Complex function maps into lines and circles

Question

We have the following function: $$f(z) = \frac{z-1}{z-i}$$

What is $f$ of the real line (so $f(\mathbb{R}))$?

What we did in the solutions was:

$$f(\infty) = 1 \\ f(0) = 1/i = -i\\ f(1) =0\\$$

Then we said: $$|z-(a+bi)| = r\\ |1-a-bi| = r\implies (1-a)²+b²=r² \implies a = 1/2 \\ |-i-a-bi| = r\implies a²+(b+1)^2 = r²\implies b = -1/2\\ |a+bi| = r \implies\\ \implies a²+b² = r² \implies r = \sqrt{1/4 + 1/4} = \\ =\sqrt{2}/2 \implies |z-(1/2-i/2)| = \sqrt{2}/2\\$$

However I dont understand what we just did above. Can somebody explain what the procedure was and why. What would $f$ do to the imaginary line $\mathbb{I}$ or maybe the outside of the unit circle?

I really need help with understanding the procedure for exercises like this one.

Answer 1

The solution outlined in OP's question relies on understanding of the Möbius transformation, which implies that $\,f(\mathbb R)\,$ is a generalized circle. To find it, it's enough to find $3$ points on it, which in this case were determined to be $1, -i, 0$. Its center lies at the intersection of the perpendicular bisectors of segments $(0,1)$ and $(0,-i)$ which is the point $\frac{1-i}{2}$, and its radius is $\left|\frac{1-i}{2}\right|=\frac{\sqrt{2}}{2}$.

The following is an alternative solution, without assuming prior knowledge of Möbius transformations.

Let the image of $\,z\,$ be $\,w=f(z)=\frac{z-1}{z-i} \iff z = \frac{i w-1}{w-1}\,$. When $z \in \mathbb R \iff z = \bar z$:

$$ \require{cancel} \frac{i w-1}{w-1} = \frac{-i \bar w-1}{\bar w-1} \\ \iff\quad\quad (iw-1)(\bar w - 1) = (w-1)(-i\bar w- 1) \\ \iff\quad\quad iw\bar w - iw - \bar w + 1 = -i w \bar w - w + i \bar w + 1 \\ \iff\quad\quad 2iw\bar w + (1-i) w - (1+i)\bar w = 0 \\ \iff\quad\quad 2w\bar w - (1+i) w - (1-i)\bar w = 0 \\ \iff\quad\quad \big((1+i)w - 1\big)\big((1-i)\bar w - 1\big) - 1 = 0 \\ \iff\quad\quad \big|(1+i)w - 1\big|^2 = 1 \\ \iff\quad\quad \big|2w - (1-i)\big|^2 = 2 \\ \iff\quad\quad \left|w - \frac{1-i}{2}\right| = \frac{\sqrt{2}}{2} $$

The latter is the equation of a circle of radius $\frac{\sqrt{2}}{2}$ centered at $\frac{1-i}{2}$ (which, for verification, matches the result obtained using the other method).

Answer 2

Let $z=t$ on $\mathbb{R}$ and then $$ f(z)=\frac{t-1}{t-i}=\frac{(t-1)(t+i)}{t^2+1}=\frac{t(t-1)}{t^2+1}+\frac{t-1}{t^2+1}i. $$ Let $$ x=\frac{t(t-1)}{t^2+1}, y=\frac{t-1}{t^2+1} $$ and then $$ (x-\frac12)^2+(y+\frac12)^2=\frac12 $$ which is a circle centered at $(\frac12,-\frac12)$ with radius $\frac1{\sqrt2}$.