Here's what I did to graph
$$f(z) = \frac{z-1}{z+1}$$
where $\operatorname{Re}z=0$. Given $\operatorname{Re}z=0$, $z$ can be written as $z = iy$ and the function becomes
$$f(y) = \frac{iy-1}{iy+1}$$
If we separate into real and imaginary parts we get $$f(y) = \frac{y^2-1}{y^2+1}+\frac{2iy}{y^2+1} = \frac{(y+i)^2}{y^2+1},$$ even though the last equality didn't give anything to me.
From that we get two functions: the real part $u(y) = \frac{y^2-1}{y^2+1}$ and the imaginary part $v(y) = \frac{2y}{y^2+1}$.
What I did next is to plot the points in the Argand plane. It appears to be a circle of radius 1, with the center at the origin. I guess its equation is something like $(\operatorname{Re}z)^2 + (\operatorname{Im}z)^2 = 1$, but excluding the point (1,0). However, I don't know how to obtain it analytically.
The question is: how can I prove that $f(y) = \frac{iy-1}{iy+1}$represents a circle of radius 1 and centered at origin. Is there a general method for the proof? Plotting the function with simple points is helpful, but not a proof.
Best Answer
It is actually straightforward to show that
$$w=\frac{iy-1}{iy +1}$$
represents a circle. Evaluate
$$|w|^2=w\bar{w}=\frac{iy-1}{iy +1}\cdot \frac{-iy-1}{-iy +1} =1$$
which leads to the solution $w=e^{i \theta}$. Thus, $w$ describes a unit circle with center at the origin. The exclusion of the point $w=1$ is seen from
$$\lim_{y\rightarrow \pm \infty} w = \lim_{y\rightarrow \pm \infty} \frac{iy-1}{iy +1} =1$$
Edit:
Let $w=u+iv$. Then $|w|^2 = 1$ leads to
$$w\bar{w}=(u+iv)(u-iv)=1$$
or, in the form of equation for the real part,
$$u^2+v^2=1$$
which represents a unit circle with the center at the origin.