Question

The complex number \(z_1, z_2\) and origin, form an equilateral triangle only if:

• A. \( z_1^2 - z_2^2 + z_1 z_2 = 0 \)
• B. \( z_1^2 + z_2^2 - z_1 z_2 = 0 \)
• C. \( z_1^2 + z_2^2 + 3z_1 z_2 = 0 \)
• D. \( z_1^2 + z_2^2 + z_1 - z_2 = 0 \)

Hint:

Remember the general condition for three points \(z_1, z_2, z_3\) forming an equilateral triangle: \(z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1\). If one of the points is the origin, you can simply set it to zero in this formula to get the required condition quickly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For three complex numbers \(z_1, z_2, z_3\) to be the vertices of an equilateral triangle, they must satisfy the condition \( z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \). In this problem, the vertices are \(z_1\), \(z_2\), and the origin, so we take \(z_3 = 0\).
Step 2: Key Formula or Approach:
An alternative approach is using the geometric property of rotation. If 0, \(z_1\), and \(z_2\) form an equilateral triangle, then \(z_2\) can be obtained by rotating \(z_1\) about the origin by an angle of \(\pm 60^\circ\) (\(\pm \pi/3\) radians).
\[ z_2 = z_1 e^{\pm i\pi/3} \] Step 3: Detailed Explanation:
From the rotation 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 \( \omega = \frac{z_2}{z_1} \). We can form a quadratic equation with these roots. 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 quadratic equation is \( \omega^2 - (\text{sum})\omega + (\text{product}) = 0 \).
\[ \omega^2 - \omega + 1 = 0 \] Substitute back \( \omega = \frac{z_2}{z_1} \):
\[ \left(\frac{z_2}{z_1}\right)^2 - \left(\frac{z_2}{z_1}\right) + 1 = 0 \] Multiply the entire equation by \(z_1^2\) (since \(z_1 \neq 0\)):
\[ z_2^2 - z_1 z_2 + z_1^2 = 0 \] Rearranging this gives:
\[ z_1^2 + z_2^2 - z_1 z_2 = 0 \] Step 4: Final Answer:
The condition for the vertices 0, \(z_1\), and \(z_2\) to form an equilateral triangle is \( z_1^2 + z_2^2 - z_1 z_2 = 0 \).