Here is a description of constructing Möbius map sending $1$, $i$, $-1$ to 0, 1, $\infty$, respectively

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

I do not know how it was converted to :

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


  • 2
    $\begingroup$ $(i+1) = -i\cdot (i-1)$. $\endgroup$ Commented Jan 1, 2014 at 13:22
  • $\begingroup$ @Daniel Fisher Thx, it is more answer then comment. $\endgroup$
    – Adam
    Commented Jan 1, 2014 at 13:29

1 Answer 1


When you want to simplify a ratio of two complex numbers such as


you should multiply the top and bottom by the complex conjugate of the denominator (this is a complex analogue of rationalising the denominator). By doing that we get

$$\frac{a+bi}{c+di}\frac{c-di}{c-di} = \frac{(a+bi)(c-di)}{c^2+d^2}$$

which can then be written in the form $x + yi$ for some real $x$ and $y$.

If you do this for the expression


you will find that it reduces to $-i$.


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