complex numbers

[edi10000]

if z1*z2!=-1 and |z1|=|z2|=1 then (z1+z2)/(1+z1*z2) real
how can i prove it?
i try to prove X=X' but i lost...
thank you

 

[pka]

if z1*z2!=-1 and |z1|=|z2|=1 then (z1+z2)/(1+z1*z2) real
how can i prove it?

It seems to me the there are contradictions in the post.
Please review what you have posted. If you would use \(Z~\&~W\) instead of \(Z_1~|\&~Z_2\) I think it would be better.
For example in th OP you wrote Z1*Z2!=-1. What the heck does that mean?

 

[edi10000]

It seems to me the there are contradictions in the post.
Please review what you have posted. If you would use \(Z~\&~W\) instead of \(Z_1~|\&~Z_2\) I think it would be better.
For example in th OP you wrote Z1*Z2!=-1. What the heck does that mean?

Given:
z1*z2 not equal -1 & |z1|=|z2|=1

Prove:
(z1+z2)/(1+z1*z2) is real number

 

[pka]

Given: z1*z2 not equal -1 & |z1|=|z2|=1
Prove: (z1+z2)/(1+z1*z2) is real number

Well if you do not want to learn to use LaTeX, at least learn use Z_1 for \(Z_1\).
If \(Z_1=a+bi~\&~Z_2=c+di\) then \(Z_1\cdot Z_2=(ac-bd)+i(ad+bc)\ne-1\).
From which we know that \((ac-bd)\ne-1\text{ if }(ad+cd)=0\).
Because \((\forall W\ne 0)\left[\dfrac{1}{W}=\dfrac{\overline{W}}{|W|^2}\right]\)
we have \(\dfrac{Z_1+Z_2}{1+Z_1\cdot Z_2}= \)\( \dfrac{(Z_1+Z_2)(1+\overline{Z_1\cdot Z_2}))}{|1+Z_1\cdot Z_2|^2}\)
To show the it is real all we must do s shoe that the generator \( (Z_1+Z_2)(1+\overline{Z_1\cdot Z_2}))\) is real.