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?
 A: Hint: $z=x+iy$
First transform the fraction so that the denominator is real, by multiplying above and belove by the conjugate of the denominator.
Then check if the real and imaginary parts of your z satisfy the equation of a circle... Which one? The easiest you can think of...
A: 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.
A: Otherwise, put in polar form, to get $z=e^{i2\arctan(t)}=e^{i\alpha}$, with $\pi <\alpha \le \pi$
A: Express $t$ as in terms of $z$,
$$t= \frac1i \frac{z-1}{z+1}$$
Since $t=t^*$ for real $t$, we have
$$ \frac1i \frac{z-1}{z+1}=\left(\frac1i \frac{z-1}{z+1}\right)^*$$
Simplify above expression,
$$zz^*=|z|^2=1$$
which describes a unit circumference in the complex plane.
