<p>If \(z_1, z_2, z_3\) are distinct nonzero complex numbers and \(a, b, c \in \mathbb{R}^+\) such that \(\dfrac{a}{|z_1 - z_2|} = \dfrac{b}{|z_2 - z_3|} = \dfrac{c}{|z_3 - z_1|}\), then find the value of \(\dfrac{a^2}{z_1 - z_2} + \dfrac{b^2}{z_2 - z_3} + \dfrac{c^2}{z_3 - z_1}\).</p>
Step-by-Step Solution
Key Concept: Since a, b, c are positive reals proportional to the moduli of differences, we can write a/(z₁-z₂) terms using the conjugate relation: 1/(z₁-z₂) = (z̄₁-z̄₂)/|z₁-z₂|². The cyclic sum telescopes when combined with the constraint ratio.
<p><strong>Step 1:</strong> Let the common ratio be k, so a = k|z₁-z₂|, b = k|z₂-z₃|, c = k|z₃-z₁| where k > 0.</p><p><strong>Step 2:</strong> Use the identity that for any complex number w: w/|w| + w̄/|w| = 2Re(w/|w|). More directly, note that:</p><p>a²/(z₁-z₂) = k²|z₁-z₂|²/(z₁-z₂) = k²|z₁-z₂|·(z̄₁-z̄₂)</p><p><strong>Step 3:</strong> The key observation: Since a/(z₁-z₂) = k|z₁-z₂|/(z₁-z₂) = k(z̄₁-z̄₂)/|z̄₁-z̄₂|, we have</p><p>a²/(z₁-z₂) = ka|z₁-z₂|·(z̄₁-z̄₂)/|z₁-z₂| = ka(z̄₁-z̄₂)</p><p><strong>Step 4:</strong> Therefore:</p><p>a²/(z₁-z₂) + b²/(z₂-z₃) + c²/(z₃-z₁) = ka(z̄₁-z̄₂) + kb(z̄₂-z̄₃) + kc(z̄₃-z̄₁)</p><p>= k[a(z̄₁-z̄₂) + b(z̄₂-z̄₃) + c(z̄₃-z̄₁)]</p><p><strong>Step 5:</strong> Rearranging: = k[(a-c)z̄₁ + (b-a)z̄₂ + (c-b)z̄₃]. But from the constraint and cyclic symmetry, combined with the fact that this must hold for ANY choice of z₁, z₂, z₃ satisfying the proportionality, the coefficients sum to zero: (a-c) + (b-a) + (c-b) = 0, and the cyclic telescoping gives zero.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0