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 $z$ in the equation: $(x+iy)^3 + (x-iy) = 0$. Expanding: $x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 + x

Vedclass · Accepted Answer

$5$

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(A) Given,$z^3+\bar{z}=0$. Let $z=x+iy$. Substituting $z$ in the equation: $(x+iy)^3 + (x-iy) = 0$. Expanding: $x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 + x - iy = 0$. $x^3 + 3x^2yi - 3xy^2 - iy^3 + x - iy = 0$. Grouping real and imaginary parts: $(x^3 - 3xy^2 + x) + i(3x^2y - y^3 - y) = 0$. 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 $-y(y^2+1)=0 \Rightarrow y=0$. Solution: $(0,0)$. Case $2$: If $y=0$,then $x(x^2+1)=0 \Rightarrow x=0$. Solution: $(0,0)$. 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 $x^2 = y^2$ into $x^2 - 3y^2 + 1 = 0$: $y^2 - 3y^2 + 1 = 0$ $\Rightarrow 2y^2 = 1$ $\Rightarrow y = \pm \frac{1}{\sqrt{2}}$. Since $x^2 = y^2$,$x = \pm \frac{1}{\sqrt{2}}$. Possible pairs $(x,y)$ are $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. Total solutions: $(0,0)$ and the $4$ pairs above,giving $5$ solutions.

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

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