Graphically showing $\operatorname{Re}\left(\frac{1+z}{1-z}\right) = 0$ for $z=\cos\theta+i\sin\theta\neq 1$ 
The given complex number is $z = \cos\theta + i\sin\theta$, where $z$ is not $1$
I have to show that $$\operatorname{Re}\left(\frac{1+z}{1-z}\right) = 0$$

Algebraically, I have managed to do that using trigonometry and Euler's equation. But I can't succeed to imagine it on an Argand diagram using basic angle rules, without solving anything. In this case division, so angle subtraction. The angle should either be $90^\circ$ or $270^\circ$ since it should end up only on the complex part, I tried drawing it on the whiteboard but perhaps I am doing something wrong since I cannot get it to add up. Conceptually this is right.
 A: One way could be to note that, not just this problem, but any linear fractional transformation $$z\mapsto\frac{az+b}{cz+d}$$ can always be seen as a composition of inversions, reflections, translations, and homotheties. You only need to reduce the fraction, to see the composition of these transformations.
$$\frac{1+z}{1-z}=1+\frac{2z}{1-z}=1+\frac{2}{1/z-1}$$
Therefore, to get the left hand side, we are composing $z\mapsto 1/z$, $z\mapsto z-1$, $z\mapsto 1/z$, $z\mapsto 2z$, $z\mapsto 1+z$, and $z\mapsto \operatorname{Re}(z)$

*

*The map $z\mapsto 1/z$ is an inversion with respect to the unit circle, followed by a reflection with respect to the real axis.

*The map $z\mapsto z-1$ is a translation one unit to the left.

*The map $z\mapsto 2z$ is a homothety with center $0$ and ratio $2$.

*The map $z\mapsto 1+z$ is a translation one unit to the right.

*The map $z\mapsto\operatorname{Re}(z)$ is the orthogonal projection to the real axis.

If you only care about what happens to the unit circle, you can follow each transformation.

*

*The inversion of the unit circle with respect to the unit circle is the identity. The reflection with respect to the real axis, well, flips the circle.

*The translation one unit to the left you know.

*Next there is another inversion with respect to the unit circle, but since now the circle passes through the origin, it becomes a straight line perpendicular to the real axis and passing through $-1/2$.

*The homothety with ratio $2$, that will turn the line to another vertical line, but passing through $-1$.

*There is a one unit translation to the right. This brings the vertical line to the imaginary axis.

*Finally, the orthogonal projection to the real axis maps this vertical line to $0$.

A: Here is perhaps another geometric intuition.
Take a complex number $z\neq 1$ on the unit circle, note $-z$ is its reflection across the origin, and adding $1$ translates them to the right by one unit. And to show $Re(\frac{1+z}{1-z})=0$ is to show the angle difference of $1+z$ and $1-z$ is $90$ or $270$ degrees.
Let us visualize it by taking a $z$ in the first quadrant for illustration:

Note $z,1+z,1,0$ and $0,-z,1-z,1$ forms two congruent parallelograms, and the two angles that you are looking for are $a$ and $b$, which are complementary, and hence $90$ degrees (after meditating on subtracting negative angles).
A: Given a complex number $z$ where $|z|=1$ and $z\ne\pm 1\,$ then the three points $\,z,1,-1\,$ form a right triangle by
Thales theorem.
The right angle is at $z$, the hypoteneuse is the segment $[-1,1]$ and the segment from $-1$ to $z$ and segment from $z$ to $1$
form the legs of the triangle. This implies that the quotient $\,\frac{1+z}{1-z}\,$ makes an angle of $90^\circ$
with the $x$-axis and has real part $0$. If $\,z=-1\,$ then the quotient is $0$ with real part $0$.
This can also be visualized by adding a fourth point $-z$ to form
a rectangle with a right angle at $-1$. Now a rigid motion
that takes the points  $-1,z,-z$ to $0,1+z,1-z$ and preserves the
right angle which is formed now at the origin by $1+z$ and $1-z$
as in another answer.
A: This is more a comment on @bonsoon's answer, remarking that $1+z$ and $1-z$ are the diagonals of the upper parallelogram in their figure, but ended up being too long:
consider the points $A$ and $B$ corresponding to $1$ and $z$ in the complex plane, and construct the parallelogram $OACB$ where $O$ is the origin. Then $C$ corresponds to $1+z$.
Since $|z|=1$, the parallelogram is actually a rhombus, so its diagonals $OC$ and $AB$ are perpendicular. But $\vec{AB}=\vec{OB}-\vec{OA}$ is parallel to the vector $\vec{OD}$ where $D$ is the point corresponding to $z-1$.
As a consequence, $\arg(1+z)$ and $\arg(z-1)$ differ by $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ (mod $2\pi$), so $Re\left(\frac{1+z}{z-1}\right)=Re\left(\frac{1+z}{1-z}\right)=0$.
(Alternatively, you could consider the parallelogram $ODEC$ where $E$ corresponds to $z-1+1+z=2z$ and remark that its diagonal have the same length).
