Question

The number of solutions of the equation $z^3+\overline{z}=0$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

Given,

$z^3+\overline{z}=0$

To find no. of solution of given equation.

Solution,

$z^3+\overline{z}=0$

Let, $z=x+iy\&z=x-iy\Rightarrow {(x+iy)}^3+x-iy=0\Rightarrow x^3+x-3x{y}^2+i(3x^{2}y-y^3-y)=0$

Real part $\&$ imaginary part both equals to 0.

$x(x^2+1-{3y}^2)=0y(y^2+1-{3x}^2)=0$

One of the root is

$x_1=y_1=0$

Other roots are

$x^2-3y^2+1=0⟶\left(1\right)y^2-3x^2+1=0⟶\left(2\right)$ By solving equation $\left(1\right)\&\left(2\right)$ we get.

$Y=\pm xandx=\pm \frac{1}{\sqrt{2}}=\pm y$

Therefore roots are

$z_1=0,z_2=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}},z_3=\frac{-1}{\sqrt{2}}-\frac{i}{\sqrt{2}}{z}_4=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}{z}_5=\frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$

Hence the no. of solution are five.