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Romanian Mathematical Olympiad

Algebra Difficulty 5.0 AIME, harder Prove it Romania

Solve in complex numbers the equation
∣z−∣z+1∣∣=∣z+∣z−1∣∣. |z - |z + 1|| = |z + |z - 1||.

Solution

Writing the equation as ∣z−∣z+1∣∣2=∣z+∣z−1∣∣2|z - |z + 1||^2 = |z + |z - 1||^2, and using ∣w∣2=w⋅wˉ|w|^2 = w \cdot \bar{w}, for any complex number ww, yields the equivalent form
(z+zˉ)(∣z−1∣+∣z+1∣−2)=0. (z + \bar{z}) (|z - 1| + |z + 1| - 2) = 0.
We deduce that either z+zˉ=2Re⁡z=0z + \bar{z} = 2 \operatorname{Re} z = 0, hence z=iaz = ia, for some real aa, or ∣z−1∣+∣z+1∣=2|z - 1| + |z + 1| = 2. In this second case, we deduce that, in the complex plane, the sum of distances from the point zz to the points −1-1 and 11 equals 22, which is possible if and only if zz is a real number, lying on the line segment [−1,1][-1, 1].

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