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|（z1+z2)/(z1−z2)|=1 Means z1/z2 Purely Imaginary | JEE

JEE Maths question with a full step-by-step solution.

Question
If z1z_1 and z2z_2 are two complex numbers satisfying ∣z1+z2z1−z2∣=1\left|\dfrac{z_1 + z_2}{z_1 - z_2}\right| = 1, then z1z2\dfrac{z_1}{z_2} is a number which is
Apositive real
Bnegative real
Czero or purely imaginarycorrect
Dnone of these
Solution
Step 1: Clear the modulus. The condition gives ∣z1+z2∣=∣z1−z2∣|z_1 + z_2| = |z_1 - z_2|. Step 2: Divide numerator and denominator inside by z2z_2 (assume z2≠0z_2 \neq 0) and let w=z1z2w = \dfrac{z_1}{z_2}:
∣z1+z2z1−z2∣=∣w+1w−1∣=1  ⟹  ∣w+1∣=∣w−1∣\left|\frac{z_1 + z_2}{z_1 - z_2}\right| = \left|\frac{w + 1}{w - 1}\right| = 1 \implies |w + 1| = |w - 1|
 Step 3: Interpret geometrically. ∣w+1∣=∣w−1∣|w + 1| = |w - 1| says ww is equidistant from the points −1-1 and +1+1 on the real axis. The set of points equidistant from −1-1 and 11 is their perpendicular bisector — the imaginary axis. Step 4: Confirm algebraically. Let w=p+iqw = p + iq. Then:
∣w+1∣2=(p+1)2+q2,∣w−1∣2=(p−1)2+q2|w + 1|^2 = (p + 1)^2 + q^2, \qquad |w - 1|^2 = (p - 1)^2 + q^2
 Setting them equal:
(p+1)2=(p−1)2  ⟹  4p=0  ⟹  p=0(p + 1)^2 = (p - 1)^2 \implies 4p = 0 \implies p = 0
 So w=iqw = iq is purely imaginary (or zero when q=0q = 0). Therefore z1z2\dfrac{z_1}{z_2} is zero or purely imaginary. Correct answer: (3)
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