Question

Let \( a = \cos \theta_1 + i \sin \theta_1 \), \( b = \cos \theta_2 + i \sin \theta_2 \), \( c = \cos \theta_3 + i \sin \theta_3 \) and \( a + b + c = 0 \), then \[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = ? \]

• A. 1
• B. -1
• C. \(\sqrt{2}\)
• D. 0

Answer 1

The Correct Option is D

Given \( a + b + c = 0 \), take reciprocal: \[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + bc + ca}{abc} \] Now since \(a + b + c = 0 \Rightarrow (a + b + c)^2 = 0 \Rightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0\) Since \( |a| = |b| = |c| = 1 \Rightarrow a^2 + b^2 + c^2 \in \mathbb{C} \), but more importantly: We get \( ab + bc + ca = -\frac{1}{2}(a^2 + b^2 + c^2) \) Thus numerator becomes zero, and so: \[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \]