Does $z^4+2z^2+z=0$ have complex roots? Does $z^4+2z^2+z=0$ have complex roots? How to find them? Besides $z=0$, I got the equation $re^{3i\theta}+2re^{i\theta}=e^{i(\pi+2k\pi)}$, $k\in \mathbb Z$. How to find the complex roots?
 A: Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
$$
\Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
$$
$$
\Delta_3=\begin{cases} >0 & \text{3 distinct real roots}\\
<0 & \text{1 real, 2 conjugate complex roots}\\
=0 & \text{3 real roots with duplicates}\\
\end{cases}
$$
In your case, it's $a=1$, $b=0$, $c=2$ and $d=1$, so $\Delta_3=-59$, hence the equation has one real and two complex conjugate roots, with all 3 distinct. As for finding the roots themselves, the Cardano formula helps check this out. The final result is:
$$
x_1=\frac{\sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{3^{2/3}}-2 \sqrt[3]{\frac{2}{3
   \left(\sqrt{177}-9\right)}}
$$
$$
x_2=\left(1+i \sqrt{3}\right)
   \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1-i \sqrt{3}\right)
   \sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\ 3^{2/3}}
$$
$$
x_3=\left(1-i
   \sqrt{3}\right) \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1+i
   \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\
   3^{2/3}}
$$
A: According to Wolfy,
the roots are
$0, ≈-0.453397651516404,
≈0.22669882575820 \pm 1.46771150871022 i
$
so the answer is yes.
Another way to see this is that,
if 
$f(z) = z^4+2z^2+z$,
then
$f'(z)
= 4z^3+4z+1
$
and
$f''(z)
= 12z^2+4
\gt 0
$
so $f(z)$
can have at most two real roots.
Since
$f(0) = 0$
and $f'(0)=1 \ne 0$,
$f(z)$ has exactly
two real roots
so has
two (conjugate) complex roots
since all its coefficients
are real.
