$m(z) = \frac{(z-1)}{(z+1)} \frac{(i+1)}{(i-1)}$
I do not know how it was converted to :
$m(z) = -i \frac{(z-1)}{(z+1)} $
TIA
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2$\begingroup$ $(i+1) = -i\cdot (i-1)$. $\endgroup$ – Daniel Fischer♦ Jan 1 '14 at 13:22
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$\begingroup$ @Daniel Fisher Thx, it is more answer then comment. $\endgroup$ – Adam Jan 1 '14 at 13:29
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When you want to simplify a ratio of two complex numbers such as
$$\frac{a+bi}{c+di}$$
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
$$\frac{1+i}{1-i}$$
you will find that it reduces to $-i$.
answered Jan 1 '14 at 13:34
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$\begingroup$ Thx. See also [wikibooks] [1]:en.wikibooks.org/wiki/Fractals/Computer_graphic_techniques/2D/… $\endgroup$ – Adam Jan 1 '14 at 14:09
