Question

The number of solutions of the equation $z^2+\overline{z}=0$ where $zϵC$ are

• A. $1$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (B)

$4$

Let $z=x+iy$
Hence
$z^2+\overline{z}=0$
$(x+iy{)}^2+(x-iy)=0$
$x^2-y^2+i2xy+x-iy=0$
$\left(\omega \right)=(x^2+x-y^2)+i(2xy-y)=0$
Hence
$Im\left(\omega \right)=2xy-y=0$
$y=0$ or $x=\frac{1}{2}$.
And $Real\left(\omega \right)=0$ implies
$x^2+x-y^2=0$. If $y=0$, $x=0,-1$
And if $x=\frac{1}{2}$, $y=\frac{\pm \sqrt{3}}{2}$.
Thus the imaginary number z can be
$z=\frac{1+\sqrt{3}i}{2}$ and $\frac{1-\sqrt{3}i}{2}$.
Or
$z=0+0i$ and $z=-1+0i$
Hence 4 solutions.