Vertices of Equilateral Triangle in Complex Plane/Sufficient Condition

Theorem

Let $z_1$, $z_2$ and $z_3$ be complex numbers.

Let $z_1$, $z_2$ and $z_3$ represent on the complex plane the vertices of an equilateral triangle.

Then:

${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$

Corollary

Let $u, v \in \C$ be complex numbers.

Then:

$0$, $u$ and $v$ represent on the complex plane the vertices of an equilateral triangle.

if and only if:

$u^2 + v^2 = u v$

Proof

Let $T$ be the equilateral triangle whose vertices are $z_1$, $z_2$ and $z_3$.

We have that $z_2 - z_1$ and $z_3 - z_1$ are two sides of $T$ which meet at $z_1$.

From the geometry of $T$ it follows that $z_2 - z_1$ is at an angle of $\pi/3$ to $z_3 - z_1$.

Similarly, $z_1 - z_3$ and $z_2 - z_3$ are two sides of $T$ which meet at $z_3$.

From the geometry of $T$ it follows that $z_1 - z_3$ is at an angle of $\pi / 3$ to $z_2 - z_3$.

From Complex Multiplication as Geometrical Transformation/Corollary:

$(1): \quad z_2 - z_1 = e^{i \pi / 3} \left({z_3 - z_1}\right)$

$(2): \quad z_1 - z_3 = e^{i \pi / 3} \left({z_2 - z_3}\right)$

Then:

\(\ds \dfrac {z_2 - z_1} {z_1 - z_3}\)

\(=\)

\(\ds \dfrac {z_3 - z_1} {z_2 - z_3}\)

$(1)$ divided by $(2)$

\(\ds \leadsto \ \ \)

\(\ds \paren {z_2 - z_1} \paren {z_2 - z_3}\)

\(=\)

\(\ds \paren {z_3 - z_1} \paren {z_1 - z_3}\)

\(\ds \leadsto \ \ \)

\(\ds {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1\)

\(=\)

\(\ds - {z_1}^2 - {z_3}^2 + 2 z_3 z_1\)

\(\ds \leadsto \ \ \)

\(\ds {z_1}^2 + {z_2}^2 + {z_3}^2\)

\(=\)

\(\ds z_1 z_2 + z_2 z_3 + z_3 z_1\)

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $51$