Q198 / 270 · NDA 2016 (Phase 1) · Mathematics

Let $z_1, z_2$ and $z_3$ be non-zero complex numbers satisfying $z^2=i \bar{z}$, where $i=\sqrt{-1}$. What is $z_1+z_2+z_3$ equal to?

  • A
    i
  • B
    –i
  • Correct answer
    C
    0
  • D
    1

Solution

Given $z^2=i \bar{z}$ Let us suppose that $z=x+i y$ $\begin{aligned} & \Rightarrow(x+i y)^2=i(x-i y) \\ & \Rightarrow x^2-y^2+2 x y i=i x+y \end{aligned}$ Comparing real and imaginary part of both sides $x^2-y^2=y$ and $2 x y=x$ Taking $2 x y=x$ $\begin{aligned} & (2 y-1) x=0 \\ & \Rightarrow x=0 \text { and } y=\frac{1}{2} \\ & \text { if } x=0 \\ & \therefore y+y^2=0 \quad\left(\because \mathrm{x}^2-\mathrm{y}^2=\mathrm{y}\right) \\ & y(y+1)=0 \\ & y=0 \text { and }-1 \\ & x=0 \\ & y=0 \text { and }-1 \end{aligned}$ If $y=\frac{1}{2}$ $\begin{aligned} & x^2-\frac{1}{4}=\frac{1}{2} \\ & \Rightarrow x^2=\frac{3}{4} \end{aligned}$ $\Rightarrow x= \pm \frac{\sqrt{3}}{2}$ Since given numbers are non zero complex numbers. So, $z_1=0+(-1) i=-i$ $\begin{aligned} & z_2=\frac{\sqrt{3}}{2}+\frac{1}{2} i \quad z_3=\frac{-\sqrt{3}}{2}+\frac{1}{2} i \\ & z_1+z_2+z_3=(-i)+\left(\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)+\left(-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)=0 \end{aligned}$ Hence $z_1+z_2+z_3=0$

Hence, the correct answer is option: 'C'.

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