The number of solutions for z^3+bar{z}=0 is | vedclass

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(A) Given,$z^3+\bar{z}=0$. Let $z=x+iy$. Substituting $z$ into the equation: $(x+iy)^3 + (x-iy) = 0$. Expanding the cube: $(x^3 - 3xy^2 + x) + i(3x^2y

Vedclass · Accepted Answer

$5$

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(A) Given,$z^3+\bar{z}=0$. Let $z=x+iy$. Substituting $z$ into the equation: $(x+iy)^3 + (x-iy) = 0$. Expanding the cube: $(x^3 - 3xy^2 + x) + i(3x^2y - y^3 - y) = 0$. Equating real and imaginary parts to zero: $1) x(x^2 - 3y^2 + 1) = 0$ $2) y(3x^2 - y^2 - 1) = 0$. Case $1$: If $x=0$,then $-y(y^2+1)=0 \Rightarrow y=0$. So,$(0,0)$ is a solution. Case $2$: If $y=0$,then $x(x^2+1)=0 \Rightarrow x=0$. (Already found). Case $3$: If $x 
eq 0$ and $y 
eq 0$,then $x^2 - 3y^2 + 1 = 0$ and $3x^2 - y^2 - 1 = 0$. Adding the two equations: $4x^2 - 4y^2 = 0 \Rightarrow x^2 = y^2$. Substituting $y^2 = x^2$ into $x^2 - 3y^2 + 1 = 0$: $x^2 - 3x^2 + 1 = 0$ $\Rightarrow 2x^2 = 1$ $\Rightarrow x = \pm \frac{1}{\sqrt{2}}$. Since $x^2 = y^2$,$y = \pm \frac{1}{\sqrt{2}}$. The solutions are $(0,0)$,$(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$,$(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$,$(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$,and $(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. Total number of solutions is $5$.

The number of solutions for $z^3+\bar{z}=0$ is

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