Find the image of $\{z\in\mathbb{C}:|z-2|<2$ and $|z-1|>1\}$ under the map $z\mapsto \frac{1}{z}$

Question

As the title explains, I'm trying to solve a question which asks me to find the image of

$\{z\in\mathbb{C}:|z-2|<2$ and $|z-1|>1\}$

under the map $z\mapsto \frac{1}{z}$.

I find it really hard to understand the intuition behind these kinds of questions, so it'd mean a lot if you could help explain the answer to this particular question and perhaps give me an overview of some strategies to solve similar problems.

Answer 1

Disclaimer: There is probably a better way to do this but I am no complex analysis kiddo

let $S=\{z\in \Bbb C : |z-2|<2 : |z-1|>1\}$ and $T=\{A\cos(\alpha)e^{i\alpha} : A \in (2,4) : \alpha \in (\frac{-\pi}{2},\frac{\pi}{2})\}$

If $A\cos(\alpha)e^{i\alpha} \in T$ then $$ |A\cos(\alpha)e^{i\alpha}-1|=\sqrt{ \left(A\cos^2(\alpha ) -1\right)^2 + \left(A\sin(\alpha)\cos(\alpha)\right)^2} $$ $$=\sqrt{(A^2-2A)\cos^2(\alpha)+1}>1$$ as $A>2$ and $\alpha \in (\frac{-\pi}{2},\frac{\pi}{2})$

Furthermore $$ |A\cos(\alpha)e^{i\alpha}-2|=\sqrt{ \left(A\cos^2(\alpha ) -2\right)^2 + \left(A\sin(\alpha)\cos(\alpha)\right)^2} $$ $$=\sqrt{(A^2-4A)\cos^2(\alpha)+4}<2$$ as $A<4$ and $\alpha \in (\frac{-\pi}{2},\frac{\pi}{2})$

Now if $z\in S$ let $z=Re^{i\alpha}$. By the triangle inequality $Re(z)>0$ thus $\alpha \in (\frac{-\pi}{2},\frac{\pi}{2})$ and $\cos(\alpha)>0$ thus we can let $R=A\cos(\alpha)$. As $|z-1|>1$ $$\sqrt{(A^2-2A)\cos^2(\alpha)+1}>1$$ which implies $ A>2$

As $|z-2|<2$ $$\sqrt{(A^2-4A)\cos^2(\alpha)+4}<2$$ which implies $A<4$

This proves that S=T now if $z=A\cos(\alpha)e^{i\alpha} \in S$ then $$1/z=\frac{1}{A\cos(\alpha)}e^{-i\alpha}=\frac{1}{A}- \frac{i \tan(\alpha)}{A}$$ as $\alpha$ varies the vertical line $x=\frac{1}{A}$ is formed and as then letting $A$ vary you get the set $W=\{ x+iy : x,y \in \Bbb R : 1/2\gt x\gt1/4\}$

Parameterisation is a nice method to solve these sort of problems btw. I found the parameteriation I showed you by considering the circles $|z-a|=a$ and subbing $z=a e^{i\theta}$ and then using the trig half angle identities to get it into a nice form.

Answer 2

This is a Möbius transformation. They take generalized circles to generalized circles.

If $z=re^{i\theta}$, then $\frac1z=\frac1r e^{-i\theta}$.

Going back to the first point, we know the boundaries $\mid x-2\mid=2$ and $\mid x-1\mid=1$ go to generalized circles. Since $0$ goes to $\infty $, it appears that both circles go to lines ($0$ lies on both circles).

Try a couple of other points. $2\to\frac12$ and $1+i\to\frac1{1+i}=\frac12-\frac12i$. So $\mid z-1\mid=1$ goes to the vertical line through $\frac12$ and $\frac12-\frac 12i$.

Next, $\mid z-2\mid=2$ goes to the vertical line through $\frac14$ and $\frac1{2+2i}=\frac14-\frac14i$.

Then as to the region between the two circles (the region of interest), just take a test point. Like $\frac52$, for instance. $\frac52\to\frac25$. This is the region between the two vertical lines. An infinite vertical strip.