Question

If the complex numbers \( z_1, z_2 \) and \( z_3 \) denote the vertices of an isosceles triangle, right angled at \( z_1 \), then \( (z_1 - z_2)^2 + (z_1 - z_3)^2 \) is equal to:

• A. \( 0 \)
• B. \( (z_2 + z_3)^2 \)
• C. \( 2 \)
• D. \( 3 \)
• E. \( (z_2 - z_3)^2 \)

Hint:

For any right-angled triangle at $z_1$, the ratio of $(z_3-z_1)/(z_2-z_1)$ is purely imaginary ($ki$). If it's isosceles, $k = \pm 1$.

Answer 1

The Correct Option is A

Concept: In the complex plane, a rotation of a vector by an angle \( \theta \) is achieved by multiplying by \( e^{i\theta} \). For an isosceles right-angled triangle at \( z_1 \), the vector \( z_3 - z_1 \) is the vector \( z_2 - z_1 \) rotated by \( \pm \pi/2 \) with equal magnitude.

Step 1: Express the relationship using rotation.
Since the sides are equal and the angle is \( 90^\circ \): \[ (z_3 - z_1) = (z_2 - z_1) \cdot e^{\pm i\pi/2} \] We know \( e^{i\pi/2} = i \) and \( e^{-i\pi/2} = -i \). \[ (z_3 - z_1) = \pm i(z_2 - z_1) \]

Step 2: Square both sides.
\[ (z_3 - z_1)^2 = [\pm i(z_2 - z_1)]^2 \] \[ (z_3 - z_1)^2 = i^2 (z_2 - z_1)^2 \] \[ (z_3 - z_1)^2 = -1 \cdot (z_2 - z_1)^2 = -(z_1 - z_2)^2 \]

Step 3: Rearrange to find the sum.
\[ (z_1 - z_2)^2 + (z_3 - z_1)^2 = 0 \] Since \( (z_3 - z_1)^2 = (z_1 - z_3)^2 \): \[ (z_1 - z_2)^2 + (z_1 - z_3)^2 = 0 \]