Vertices of Equilateral Triangle in Complex Plane/Necessary Condition
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Theorem
Let $z_1$, $z_2$ and $z_3$ be complex numbers.
Let $z_1$, $z_2$ and $z_3$ fulfil the condition:
- ${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
Then $z_1$, $z_2$ and $z_3$ represent on the complex plane the vertices of an equilateral triangle.
Proof
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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $139$