Artificial intelligent assistant

Elementary transformations of circles > Discuss the image of the circle $|z-2|=1$ and its interior under the following transformations. > > a.) $w=\frac{z-2}{z-1}$ > > b.) $w=\frac{z-4}{z-3}$ > > c.) $w=\frac1z.$ The answer that is provided are, respectively: $$\\{w:Rew\le\frac12\\},\\{w:Rew\ge \frac32\\},\\{w:|w-\frac23|\le\frac13\\}.$$ Not sure how they got that. I know each of the circles has a pole and those poles lies on the circle, so the image has to be a straight line, but how do I go about finding those lines? Is there an algebraic or a more easy recognizable method to follow?

The usual mantra is 'three points defined a circle (or line)'. This is usually the quickest way to find the answer.

We'll use the fact that $1,3,2+i$ are in the original circle.

(a) $1\mapsto \infty$ so it is a line. $3 \mapsto \frac{1}{2}$, and $2+i \mapsto \frac{i}{1+i} = \frac{1}{2}(1+i)$. Hence we have a line through $\frac{1}{2},\frac{1}{2} + \frac{i}{2}$. This is clearly the line $\mathrm{Re} w = \frac{1}{2}$. To check which side of the line is which, note that $z=2$ is in the interior, and $2 \mapsto 0$ so the region is $\mathrm{Re} w \le \frac{1}{2}$

(b) I'll leave to you.

(c) You could do this as the above too. Alternatively, note $$|z-2| = 1 \iff (z-2)(z^\star-2) = 1 \iff 1 - 2/z - 2/z^\star+3/zz^\star = 0$$ and then using $w=1/z$ we find $$3ww^\star - 2w -2w^\star + 1 = 0 \iff \left(w-\frac 2 3 \right)\left(w^\star-\frac 2 3 \right) = \frac 1 9$$ which gives the result. Again, one point is enough to tell you where the interior is.

xcX3v84RxoQ-4GxG32940ukFUIEgYdPy 77d73da280e8b6349e36f27a1b6169d0