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Sum of Squares of Square Vertices in Complex Plane | JEE

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

Question
If z1,z2,z3z_1, z_2, z_3 and z4z_4 are the consecutive vertices of a square, then z12+z22+z32+z42z_1^2 + z_2^2 + z_3^2 + z_4^2 equals
Az1z2+z2z3+z3z4+z4z1z_1 z_2 + z_2 z_3 + z_3 z_4 + z_4 z_1correct
Bz1z2+z2z3+z1z4+z2z4+z3z4z_1 z_2 + z_2 z_3 + z_1 z_4 + z_2 z_4 + z_3 z_4
C00
DNone of the above
Solution
Step 1: For a square with consecutive vertices, the side and diagonal relations give z4−z1=i(z2−z1)z_4 - z_1 = i(z_2 - z_1) and the analogous rotations. These reduce to:
(z4−z1)2+(z2−z1)2=0(z_4 - z_1)^2 + (z_2 - z_1)^2 = 0
 Step 2: Expanding the square conditions across all vertices and summing yields:
z12+z22+z32+z42=z1z2+z2z3+z3z4+z4z1z_1^2 + z_2^2 + z_3^2 + z_4^2 = z_1 z_2 + z_2 z_3 + z_3 z_4 + z_4 z_1
 Verification with the unit square 0,1,1+i,i0, 1, 1+i, i: LHS =0+1+2i+(−1)=2i= 0 + 1 + 2i + (-1) = 2i and RHS =0+(1+i)+(i−1)+0=2i= 0 + (1+i) + (i-1) + 0 = 2i. Correct answer: (1)
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