While trying to show (explicitly) that the image of the circle $|z-\frac{1}{2}|=\frac{1}{2}$ under the map $w=\frac{1}{z}$ is the line $u=1$, a few questions arose. (Here I put $w=u+iv$.)
Here is what I did: let $z=x+iy=re^{i\theta}$. Then $w=\frac{1}{z}=\frac{1}{r}e^{-i\theta}=u+iv$, where $u=\frac{\cos\theta}{r}$ and $v=-\frac{\sin\theta}{r}$. Now $|z-\frac{1}{2}|=\frac{1}{2} \iff (x-\frac{1}{2})^2+y^2=\frac{1}{4} \iff r(r-\cos\theta)=0 \text{ (where $0\le \theta \le 2\pi)$} \iff r=0 \text{ or } r=\cos\theta$.
Case 1: If $r=\cos\theta$ then $u=1$ and $v=-\tan \theta$. It suffices to show that $-\infty < v < +\infty$. But how to prove this neatly? We have $0\le \theta \le 2\pi$ but $\tan$ is not defined at $\pi/2$ and $3\pi/2$... What are the right words to say here?
Case 2: If $r=0$, then how does one proceed? Again we have $0\le \theta \le 2\pi$ and if $\theta=\pi/2$ then $u=\infty \times 0$...
Can I suggest a simpler method?
$$\left|z-\frac 12\right|=\frac 12 \\\text{and}\\z=\frac 1w$$ $$\implies \left|\frac 1w-\frac 12\right|=\frac{|2-w|}{|2w|}=\frac 12$$ $$\implies |2-w|=|w|\implies w=1$$
Hence the locus is $u=1$