UPSC NDA I 2017 Mathematics Solved Paper

Concept:Represent the complex number as $z = x + iy$ and its conjugate as $\overline{z} = x - iy$.Then equate the real and imaginary parts after substitution.Explanation:Let $z = x + iy$, so $\overline{z} = x - iy$.The given equation $z^{2} = 2\overline{z}$ becomes $(x+iy)^{2}=2(x-iy)$.Expand: $(x^{2}-y^{2}) + 2xy i = 2x - 2y i$.Now equate real and imaginary parts:Real: $x^{2} - y^{2} = 2x$ …(1)Imaginary: $2xy = -2y$ …(2)From (2): $2xy + 2y = 0 \Rightarrow 2y(x+1)=0$, so $y=0$ or $x=-1$.Case 1: $y=0$ in (1) gives $x^{2}=2x \Rightarrow x(x-2)=0$, so $x=0$ or $x=2$.Thus $z=0$ and $z=2$ are solutions.Case 2: $x=-1$ in (1) gives $(-1)^{2}-y^{2}=2(-1) \Rightarrow 1-y^{2}=-2$, so $y^{2}=3$, hence $y= \pm \sqrt{3}$.Thus $z=-1 + \sqrt{3}i$ and $z=-1 - \sqrt{3}i$ are solutions.In total we have four distinct complex numbers satisfying the equation.Answer:The number of roots is 4 (Option C).