How do I solve $z^6+z^5+z^4+z^3+z^2+z=0$?

Question

I tried to use de Moivre's formula to solve it, but I had no success.

Can you give me any tips?

Answer 1

$$z^6+z^5+z^4+z^3+z^2+z=z(z^5+z^4+z^3+z^2+z+1)=z\,\frac{z^6-1}{z-1}\quad\text{if }z\ne 1.$$ So the roots are either $z=0\:$ or the $6^{\text{th}}$ roots of unity different from $1$.

Answer 2

Hint: $$z^6+z^5+z^4+z^3+z^2+z$$ $$=z^5(z+1)+z^3(z+1)+z(z+1)$$ $$=(z+1)(z^5+z^3+z)$$ $$=z(z+1)(z^4+z^2+1)$$

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

Answer 3

Besides the trivial $z=0$ we also have $z=e^{2\pi ik/6}$ for $k=1,2,3,4,5$ the $6^{th}$ roots of unity different from $1$

Answer 4

We have

$z(z^5 + z^4 + z^3 + z^2 + z + 1) = z^6 + z^5 + z^4 + z^3 + z^2 + z = 0; \tag 1$

we see that

$z = 0 \tag 2$

is a root of (1), and that if

$z \ne 0, \tag 3$

then

$z^5 + z^4 + z^3 + z^2 + z + 1 = 0; \tag 4$

it is easy to discover the zeroes of (4) if we observe or recall that

$(z - 1)(z^5 + z^4 + z^3 + z^2 + z + 1) = z^6 - 1; \tag 5$

thus any $z$ satisfying (4) must also be a root of

$z^6 - 1 = 0, \tag 6$

which is satisfied by the $6$-th roots of unity

$\omega = e^{\pi i / 3}, \; \omega^2 = e^{2 \pi i / 3}, \; \omega^3 = e^{\pi i} = -1, \; \omega^4 = e^{4\pi i / 3}, \; \omega^5 = e^{5\pi i / 3}, \; \omega^6 = e^{2 \pi i} = 1; \tag 7$

that is, by the

$\omega^k = e^{k\pi i / 3} = \cos \dfrac{k\pi}{3} + i\sin \dfrac{k\pi}{3}, \; 1 \le k \le 6; \tag 8$

of course, in constructing (5), (6) we have admitted

$z = 1 \tag 9$

as a root (a root of (6), that is); but $1$ clearly doesn't satisfy (1) or (4), so we have to rule it out; it follows from (5), however, that every other zero of (5)-(6) is one of the $\omega^k$, with $k \ne 6$, and that these all solve (4); thus the roots of (1) must be the six complex numbers

$z = 0, \; \omega^{k\pi i / 3}, \; 1 \le k \le 5, \tag{10}$

that is,

$z = 0, \; e^{\pi i / 3}, \; \omega^2 = e^{2 \pi i / 3}, \; \omega^3 = e^{\pi i}, \; \omega^4 = e^{4\pi i / 3}, \; \omega^5 = e^{5\pi i / 3}. \tag{11}$

$OE\Delta$.

Nota Bene: It should be observed that we did in fact us de Moivre's formula in writing the explicit form (8) of the solutions to (1). End of Note.