Prove $f(z) = \frac{z-1}{z+1}$, with $\operatorname{Re}z=0$, is a unit circle

Question

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.

Answer 1

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.

Answer 2

Your general approach is right. Apply the assumptions to the function, then find real and imaginary parts and interpret that as a plot in the complex plane. Proving what shape it is depends on the problem.

Regarding your example,

$$f(z)=\frac{z-1}{z+1}$$ If the real part is always zero then $$f(z)=\frac{-1+iy}{1+iy}\frac{1-iy}{1-iy}=\frac{-1+2iy+y^2}{1+y^2}=\frac{y^2-1}{y^2+1}+i\frac{2y}{1+y^2}.$$ So you have $$u(t)=\frac{t^2-1}{t^2+1}\quad\text{and}\quad v(t)=\frac{2t}{1+t^2}.$$ I plotted this with Mathematica,

ParametricPlot[{(t^2 - 1)/(t^2 + 1), 2 t/(1 + t^2)}, {t, -11, 6}, AspectRatio -> 1]

so it looks like you get a circle since $t\in\mathbb{R}$.

These are the tangent half-angle substitution formulae a.k.a Euler/Weierstrass substitution formulae. According to this article about the geometry of these equations, the point $(-1,0)$ is never reached, which corresponds to the point $(1, 0)$ given our equations (it's negated). Also you can make the substitution $t=\tan(x/2)$ and work through the algebra you will find $u(t)=-\cos x$ ad $v(t)=\sin x$, but I guess there must be some restrictions on $x$ since the point $(1,0)$ is never reached.