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

The number of solutions for $z^3+\bar{z}=0$, is
  1. $5$
  2. $1$
  3. $2$
  4. $3$

Solution

Given, $z^3+\bar{z}=0$ Let $\quad z=x+i y$ $ \begin{aligned} & \therefore \quad(x+i y)^3+x-i y=0 \\ & \Rightarrow x^3+(i y)^3+3 x^2 i y+3 x(i y)^2+x-i y=0 \\ & \Rightarrow x^3-y^3 i+3 x^2 y i-3 x y^2+x-i y=0 \\ & \Rightarrow\left(x^3-3 x y^2+x\right)+\left(-y^3+3 x^2 y-y\right) i=0 \end{aligned} $ On equating real and imaginary parts, we get $ \begin{aligned} & x^3-3 x y^2+x=0 \\ & \text { or } \quad-y^3+3 x^2 y-y=0 \\ & \Rightarrow \quad x\left(x^2-3 y^2+1\right)=0 \\ & \text { or } \quad-y\left(y^2-3 x^2+1\right)=0 \\ & \Rightarrow \quad x=0 \text { and } x^2-3 y^2+1=0 \\ & \text { or } y=0 \text { and } y^2-3 x^2+1=0 \\ & \text { Now, } \\ & x^3-3 y^2+1=y^2-3 x^2+1 \\ & 4 x^2=4 y^2 \\ & x= \pm y \\ & y^2-3 y^2+1=0 \\ & \Rightarrow \quad 2 y^2=1 \Rightarrow y= \pm \frac{1}{2} \\ & \Rightarrow \quad x= \pm \frac{1}{2} \\ & \end{aligned} $ Hence, their solutions are $(0,0)$ and $\left( \pm \frac{1}{2}, \pm \frac{1}{2}\right)$ Hence, number of solutions is 5

Asked in: AP EAMCET 2014

Practice more Complex Number questions on Aicharya