Compute complex equation $\rho^2 e^{2i\theta}+ 3\rho e^{-i\theta}=0$

Question

I tried compute equation: $$ z^{2} + 3\overline z = 0 $$ using method with polar reprezentation.
$$ z = re^{i\theta} $$

$$\rho^2 e^{2i\theta}+ 3\rho e^{-i\theta}=0$$ $\color{red}{\text{assume that } \rho > 0}$ $$\color{red}\rho=-3e^{-3i\theta}$$ As $\rho$ is positive and real, and exponential of an imaginary argument is on a unit circle, you know that the only solution is $\color{red}{\rho=3}$ and $e^{-3i\theta}=-1=e^{i\pi}$ meaning $$-3i\theta = i\pi k$$ where $k \in \mathbb{Z}$ and $$\theta = -\pi/3k$$ so $$\theta\in\lbrace \pi/3, 2\pi/3 \rbrace$$

and get the solution $z=0$ (from $\rho=0$),

Please check my solution. Thanks

Answer 1

From $$\rho^2 e^{2i\theta}+ 3\rho e^{-i\theta}=0$$

You should have concluded

$$\rho = 0 \quad \text{or} \quad \rho = -3e^{-3i\theta}$$

Answer 2

Let $z=x+iy$ then we get $$x^2-y^2+3x+i(2xy-3y)=0$$ so $$x^2-y^2+3x=0$$ and $$2xy-3y=0$$

From the second equation we get $$y=0$$

or $$2x-1=0$$