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How to simplify the following expression:

$$\frac{z-1}{z+1}~, \quad \text{where} z\in \mathbb{C}\setminus \{-1\}$$

There is just nothing i can come up with, neither in cartesian, nor in polar.

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closed as off-topic by José Carlos Santos, Lee David Chung Lin, Shailesh, Leucippus, Alexander Gruber May 1 at 6:15

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    $\begingroup$ You mean, write it in $a+bi$ form? What does 'simplify' mean? It looks fairly simple to me. $\endgroup$ – The Count Apr 30 at 22:30
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    $\begingroup$ Be more clear. What they exactly ask you to do. Simplifying this expression is senseless unless you give more precision. $\endgroup$ – HAMIDINE SOUMARE Apr 30 at 22:30
  • $\begingroup$ I presume you would like to realise the denominator? Write $z = a+bi$, and try multiplying top and bottom by $\overline{z}$ $\endgroup$ – bounceback Apr 30 at 22:34
  • $\begingroup$ I made a mistake in my scratchwork. I've corrected the answer I posted. $\endgroup$ – R. Burton Apr 30 at 23:25
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Your expression is about as simplified as it can get. That being the case, I assume that when you say "simplify" you mean that you wish to evaluate the expression and need to get rid of the complex denominator. To this end:

Let...

$$f(z)=\frac{z-1}{z+1}$$

Let $v=z-1$, and $w=z+1$. Then...

$$f(z)=\frac{v}{w}=\frac{\bar{w}v}{|w|^2}$$

Substitute $z-1$ and $z+1$ back in for $v$ and $w$, respectively...

$$f(z)=\frac{(\bar{z}+1)(z-1)}{|z+1|^2}$$

The denominator is now a real number, and you may proceed as usual from here.

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  • $\begingroup$ How you get $(z-1)^2$ $\endgroup$ – HAMIDINE SOUMARE Apr 30 at 23:15
  • $\begingroup$ @HAMIDINESOUMARE Sorry, that's a mistake. I will correct it. $\endgroup$ – R. Burton Apr 30 at 23:20
  • $\begingroup$ And you can symplify further. $f(z)=\frac{|z|^2+2i\operatorname{Im}(z)-1}{|z+1|^2}$ $\endgroup$ – HAMIDINE SOUMARE Apr 30 at 23:28

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