Does $z^4+2z^2+z=0$ have complex roots?

Question

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?

Answer 1

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: $$ x_1=\frac{\sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{3^{2/3}}-2 \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}} $$ $$ x_2=\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\ 3^{2/3}} $$ $$ x_3=\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\ 3^{2/3}} $$

Answer 2

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.