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

${zinmathbb{C}:|z-2|<2$ and $|z-1|>1}$

under the map $zmapsto frac{1}{z}$.

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.

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

If $z=re^{itheta}$, then $frac1z=frac1r e^{-itheta}$.

Going back to the first point, we know the boundaries $mid x-2mid=2$ and $mid x-1mid=1$ 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. $2tofrac12$ and $1+itofrac1{1+i}=frac12-frac12i$. So $mid z-1mid=1$ goes to the vertical line through $frac12$ and $frac12-frac 12i$.

Next, $mid z-2mid=2$ 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. $frac52tofrac25$. This is the region between the two vertical lines. An infinite vertical strip.

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

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

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

Furthermore $$ |Acos(alpha)e^{ialpha}-2|=sqrt{ left(Acos^2(alpha ) -2right)^2 + left(Asin(alpha)cos(alpha)right)^2} $$
$$=sqrt{(A^2-4A)cos^2(alpha)+4}<2$$
as $A>4$ and $alpha in (frac{-pi}{2},frac{pi}{2})$

Now if $zin S$ let $z=Re^{ialpha}$. By the triangle inequality $Re(z)>0$ thus $alpha in (frac{-pi}{2},frac{pi}{2})$ and $cos(alpha)>0$ thus we can let $R=Acos(alpha)$. 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 $z=Acos(alpha)e^{ialpha} in S$ then $$1/z=frac{1}{Acos(alpha)}e^{-ialpha}=frac{1}{A}- frac{i tan(alpha)}{A}$$
as $alpha$ varies the vertical line $x=frac{1}{A}$ is formed and as then letting $A$ vary you get the set $W={ x+iy : x,y in Bbb R : 1/2gt xgt1/4}$

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 $z=a e^{itheta}$ and then using the trig half angle identities to get it into a nice form.