Unit Vectors in Complex Plane which are Vertices of Equilateral Triangle

Theorem

Let $\epsilon_1, \epsilon_2, \epsilon_3$ be complex numbers embedded in the complex plane such that:

$\epsilon_1, \epsilon_2, \epsilon_3$ all have modulus $1$

$\epsilon_1 + \epsilon_2 + \epsilon_3 = 0$

Then:

$\paren {\dfrac {\epsilon_2} {\epsilon_1} }^3 = \paren {\dfrac {\epsilon_3} {\epsilon_2} }^2 = \paren {\dfrac {\epsilon_1} {\epsilon_3} }^2 = 1$

Proof

We have that:

\(\ds \epsilon_1 + \epsilon_2 + \epsilon_3\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \epsilon_1 - \paren {-\epsilon_2}\)

\(=\)

\(\ds -\epsilon_3\)

Thus by Geometrical Interpretation of Complex Subtraction, $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ form the sides of a triangle.

As the modulus of each of $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ equals $1$, $\triangle \epsilon_1 \epsilon_2 \epsilon_3$ is equilateral.

By Complex Multiplication as Geometrical Transformation:

\(\ds \arg \epsilon_1\)

\(=\)

\(\ds \arg \epsilon_2 \pm \dfrac {2 \pi} 3\)

Complex Multiplication as Geometrical Transformation

\(\ds \leadsto \ \ \)

\(\ds \arg \paren {\dfrac {\epsilon_2} {\epsilon_1} }\)

\(=\)

\(\ds \pm \dfrac {2 \pi} 3\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\epsilon_2} {\epsilon_1}\)

\(=\)

\(\ds \cos \dfrac {2 \pi} 3 \pm i \sin \dfrac {2 \pi} 3\)

\(\ds \)

\(=\)

\(\ds \omega \text { or } \overline \omega\)

where $\omega$, $\overline \omega$ are the complex cube roots of unity

\(\ds \leadsto \ \ \)

\(\ds \paren {\dfrac {\epsilon_2} {\epsilon_1} }^3\)

\(=\)

\(\ds 1\)

The same analysis can be done to the other two pairs of sides of $\triangle \epsilon_1 \epsilon_2 \epsilon_3$.

Hence the result.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Example $2$.