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How to show that the modulus of $\frac{z-w}{1-\bar{z}w}$ is always $1$?

I wish to prove:

$$|\frac{z_{1}-z_{2}}{1-\overline{z_{1}}z_{2}}|=1$$ when $$|z_1|=1$$.

This is what I have tried:

$$|\frac{z_{1}-z_{2}}{1-\overline{z_{1}}z_{2}}|=1\iff|z_{1}-z_{2}|=|1-\overline{z_{1}}z_{2}|$$ Denote $z_{1}=a+bi,z_{2}=c+di$. $$ |z_{1}-z_{2}|=|1-\overline{z_{1}}z_{2}| $$

$$ \iff(a-c)^{2}+(b-d)^{2}=(1-ac-bd)^{2}+(ac+bd)^{2}$$

$$\iff a^{2}-2ac+c^{2}+b^{2}-2bd+d^{2}=1-2ac+a^{2}c^{2}-2bd+2abcd+b^{2}d^{2}+a^{2}c^{2}+b^{2}d^{2}+2abcd$$

$$\iff c^{2}+d^{2}=2a^{2}c^{2}+4abcd+2b^{2}d^{2}$$

This is where I am stuck, the last line doesn't even seem correct, though I checked my calculations multiple times. Can someone please provide a hint/help on how to continue or how to solve this problem ? I would appriciate it very much!

Community
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Belgi
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3 Answers3

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$$|\frac{z_{1}-z_{2}}{1-\overline{z_{1}}z_{2}}|=|\frac{z_{1}-z_1\overline{z_1}z_{2}}{1-\overline{z_{1}}z_{2}}|$$

Make $z_1$ a common factor in the numerator and you are done.

N. S.
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\begin{align} \vert z_1 - z_2 \vert^2 & = (z_1 - z_2)(\bar{z}_1 - \bar{z}_2) & \\ & = \vert z_1 \vert^2 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + \vert z_2 \vert^2 & \\ & = 1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + \vert z_2 \vert^2 & (\because \vert z_1 \vert = 1)\\ & = 1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + \vert \bar{z}_1 z_2 \vert^2 & (\because \vert z_1 \vert = 1)\\ & = \vert 1 - \bar{z}_1 z_2 \vert^2 \end{align}

  • Thank you very much, I don't think I would of noted that, but next time I might know better to look for this kind of things. (+1) – Belgi Nov 02 '12 at 03:06
  • I have to say that squaring the norm and then writing it as in the first line have helped my with another question. thank you for showing me this approach! – Belgi Nov 02 '12 at 10:50
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Your approach is longer than the proposed answers, but it certainly works.

Your only error is when you go to $a,b,c,d$.

The imaginary part of $1-\overline{z_1}z_2$ is $-ad+bc$, not $ab+cd$.

At the end, you would have to use $a^2+b^2=c^2+d^2=1$ to finish. (Either by factorizing the right-hand side first, or by simply substituting $b^2$ and $d^2$ in the equation.)

Phira
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  • Thank you Phira, I have accepted your answer as I feel this is a better answer given that this is a homework problem hence the way of doing this was the imprtant thing, not the result – Belgi Nov 02 '12 at 10:51