# How can I demonstrate the image of a contour in $\mathbb{C}$?

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 $x=-\frac{1}{2}$.

-

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 $-\infty$.
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.
If I invert $\omega$ I get $z = \frac{1}{\omega} + 1$ but I do not see how this helps me... – tacos_tacos_tacos Nov 9 '12 at 10:42
The map $\omega: z \to \frac{1}{z-1}$ is a composition of two maps: $\theta: z \to z-1$ and $\eta: z \to \frac{1}{z}$. Under $\theta$ your contour will simply shift to the left. Then, to understand what happens to it under $\eta$, you can use the fact that $\eta$ is a composition of a reflection and an inversion. The reflection doesn't really matter, and the inversion will turn your circle into a straight line. – Dan Shved Nov 9 '12 at 11:02