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It is a well-known fact that the points $1,\omega,\omega^2$ represent an equilateral triangle in the complex plane (where $\omega, \omega^2$ are the complex cube roots of unity). $$ \omega=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2} $$ We observe that the origin is the centroid of the triangle formed by these points. To shift this triangle to another location in the complex plane, we can simply shift the centroid (let it be $z_0)$. So, the vertices become :

$$ z_0+1, z_0+\omega, z_o+\omega^2 $$

Now, considering an equilateral triangle of any side length, we introduce another parameter $a$ which is a complex number as it adds the element of rotation to our equilateral triangle besides acting as a parameter of the side length. So, our final result becomes :

$$ z_0+a, z_0+a\omega, z_o+a\omega^2 $$ where both $z_0$ and $a$ are complex numbers.

By our reasoning, this should represent any equilateral triangle in the complex plane... I was just wondering if this reasoning is accurate and if these vertices do represent all equilateral triangles in the complex plane ?

Vishnu V.S
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    http://math.stackexchange.com/questions/481133/equilateral-triangle-in-complex-plane?rq=1 says

    Prove that the points $a_1,a_2,a_3$ are vertices of an equilateral triangle if and only if $a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_3a_1$.

    with simple calculation we see $z_0+a, z_0+a\omega, z_o+a\omega^2$ satisfy this condition, so they are vertices of a equilateral triangle. Also [http://math.stackexchange.com/questions/102458/how-does-this-equality-on-vertices-in-the-complex-plane-imply-they-are-vertices] shows all equilateral triangles in the complex plane are with this form.

     – Nosrati Mar 27 '17 at 16:57

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