complex numbers

edi10000

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May 31, 2020
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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
 
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 & WZ~\&~W instead of Z1 & Z2Z_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?
 
It seems to me the there are contradictions in the post.
Please review what you have posted. If you would use Z & WZ~\&~W instead of Z1 & Z2Z_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
 
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 Z1Z_1.
If Z1=a+bi & Z2=c+diZ_1=a+bi~\&~Z_2=c+di then Z1Z2=(acbd)+i(ad+bc)1Z_1\cdot Z_2=(ac-bd)+i(ad+bc)\ne-1.
From which we know that (acbd)1 if (ad+cd)=0(ac-bd)\ne-1\text{ if }(ad+cd)=0.
Because (W0)[1W=WW2](\forall W\ne 0)\left[\dfrac{1}{W}=\dfrac{\overline{W}}{|W|^2}\right]
we have Z1+Z21+Z1Z2=\dfrac{Z_1+Z_2}{1+Z_1\cdot Z_2}= (Z1+Z2)(1+Z1Z2))1+Z1Z22 \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 (Z1+Z2)(1+Z1Z2)) (Z_1+Z_2)(1+\overline{Z_1\cdot Z_2})) is real.
 
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