Step 1: Foundational Condition:
For three complex numbers \(z_1, z_2, z_3\) to form an equilateral triangle, the relation \( z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \) must hold. For this problem, with vertices \(z_1\), \(z_2\), and the origin (\(z_3 = 0\)), the condition simplifies.
Step 2: Rotational Approach:
An alternative method utilizes the geometric property of rotation. If the origin, \(z_1\), and \(z_2\) form an equilateral triangle, then \(z_2\) is equivalent to \(z_1\) rotated around the origin by an angle of \(\pm 60^\circ\) (\(\pm \pi/3\) radians).
\[ z_2 = z_1 e^{\pm i\pi/3} \]
Step 3: Derivation Details:
Based on the rotational property:
\[ \frac{z_2}{z_1} = e^{\pm i\pi/3} = \cos(\pi/3) \pm i\sin(\pi/3) = \frac{1}{2} \pm i\frac{\sqrt{3}}{2} \]
Let \( \omega = \frac{z_2}{z_1} \). These values are roots of a quadratic equation.
Sum of roots: \( (\frac{1}{2} + i\frac{\sqrt{3}}{2}) + (\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 1 \).
Product of roots: \( (\frac{1}{2} + i\frac{\sqrt{3}}{2})(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = (\frac{1}{2})^2 - (i\frac{\sqrt{3}}{2})^2 = \frac{1}{4} - (-\frac{3}{4}) = 1 \).
The general form of a quadratic equation is \( \omega^2 - (\text{sum of roots})\omega + (\text{product of roots}) = 0 \).
\[ \omega^2 - \omega + 1 = 0 \]
Substituting \( \omega = \frac{z_2}{z_1} \) back:
\[ \left(\frac{z_2}{z_1}\right)^2 - \left(\frac{z_2}{z_1}\right) + 1 = 0 \]
Multiplying by \(z_1^2\) (given \(z_1 eq 0\)):
\[ z_2^2 - z_1 z_2 + z_1^2 = 0 \]
Rearranging yields:
\[ z_1^2 + z_2^2 - z_1 z_2 = 0 \]
Step 4: Concluding Relation:
The condition for the vertices 0, \(z_1\), and \(z_2\) to constitute an equilateral triangle is \( z_1^2 + z_2^2 - z_1 z_2 = 0 \).