Q67 / 180 · KCET 2012 · Mathematics

The number of solutions of the equation $\mathrm{z}^{2}+\overline{\mathrm{z}}=0$, where $\mathrm{z} \in \mathrm{C}$, are

  • A
    1
  • Correct answer
    B
    4
  • C
    5
  • D
    6

Solution

Let $z=x+i y$ $$ \begin{aligned} &\Rightarrow \quad z^{2}=x^{2}-y^{2}+2 i x y \\ &\because \quad z^{2}+\bar{z}=0 \\ &\therefore \quad x^{2}-y^{2}+2 i x y+x-i y=0 \\ &\Rightarrow \quad\left(x^{2}+x-y^{2}\right)+i(2 x y-y)=0 \end{aligned} $$ Equating the real and imaginary parts, we get $$ x^{2}+x-y^{2}=0 $$ and $\quad 2 x y-y=0 \quad \ldots$ (ii) By Eq. (ii), we get $y(2 x-1)=0$ $\Rightarrow \quad \mathrm{y}=0 \quad$ or $\quad \mathrm{x}=\frac{1}{2}$ On solving these two with first equation, we get four solutions $(0,0),(-1,0)$ and $\left(\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)$.

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

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