How can I solve $\bar{z}^2+3z+1=0$?

Question

I do not manage to solve it as a real equation nor to find a solution with the algebraic form.

Answer 1

Pair the original equation

$\overline z^2+3z+1=0$

with its conjugate which must be simultaneously satisfied:

$z^2+3\overline z+1=0$

From the conjugate equation infer $\overline z=-(z^2+1)/3$. Substitute into the first equation thus

$(z^2+1)^2/9+3z+1=0$

$z^4+2z^2+27z+10=0$

If you try rational roots to this equation you will fail. But there is a better way to factor it: if the equation without conjugation

$z^2+3z+1=0$

has real roots then these also satisfy the complex equation you are trying to solve because then $z=\overline z$. That works because the discriminant is positive. So we are sure that $z^2+3z+1$ is a factor of the quartic and we can divide it out to get the factored equation

$(z^2+3z+1)(z^2-3z+10)=0$

Each of these quadratic equations is solved to give the roots

$(-3\pm\sqrt5)/2$

$(3\pm i\sqrt{31})/2$

Answer 2

Put $z=x+iy$ and $\bar z=x-iy$ $$\bar z^2+3z+1=0$$ $$\implies (x-iy)^2 +3(x+iy) +1=0$$ $$\implies (x^2-y^2+3x +1) +i(3y-2xy)=0$$

Now you have two variables (x and y) and two equations. Solve them to get the value of x and y and eventually z. $$x^2-y^2+3x +1 =0$$ $$3y-2xy=0$$