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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?

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    $\begingroup$ By the fundamental theorem of algebra, every polynomial of degree $n$ has exactly $n$ roots in $\mathbb{C}$. As for finding yours, though, I'm not sure. $\endgroup$ – Eevee Trainer Nov 29 '18 at 3:01
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    $\begingroup$ The polynomial is $zP$ where $P = z^3+2z+1$. Since $\frac{dP}{dz} = 3z^2 + 2$ is positive for all $z$, it follows that $P$ is an increasing function and can have only one real root. $\endgroup$ – Théophile Nov 29 '18 at 3:17
  • $\begingroup$ Theophile's answer is by far the best here. Note we can get even more via a quick application of Rouche's Theorem, which tells us each root must be the disk $|z| \le 2$ (since on the circle of radius two, $|2z^2 + z| \le 2|z|^2 + |z| = 10 and $|z^4| = 16,$ we conclude all four roots lie in the disk. $\endgroup$ – Brevan Ellefsen Nov 29 '18 at 9:26
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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: $$ x_1=\frac{\sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{3^{2/3}}-2 \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}} $$ $$ x_2=\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\ 3^{2/3}} $$ $$ x_3=\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\ 3^{2/3}} $$

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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.

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  • $\begingroup$ I think OP is asking for complex, not real roots. $\endgroup$ – YiFan Nov 29 '18 at 13:14
  • $\begingroup$ I showed that it has two real and two complex roots. This exactly answers the question. Did you not read my last sentence? $\endgroup$ – marty cohen Nov 29 '18 at 16:20
  • $\begingroup$ The question asks, "Does $z^2+2z^2+z=0$ have complex roots? [...] How to find the complex roots?" Your answer addresses only the first part and the rest is done with Wolfy. $\endgroup$ – YiFan Nov 29 '18 at 20:34

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