UPSC NDA I 2016 Mathematics Solved Paper

Concept:We solve the equation $z^2 = i \bar{z}$ for non-zero complex numbers and then sum the resulting distinct solutions.Explanation:Let $z = x + iy$.Then $\bar{z} = x - iy$ and $z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$.The given equation becomes $x^2 - y^2 + 2ixy = i(x - iy) = ix + y$.Equate real and imaginary parts:Real: $x^2 - y^2 = y$ ...(1)Imaginary: $2xy = x$ → $x(2y-1)=0$ → either $x=0$ or $y=\frac{1}{2}$.If $x=0$, from (1): $-y^2 = y$ → $y(y+1)=0$ → $y=0$ or $y=-1$.Thus $(x,y)=(0,0)$ gives $z=0$, but $z$ is non-zero, so discard.$(x,y)=(0,-1)$ gives $z_1 = -i$.If $y=\frac{1}{2}$, from (1): $x^2 - \frac{1}{4} = \frac{1}{2}$ → $x^2 = \frac{3}{4}$ → $x = \pm\frac{\sqrt{3}}{2}$.Thus $z_2 = \frac{\sqrt{3}}{2} + i\frac{1}{2}$ and $z_3 = -\frac{\sqrt{3}}{2} + i\frac{1}{2}$.Now sum them:$z_1+z_2+z_3 = -i + \left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) + \left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)$= $(-i) + \left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) + i\left(\frac{1}{2}+\frac{1}{2}\right)$= $-i + 0 + i = 0$.Answer:$0$ (Option C).