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?

Otherwise, put in polar form, to get $z=e^{i2\arctan(t)}=e^{i\alpha}$, with $\pi <\alpha \le \pi$
