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.
Best Answer
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 $z\mapsto \operatorname{Re}(z)$
If you only care about what happens to the unit circle, you can follow each transformation.