Show that $z(t) = \frac{a+bt}{c+dt}$ with $ad-bc\neq 0$ describes a circumference or a line.

Question

I just asked a question which is related to this one, but the problem seems to be different. In this case, one has to show that $$z(t)=\frac{a+bt}{c+dt}$$

Describes a straight line or a circumference, given that $t$ takes every value in the extended real number line. That is, $\mathbb{R} \cup\{\pm\infty\}$. Also, $a,b,c,d\in\mathbb{C}$ and $ad-bc \neq 0$.

How can I prove this? I don't know how one can prove at the same time that $z(t)$ describes a circumference and/or a line. Do I need to prove each thing separately? Or is it that it only describes one of those things? Can someone point me in the right direction?

Answer 1

You need to show that for any $a,b,c,d$ with $ad-bc$, the locus of points generated by your function is EITHER a line or a circumference. For some $(a,b,c,d)$, you'll get a line. (For instance, for $a =0, b = 1, c = 0, d = 1$, you'll get exactly the real line. For others you'll get a circumference. The "right direction" is to experiment a bit and discover which $(a,b,c,d)$ quadruples generate lines (not TOO hard to guess), and then to show that for all the others, you get circumferences (considerably harder).

Answer 2

If $$z=\frac{a+bt}{c+dt}$$ then $$t=\frac{cz-a}{-dz+b}.$$ So the condition that $z$ lies on the locus is that that $t$ be real: $t=\overline t$. That is $$\frac{cz-a}{-dz+b}=\overline{\frac{cz-a}{-dz+b}}$$ or $$(cz-a)(\overline{-dz+b})=(\overline{cz-a})(-dz+b).$$ Dividing by $i$, this reduces to an equation of the form $$\alpha|z|^2+\beta x+\gamma y+\delta=0$$ where $z=x+iy$ and the coefficients are real. This is a circle when $\alpha\ne0$ and a line otherwise.

Answer 3

You might like to investigate Mobius transformations on Wikipedia.

Your question asks for the image of the extended real axis. Since Mobius transformations map lines/circles to lines/circles the result you request is simply this standard property of Mobius transformations.