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, so that $\overline{z}$ = x - iy, therefore

$z^2+\overline{z}=0\iff (x^2-y^2+x)+i(2xy-y)$ = 0

Equating real and imaginary parts , we get

$x^2-y^2+x$ = 0 .......(i)

And 2xy - y = 0 ⇒ y = 0 or x = $\frac{1}{2}$

if y = 0 , then (i) gives $x^2$ + x = 0 ⇒ 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.