5.7 The Point at Infinity
Under the mapping \(\, w = \frac {1}{z}\,\) or \(\, \rho e^{i\phi } = \frac {1}{r}e^{-i\theta },\,\) the points \(z\) interior to the circle \(\, r = R\,\) map into points \(w\) interior to the circle
\(\, \rho = \frac {1}{R}\).
The point \(\, w = 0\,\) is not an image.
By making \(R\) large enough all points \(z\) outside the large circle fall within an arbitrary small
neighbourhood of \(\, w = 0\)
We can take of talk of the points of infinity, i.e \(\, w = \infty .\)
This is the image of \(\, z = 0\,\) in the \(z-\)plane.
Example 5.9. The function \(\, w =\frac {4z^2}{(1 - z)^2}\,\) maps the point \(\, z = \infty \,\) into the point \(\, w = 4\) in the \(w-\)plane.
For, if we write \(\, z = \frac {1}{z'},\,\) so that
\[ w = \frac {\frac {4}{z'^2}}{\Big (1 - \frac {1}{z'}\Big )^2} = \frac {4}{(z' - 1)^2}\]
Setting \(\, z' = 0 \, (z =\infty )\,\) get \(w = 4\).
One checks that \(\, w = \infty \,\) is the image of the point \(z = 1\) in the \(z-\)plane. \(\quad \Big (\) set \(w = \frac {1}{w'}\Big )\).
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