Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number 
Prove: if $|z|=1$ than $\frac{z+1}{z-1}$ is an imaginary number 

I have tried to look at $$\frac{z+1}{z-1}=\frac{z+1}{z-1}\cdot \overline{\frac{z-1}{\overline{z-1}}}=\frac{z+1}{z-1}\cdot \frac{\overline{z}-1}{\overline{z}-1}$$
But did not get far
 A: Hint
$$\frac{z+1}{z-1}\cdot\frac{\overline{z}-1}{\overline{z}-1}=\frac{z\cdot\overline{z}+\overline{z}-z-1}{z\cdot\overline{z}-\overline{z}-z+1}$$
but $$z\cdot \overline{z}=|z|^2=1$$ $$z-\overline{z}=2i\text{ Im}(z)$$  $$z+\overline{z}=2\text{ Re}(z)$$
Can you finish?
A: A number is imaginary if and only if it is the opposite of its conjugate. Now we have :
$$\overline{\left(\frac{z+1}{z-1}\right)}=\frac{\overline{z+1}}{\overline{z-1}}=\frac{\overline{z}+1}{\overline{z}-1}=\frac{z\overline{z}+z}{z\overline{z}-z}=\frac{1+z}{1-z}=-\frac{z+1}{z-1}.$$
A: Just in addition to the other answers:  $z \mapsto \frac{z+1}{z-1}$ is a Möbius transform, and Möbius transforms have the general property that they map circles to lines or circles. Since the function maps $-1$ to $0$, $i$ to $-i$ and $-i$ to $i$, the image of the circle $|z| = 1$ has to be a line, the imaginary line.
A: Let $z=e^{it}$ then 
$$\frac{e^{it}+1}{e^{it}-1} = \frac{e^{it}+1}{e^{it}-1}\cdot \frac{e^{-it}-1}{e^{-it}-1}= \frac{-2i\sin t}{2-2 \cos t}=\frac{i\sin t}{\cos t-1}$$
A: You could try looking at it geometrically:
"Suppose $z$ lies on the unit circle. Consider the line $l_1$ from $-1$ to $z$ and $l_2$ from $1$ to $z$. Are $l_1$ and $l_2$ orthogonal?"
I'll leave you to figure out how this is equivalent, and the right theorem from circle geometry to apply.
A: After reading the answers I thought about another way
$$\frac{z+1}{z-1}=\frac{\frac{1}{\overline{z}}+1}{\frac{1}{\overline{z}}-1}=\frac{1+\overline{z}}{1-\overline{z}}=-\frac{\overline{z}+1}{\overline{z}-1}$$
A: When you see $z+1$ or $z-1$, where $|z|=1$, it's usually convenient to take a square root of $z$.
Write $z=e^{2ix}$; then
$$
\frac{z+1}{z-1}=\frac{e^{2ix}+1}{e^{2ix}-1}=
\frac{e^{ix}}{e^{ix}}\frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}=
\frac{2\cos x}{2i\sin x}=-i\cot x
$$
A: The geometric interpretation of
$\,w=(z-1)/(z+1)\,$ is that $\,z\,$ corresponds
a point on the unit circle, $\,z-1\,$ to
a chord from the $\,z\,$ point to the $1$
point, and $\,z+1\,$ to a chord to the $-1$
point, respectively. The argument of $\,w\,$ is the angle
between the two chords, but Thales's theorem states that
the diameter from $-1$ to $1$ and the
two chords form a right triangle with the
right angle at the point $\,z.\,$
The right angle is $\,90^\circ\,$ which corresponds to
an argument of a purely imaginary complex
number on the positive $\,y$-axis of the Cartesian
plane. This is the case if the real part of $\,z\,$
is positive. If the real part of $\,z\,$ is negative,
then the imaginary part of $\,w\,$ is also negative.
In other words, the imaginary parts of $\,z\,$ and
$\,w\,$ have the same signs.
In general, if $\,|z_1| = |z_2| \ne 0\,$ then
$\,(z_1-z_2)/(z_1+z_2)\,$ is purely imaginary and
this is the complex number equivalent of
Thales's theorem.
