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