Mapping Property of $f(z)=\frac{z-i}{z+i}$.

Question

Consider the map $f(z)=\dfrac{z-i}{z+i}$.

What is the image of Circle with radius $ r $ under $f$

I got $x=\dfrac{r^2 -1 }{r^2 -2r\cos(\theta) +1} $ and $y=\dfrac{-2r\sin(\theta)}{r^2 -2r\cos(\theta) +1}$.

I don't know what to do further.

Answer 1

Let $f(z)=w=u+iv$. $w=\frac{z-i}{z+i}$ This implies $z=\frac{i(1+w)}{1-w}$. Image of $|z|\le r$ and if $r=1$ then$|1+w|\le|1-w|$ Putting $w=u+iv$ in above equation and solving, we get $u \gt 0$.Hence f map unit circle to {$u+iv:u$$\gt$ 0} And for $r\gt 1$ same way we get
{$u+iv:$$(u-1)^2+v^2\gt 0$}

Answer 2

I assume the circle $\mathcal C$ is centered at the origin. Then $\mathcal C$ is symmetric wrt the imaginary axis, therefore $f(\mathcal C)$ is symmetric wrt the image of the imaginary axis, which is the real axis. If $r = 1$, $f(\mathcal C)$ goes through $0$ and $\infty$, therefore it's the imaginary axis. Otherwise $f(\mathcal C)$ is the circle for which $[f(-i r), f(i r)]$ is a diameter.