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If I am given a simple contour, say $\|z\|=1$, how can I find it's image under a mapping $\omega$? I can see how you arrive at the result for something like $\omega = \frac{1}{z}$, for since $\|z\|=1$ we have $z = e^{i\theta}$, and $e^{-i\theta}$ spans the exact same set for $\theta \in [0, 2\pi)$, so it is somewhat easy for me to see that the image of $\omega$ in $\|z\|=1$ is equivalently $\|z\|=1$.

How would I go about showing something a bit more difficult, say, under the mapping $\omega = \frac{1}{z-1}$? If I go through a similar method I get $\omega = \frac{1}{e^{i\theta} -1}$ but I do not see where to go from here. The goal is to get to the fact that the image is equivalent to the real part of the image being $-\frac{1}{2}$ and the imaginary part being unbounded, ie $x=-\frac{1}{2}$.

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If you work it out you get ${1\over {\rm e}^{{\rm i}\theta}-1}={{\rm e}^{-{\rm i}\theta}-1 \over ({\rm e}^{{\rm i}\theta}-1)({\rm e}^{-{\rm i}\theta}-1)}={\cos(\theta)-1+{\rm i}\sin(\theta)\over 2-2\cos(\theta)}=-{1\over 2}+{{\rm i}\over 2}{\sin(\theta)\over 1-\cos(\theta)}$ When $\theta$ runs from $0$ to $2\pi$, the imaginary part will run from $+\infty$ to $-\infty$.

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In situations like this, it can be quite useful to know a thing or two about inversions. You can prove lots of facts about inversions without even knowing what a complex number is, it's simple geometry.

If you feel comfortable with inversions, then understanding what $\omega(z)=1/z$ does becomes much easier, because $\omega$ is simply a superposition of an inversion and a reflection.

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    Then again, your problem can be solved without explicitly using inversions, but inversions are my favorite way of thinking about Möbius transformations.2012-11-09