A non-real complex number z has modulus |z| = 2, and the three points…

Question

A non-real complex number z has modulus |z| = 2, and the three points z, conjugate(z), and 1 form the vertices of an equilateral triangle in the complex plane. Writing z = x + iy, there are exactly two possible values of the real part x. Compute 4*(x_1^2 + x_2^2), where x_1 and x_2 are those two values. Report the exact integer.

Accepted Answer

Let z = x + iy with y != 0. The three vertices are P = (x, y), Q = (x, -y) (its conjugate), and R = (1, 0). The triangle is equilateral iff all three sides are equal. By symmetry |R - P| = |R - Q| automatically, so the condition reduces to |P - Q| = |R - P|. Here |P - Q| = 2|y| and |R - P|^2 = (x - 1)^2 + y^2. Setting (2|y|)^2 = (x - 1)^2 + y^2 gives 4y^2 = (x - 1)^2 + y^2, i.e. (x - 1)^2 = 3y^2. Now impose |z| = 2: x^2 + y^2 = 4, so y^2 = 4 - x^2. Substitute: (x - 1)^2 = 3(4 - x^2), i.e. x^2 - 2x + 1 = 12 - 3x^2, giving 4x^2 - 2x - 11 = 0. The two roots x_1, x_2 satisfy x_1 + x_2 = 1/2 and x_1 x_2 = -11/4, both lying in (-2, 2) so both are valid. Then x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2 x_1 x_2 = 1/4 + 11/2 = 23/4. Hence 4(x_1^2 + x_2^2) = 23.
Final answer: 23

Final answer: 23

A non-real complex number z has modulus |z| = 2, and the three points…

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