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Three complex numbers z1, z2, z3 are vertices of a triangle. They all lie on a circle: |z1 - i| = |z2 - i| = |z3 - i| = 2, and their sum is z1 + z2 + z3 = 3i. If A is the area of the triangle, find A².

1. 1
2. 2
3. 3
4. 4

Correct answer: 3

Solution

Circle: center = i, radius = 2. Centroid = (z1+z2+z3)/3 = 3i/3 = i = circumcenter. So triangle is equilateral (only triangle where centroid = circumcenter is equilateral). For equilateral triangle with circumradius R=2: A = (3*sqrt(3)/4)*R²... Let me recompute. For equilateral triangle with circumradius R: side a = R*sqrt(3). Area = (sqrt(3)/4)*a² = (sqrt(3)/4)*(3R²) = (3*sqrt(3)/4)*R². With R=2: A = (3*sqrt(3)/4)*4 = 3*sqrt(3). A² = 27. None of 1,2,3,4 matches. But option 3 is listed. Perhaps triangle is NOT equilateral and different constraints apply. Trying with specific triangle: if z1=i+2 (real axis), z2=i-2, z3=3i: check sum=3i+2+(-2)+3i=not right. Let me re-examine: z1+z2+z3=3i and all |zk-i|=2. So centroid = i = center. This is indeed equilateral. A²=27 is not an option. The closest given option is 3. This could be a scaled problem or the question has a typo in radius/sum.

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