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 $u=\infty \times 0$...


1 Answer 1


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 $u=1$

  • $\begingroup$ Thanks. However this method doesn't carry over to more complicated Moebius transformations whereas the method with polar coordinates does... That's why I'd like to understand how does one carry it out rigorously. $\endgroup$
    – user557
    Feb 12, 2017 at 21:12

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