What is the image of
- the real line
- the imaginary line
- the unit circle
Under the map $f(z)=\frac{2z-1}{2-z}$
Assume $z=x+iy$. Then setting $$w=f(z)=\frac{2(x+iy)-1}{2-iy-x}=\frac{(2x-1+2iy)(2-x+iy)}{(2-x)^2+y^2}$$.
Since the real line has $y=0$ we get the image:
$$w=\frac{(2x-1)(2-x)}{(2-x)^2}$$
Now this is in terms of $x$, when ideally I would have variables $w=u+iv$, does it mean I simply need to call $x=u$?
Similarly for the imaginary line by setting $x=0$ this time.
What happens for a unit circle?
Let $|z|=1=e^{it}$ for $t\in [0,2\pi]$. Then the image is $$f(z)=\frac{2e^{it}-1}{2-e^{it}}$$
However, what do these images actually look like? Are they lines, circles etc.? I know that circles map to circles and since $|z|\ne 2$ for the unit circle, it doesn't have a point at infinity $\rightarrow$ it's a circle.
$$w=\frac{2x-1}{2-x}$$
and we know that $x \rightarrow w$ is bijective from $\mathbb{R}-2$ to $\mathbb{R}-\{-2\}$: the real axis minus a single point.
$$z=2\dfrac{Z+\dfrac12}{Z+2} \ \ \ (1)$$
Thus, expressing that $z \in C(0,1)$ (unit circle), a consequence of (1) is:
$$ |z|=1 \Leftrightarrow 2\dfrac{|Z+\dfrac12|}{|Z+2|}=1 \Leftrightarrow |Z-(-2)|=2|Z-(-\frac12)|\Leftrightarrow dist(M,A)=2 dist(M,B)$$
(by setting $A(-2,0)$ and $B(-\frac12,0)$, $M$ corresponding to $Z$).
We are now able to apply a classical result: the locus of points $M$ such that $dist(M,A) \ = \ 2 \ dist(M,B)$ with fixed points $A$ and $B$ (https://en.wikipedia.org/wiki/Circles_of_Apollonius) is a circle with center on axis $AB$; its characteristics are rather easy to derive (center in $(-9/4,0)$, radius $5/2$).
Remark: This method can be applied to the image of the unit circle by any (homographic) transformation $Z=\dfrac{az+b}{cz+d}$. One will find always $dist(M,A)=k dist(M,B)$ for some $A$, $B$ and $k$. In some cases, one can find $k=1$ : in such a case, it is no longer a circle, but a straght line.