Vertices of Equilateral Triangle in Complex Plane

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Theorem

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


Then:

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

if and only if:

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


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

EquilateralTriangleInComplexPlane.png


Sufficient Condition

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 $(2)$
\(\ds \leadsto \ \ \) \(\ds \paren {z_2 - z_1} \paren {z_2 - z_3}\) \(=\) \(\ds \paren {z_3 - z_1} \paren {z_1 - z_3}\)
\(\ds \leadsto \ \ \) \(\ds {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1\) \(=\) \(\ds - {z_1}^2 - {z_3}^2 + 2 z_3 z_1\)
\(\ds \leadsto \ \ \) \(\ds {z_1}^2 + {z_2}^2 + {z_3}^2\) \(=\) \(\ds z_1 z_2 + z_2 z_3 + z_3 z_1\)

$\Box$


Necessary Condition

Let:

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

Then:

\(\ds {z_1}^2 + {z_2}^2 + {z_3}^2\) \(=\) \(\ds z_1 z_2 + z_2 z_3 + z_3 z_1\)
\(\ds \leadsto \ \ \) \(\ds {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1\) \(=\) \(\ds - {z_1}^2 - {z_3}^2 + 2 z_3 z_1\)
\(\ds \leadsto \ \ \) \(\ds \paren {z_2 - z_1} \paren {z_2 - z_3}\) \(=\) \(\ds \paren {z_3 - z_1} \paren {z_1 - z_3}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {z_2 - z_1} {z_1 - z_3}\) \(=\) \(\ds \dfrac {z_3 - z_1} {z_2 - z_3}\)

Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_3$ and $z_2 - z_1$.


Similarly:

\(\ds {z_1}^2 + {z_2}^2 + {z_3}^2\) \(=\) \(\ds z_1 z_2 + z_2 z_3 + z_3 z_1\)
\(\ds \leadsto \ \ \) \(\ds - {z_2}^2 - {z_1}^2 + 2 z_1 z_2\) \(=\) \(\ds {z_3}^2 - z_1 z_3 - z_2 z_3 + z_1 z_2\)
\(\ds \leadsto \ \ \) \(\ds \paren {z_2 - z_1} \paren {z_1 - z_2}\) \(=\) \(\ds \paren {z_3 - z_1} \paren {z_3 - z_2}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {z_2 - z_1} {z_3 - z_2}\) \(=\) \(\ds \dfrac {z_3 - z_1} {z_1 - z_2}\)

Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_2$ and $z_3 - z_2$.


Thus all three angles:

$\angle z_2 z_1 z_3$
$\angle z_1 z_3 z_2$
$\angle z_3 z_2 z_1$

are equal.


By definition, therefore, $\triangle z_1 z_2 z_3$ is equilateral.

$\blacksquare$


Sources

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