Question

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

• A. 1
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (B)

4

Answer 2

The correct option is(B): 4.

Let $z = x + iy$
$\Rightarrow z^{2} =x^{2} -y^{2} + 2ixy$
$\because z^{2} + \bar{z} = 0$ (given)
$\therefore x^{2} -y^{2} + 2ixy+x-iy=0$
$\Rightarrow \left(x^{2} + x - y^{2}\right) + i \left(2.xy - y\right) = 0$
Equating the real and imaginary parts, we get
$x^{2} +x - y^{2} =0\,\,\,\,\,\dots(i)$
and $2xy - y = 0\,\,\,\,\,\,\dots(ii)$
By E(ii), we gety
$\left(2x - 1\right) = 0$
$\Rightarrow y=0$ or $x = \frac{1}{2}$

so at y=0,

x2+x=0

x=0,-1.

again for $x = \frac{1}{2}$ from 2 we get

y2= $x = \frac{-1}{y}$+$\frac{1}{2}$=$\frac{3}{4}$

$y =\frac{+√2}{2}$

therefor the solution for the given equation are: (a,y), are (0,0), (-1,0), (±$\frac{√3}{2}$. $\frac{1}{2}$.)

hence correct answer is 4.