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 property of an equilateral triangle.
Since \(0\), \(z_1\), and \(z_2\) are vertices of an equilateral triangle, the angle between the vectors represented by \(z_1\) and \(z_2\) is \[ \frac{\pi}{3} \] Hence, \[ z_2=z_1\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right) \] or \[ z_2=z_1\omega \] where \[ \omega=\frac{1+i\sqrt3}{2} \] and \[ \omega^2-\omega+1=0 \]

Step 2: Express the required quantity.
\[ z_1^2+z_2^2 = z_1^2+z_1^2\omega^2 \] \[ = z_1^2(1+\omega^2) \] Using \[ \omega^2-\omega+1=0 \] we get \[ 1+\omega^2=\omega \] Therefore, \[ z_1^2+z_2^2 = z_1^2\omega \]

Step 3: Find \(z_1z_2\).
\[ z_1z_2 = z_1(z_1\omega) \] \[ = z_1^2\omega \] Thus, \[ z_1^2+z_2^2=z_1z_2 \]

Step 4: Final conclusion.
Hence, \[ \boxed{z_1z_2} \]