Question

The number of solutions of the equation $z^2+\bar{z}=0$, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

Let $z=x+iy$, so that $\bar{z}=x+iy$, therefore
$z^2+\bar{z}=0\iff \left(x^2-y^2+x\right)+i(2xy-y)=0$
equating real and imaginary parts, we get
$x^2-y^2+x=0$
and $2xy-y=0$
$\Rightarrow y=0$
or $x=\frac{1}{2}$
If $y=0$, then (1) gives $x^2+x=0$
$\Rightarrow x=0$
or $x=-1$
If $x=\frac{1}{2}$,
Then $x^2-y^2+x=0$
$\Rightarrow y^2=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$
$\Rightarrow y=\pm \frac{\sqrt{3}}{2}$
Hence, there are four solutions in all.