If z is a complex number such that z2 = (¯z)2,then
Either z is purely real or purely imaginary
Let z = x+iy, then its coonjucate ¯z = x-iy
Given that z2=(¯z)2
⇒ x2 - y2 + 2ixy = x2 - y2 - 2ixy ⇒ 4ixy = 0
if x ≠ 0 then y = 0 and if y ≠ 0 then x = 0