Show that $z(t)=\frac{1+it}{1-it}$ describes a circumference on the complex plane. I am asked to prove that $z(t)=\frac{1+it}{1-it}$ describes a circumference on the complex plane when $t$ takes every value in the extended real number line. That is, $\mathbb{R} \cup\{\pm\infty\}$. I don't have any idea how to proceed with this. Furthermore, Wolfram Alpha does not plot anything resembling a circumference when the function is plugged in. Am I not understanding the problem well enough?

Method 1
Observe that $$|z|=\frac{|1+it|}{|1-it|}$$
$$=1$$
because denominator and numerator are conjugates. This equation of a circle centred at origin and of radius $1$.
Method 2
Put $$t=\tan \frac{\theta}{2}, 0\leq \theta<2π, \theta\neq π$$
Now,
$$z=\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}$$
$$=\cos\theta+i\sin\theta$$
which is equation of circle with radius $1$. But this method has a nice property that it gives you polar coordinates of the point directly. Also, this method reveals an important information, that if $t$ varies over real numbers only, the circle would not be complete, since point $(-1,0)$ is not in the locus. So you need extended real number system to complete the circle.
