Question

If the complex numbers \(z_1\), \(z_2\), \(0\) are vertices of an equilateral triangle, then \[ z_1^2+z_2^2= \] is equal to:

• A. \(2z_1^2z_2^2\)
• B. \(z_1^2z_2^2\)
• C. \(2z_1z_2\)
• D. \(z_1z_2\)

Hint:

For equilateral triangle problems in the complex plane, use the rotation factor \[ \omega=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}. \] This converts geometric relations into algebraic identities.

Answer 1

The Correct Option is D

Step 1: Use the equilateral triangle rotation property.
If \(0, z_1, z_2\) form an equilateral triangle, one vertex is obtained from another by rotation of \(60^\circ\): \(z_2=z_1 e^{\pm i\pi/3}\).

Step 2: Compute z_1^2 + z_2^2.
\(z_1^2+z_2^2=(z_1+z_2)^2-2z_1z_2\). Since \(z_2=z_1 e^{i\pi/3}\), we get \(z_1 z_2=z_1^2 e^{i\pi/3}\) and \(z_1+z_2=z_1(1+e^{i\pi/3})\). A cleaner route: the equilateral condition gives \(z_1^2+z_2^2-z_1z_2=0\), so \(z_1^2+z_2^2=z_1z_2\). \[ \boxed{z_1z_2} \]