Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Let $z=x+iy$.

Given, $z^2+\overline{z}=0$

or, $x^2-y^2+2ixy+x-iy=0$

or, $(x^2-y^2+x)+i(2xy-y)=0$

Comparing the real and imaginary part we get,

$x^2-y^2+x=0$.......(1) and $2xy-y=0$.....(2).

From (2) we get, $2xy-y=0$

or, $y(2x-1)=0$

or, $y=0$ and $x=\frac{1}{2}$.

When $y=0$ then from (1) we get, $x^2-x=0$ or, $x=0,-1$.

Again $x=\frac{1}{2}$ then from (1) we get, $y^2=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$ or, $y=\pm \frac{\sqrt{3}}{2}$.

So the solutions, in the form $(x,y)$ are $(0,0),(-1,0),(\pm \frac{\sqrt{3}}{2},\frac{1}{2})$.

So we have $4$ solutions.