nadia
    Commented Apr 10, 2016 at 16:21
  • $\begingroup$ Or at least it can be generalized in the way that you attempted. It's possible that the other generalization I suggested still works, although it needs to be proven. $\endgroup$
    – browngreen
    Commented Apr 10, 2016 at 21:24
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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.

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  • $\begingroup$ Can you explain the notation $N_2,_3 (a_1,a_2,a_3)$ please $\endgroup$
    – Nadia
    Commented Apr 10, 2016 at 18:08
  • $\begingroup$ Just to indicate the sum of the product of 2 terms (distinguished) over a total of 3 numbers $\endgroup$
    – G Cab
    Commented Apr 10, 2016 at 18:24
  • $\begingroup$ That confirms @browngreen comment that you can extend your formulation to $N_{\,n - 1,\,n} $, for all $2<=n$ $\endgroup$
    – G Cab
    Commented Apr 10, 2016 at 18:33

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