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Prove: $\left|\frac{z-w}{1-\bar w z}\right|=1$ if $|z|=1$ and $|w| \neq 1$

My attempt:

I reasoned that either $(z-w)$=$(1-\bar w z)$ or $(z-w)$=$-(1-\bar w z)$ must be true for the whole thing to be true. Going with the first option:

$$(z-w)=(1-\bar w z)$$ $$z-w +\bar wz = 1 $$ $$z(1+\bar w)-w=1$$ $$z=\frac{1+w}{1+ \bar w}$$

At this point, I feel like there is some property of conjugates that I should use, but I'm not sure...

Thomas Andrews
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whatwhatwhat
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  • Note: for complex numbers $a,b$, if you have $|a|=|b|$, this does not mean $a=\pm b$. For example, $|i|=|1|$, this does not mean $i=\pm1$. – David Feb 22 '16 at 01:00

2 Answers2

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Hint: $$z-w=z\left(1-wz^{-1}\right)=z\left(1-w\overline{z}\right)$$

Thomas Andrews
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  • Wait, what? The inverse of a complex number is its conjugate? The inverse of $1+i$ is $\frac{1}{1+i}$. Moving the $i$ to the numerator yields $\frac{1-i}{2}$. The conjugate of $1+i$ is $1-i$. How did you arrive at your last step? – whatwhatwhat Feb 22 '16 at 01:13
  • @whatwhatwhat Is $|1+i|=1$? If $|z|=1$ then $z\overline{z}=1$. – Thomas Andrews Feb 22 '16 at 01:13
  • No, it's $\sqrt{2}$. $|1+i|$ = $\sqrt{1^2+1^2}=\sqrt{2}$... – whatwhatwhat Feb 22 '16 at 01:15
  • Ah, I think I see what you're saying. If $|x+iy|=\sqrt{x^2+y^2}=1$, then $x^2+y^2=1$ and because $(x+iy)(x-iy) =x^2+y^2$, then by that method we also get $1$, which proves that $z\bar z =1$ (because $1*1=1$). – whatwhatwhat Feb 22 '16 at 01:28
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$$\left|\frac{z-w}{1-\bar w z}\right|^2=\frac{(z-w)(\bar z-\bar w)}{(1-\bar w z)(1- w \bar z)}=\frac{1-z\bar w-w\bar z+|w|^2}{1-\bar w z- w \bar z+|w|^2}=1$$

John B
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