Proof that unit complex numbers 1, z and w form an equilateral triangle

Question

Let $w$ and $z$ be unit complex numbers such that $Arg(z) = \theta$. Knowing that $1+z+w=0$, show that the geometrical images of $1,z$ and $w$ form an equilateral triangle.

This should be easy but I do not find a solution within your previous questions. I tried to measure the distance between the vertices.

Knowing that they are all unit complex numbers, we have $|z| = |w| = |1| = 1$. Knowing that $Arg(z) = \theta$, we can say immediately that $z= e^{i\theta} = \cos \theta + i \sin \theta$.

We also know that

$1+z+w=0 <=> w=-1-z$

Therefore, we have

$ |z-w| = |z-(-1-z)| = |z+1+z| = |2z+1| = |2(\cos \theta + i \sin \theta) + 1| = |(2 \cos \theta + 1) + i (2\sin \theta) | = \sqrt{(2 \cos \theta +1)^2 + 4 \sin^2 \theta} = \sqrt{4 \cos^2 \theta + 4 \cos \theta + 1 + 4 \sin^2 \theta } = \sqrt{4 (\cos^2 \theta + \sin^2 \theta) +4 \cos \theta + 1} = \sqrt{4 + 1 + 4 \cos \theta} = \sqrt{5+4 \cos \theta} $

$ |w-1| = |-1-z-1| = |-2-z| = |-2 - \cos \theta - i \sin \theta| = \sqrt{(-2- \cos \theta)^2 + (-\sin \theta)^2} = \sqrt{4 + 4 \cos \theta + \cos^2 \theta + sin^2 \theta} = \sqrt{4 + 4 \cos \theta + 1} = \sqrt{5 + 4 \cos \theta}$

So far so good, I have two equal sides. We have $|z-1|$ left though,

$ |z-1| = |\cos \theta + i \sin \theta - 1| = | \cos \theta - 1 + i \sin \theta| = \sqrt{(cos \theta - 1)^2 + \sin^2 \theta} = \sqrt{\cos^2 \theta - 2 \cos \theta + 1 + \sin^2 \theta} = \sqrt{\cos^2 \theta + \sin^2 \theta + 1 - 2 \cos \theta} = \sqrt{1 + 1 - 2 \cos \theta} = \sqrt{2 - 2 \cos \theta} $

Therefore, this side appears to have different length than the other two, which makes of this an isosceles triangle, not equilateral. Why is this rationale wrong and how could I prove this using only these data?

Thank you very much.

Answer 1

An easy computation shows that $1=|w|^2=|1+z|^2=2 +2 \cos \theta$, hence $\cos \theta=-1/2$. It follows that $5+ 4 \cos \theta=2-2 \cos \theta$ and everything is fine !

Answer 2

It is sufficient but a bit tedious to explicitly calculate the real and imaginary parts.

Write $w=a+bi$ and $z=c+di$.

Then from $1+w+z=0$ we have $a+c=-1$ and $b+d=0$ so $b=-d$.

As these are unit complex numbers $a^2+b^2=1=c^2+d^2$

Since $b=-d$ we have $a^2+(-d)^2=c^2+d^2$ hence $a^2=c^2$ so $a=\pm c$.

As $a+c=-1$, $a=-c$ is impossible. Thus $a=c=-1/2$.

Since $a^2+b^2=1$ substituting $a=-1/2$ gives $b^2 = 3/4$ so $b=\pm \sqrt{3}/2$.

The points $1, -1/2 + \sqrt{3}/2i, -1/2 -\sqrt{3}/2i$ are exactly those on the equilateral triangle, which you can prove by calculating the distances between them or evaluating $e^{\pm 2\pi/3}$.

We see the solution using pure complex numbers (Fred's) is much cleaner.

Answer 3

The geometrical images of $\, 1, z, w \,$ form a triangle. We are given that $\, |1| = |z| = |w|,\,$ thus the unit circle is the circumcircle of this triangle and $0$ is the circumcenter. We are given that $\, 1 + z + w = 0,\,$ thus the centroid is also $0$. From the Wikipedia article Euler line

[...] the orthocenter, circumcenter and centroid are collinear. [...] In equilateral triangles, these four points coincide, but in any other triangle they are all distinct from each other.

The circumcenter and centroid coincide, thus the triangle is equilateral.