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.

  • 3
    $\begingroup$ $z$ and $-z$ would be at $\pi$ from each other, and then both are shifted to the right by $1$. Also, $|z|=1$, correct? So this seems to be saying that if you take the endpoints of any diameter on the unit circle centered at $1+0i$ and divide one by the other you end up on the imaginary line... $\endgroup$
    – abiessu
    Aug 15, 2022 at 13:42
  • $\begingroup$ Which matches up with the fact that taking a point on a circle and a diameter of that circle, connecting the point to the endpoints of the diameter gives a right triangle... $\endgroup$
    – abiessu
    Aug 15, 2022 at 13:45
  • $\begingroup$ Notice that you can equivalently state $\text{Re}\left(\frac{-1-z}{1-z}\right)=0$. The numerator is the segment connecting $z$ to $-1$ and the denominator is the segment connecting $z$ to $1$. The quotient being pure imaginary means that those two segments are orthogonal to each other. Now, what is the set of points $z$ such that the triangle with vertices $-1, 1, z$ has the right angle at $z$? $\endgroup$
    – user700480
    Aug 15, 2022 at 13:52

4 Answers 4


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.


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 $0$.
  • $\begingroup$ shouldn't the translation z-1 be one unit to the right? $\endgroup$ Aug 15, 2022 at 13:58
  • $\begingroup$ @TeodorasPaura It always depends on how you define your orientation. $\endgroup$
    – plop
    Aug 15, 2022 at 13:59
  • $\begingroup$ @TeodorasPaura For example, $5 + 0i$ gets mapped to $(5 + 0i) - 1 = 4 + 0i$. You're confusing $f(z - 1)$ with this mapping notation. $\endgroup$
    – Toby Mak
    Aug 15, 2022 at 23:48

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: enter image description here

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).


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.

  • $\begingroup$ +1. I didn't know that this theorem was called Thales' Theorem. Interestingly, Thales' Theorem in France is the name of a different theorem (about similar triangles: wikiwand.com/fr/Th%C3%A9or%C3%A8me_de_Thal%C3%A8s). $\endgroup$
    – Taladris
    Aug 16, 2022 at 4:19
  • 1
    $\begingroup$ @Taladris Thanks for that comment! Not only France but in Spain. There are two theorems of Thales. The English Wikipedia article does not even hint at the other theorem attributed to Thales. Now I know. $\endgroup$
    – Somos
    Aug 16, 2022 at 11:02

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).


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