An Unexpected Circle...

I played around with

$$z=\frac{-1+e^{it}}{\phantom{-}2+e^{it}}$$

and found that, when I draw the real against the imaginary of $z$, it pretty much looks like a circle.

But neither ${\frak{R}} z $, nor ${\frak I} z$ look like $\cos $ or $\sin$. Is it due to a kind of transformed argument of $\cos $ and $\sin$? Something like $\cos(f(t))$?

Solution 1:

Let $a,b,c,d$ be complex numbers and suppose c,d are not both 0 and a is not 0 ${}\ \ ad-bc\ne0$.

Let $f(z) = \dfrac{az+b}{cz+d}$. Functions of this kind are called "linear fractional transformations". Then if $z$ moves around a circle in the complex plane, then $f(z)$ follows either a circle, although usually it's not the same circle, or a straight line. It will be a straight line if and only if $z$ passes through some point that makes the denominator $0$. Except when $c=0$, you'll find that it doesn't move at the same angular rate, i.e. when $z$ goes through a quarter of a circle, then $f(z)$ might go through a half circle.

In your example, $a=1$, $b=-1$, $c=1$, and $d=2$.

Google the terms "linear fractional transformation" and "Riemann sphere".

PS: It is also the case that if $z$ moves along a straight line then $f(z)$ moves along either a straight line or a circle.