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Consider $P(z) = z^5+z^4+z^3+z^2+z^1+1$.

a) The polynomial has one, integer real root. Find this.

This gives $P(-1) = 0$

b) Find all complex roots of the polynomial. How do I go from here?

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    $\begingroup$ Hint: Multiply both sides of the equation by $z-1$. $\endgroup$ Sep 6, 2022 at 19:56
  • $\begingroup$ There are various ways of reducing this using the symmetry, and because of the low degree they work and can be solved. But there is also a generalisable method which doesn't depend on the low degree of this particular polynomial, and that is the one you should look out for in comments and answers, and make sure you understand it. $\endgroup$ Sep 6, 2022 at 20:52
  • $\begingroup$ Does $P(z) = z^5+z^4+z^3+z^2+z^1$, or that $+1$? Are you solving $P(z) = 0$ or $P(z) = -1$? $\endgroup$
    – peterwhy
    Sep 6, 2022 at 21:26

1 Answer 1

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Do polynomial long division to get that $z^5+z^4+z^3+z^2+z+1=(z+1)(z^4+z^2+1)$

Now solve $z^4+z^2+1=0$.

This is quadratic equation with respect to $z^2$.

Discriminant: $(-1)^2-4=-5$

Thus, $z^2=\frac{-1-\sqrt{5}i}{2}$ and $z^2=\frac{-1+\sqrt{5}i}{2}$.

Therefore the other four roots are $\sqrt{\frac{-1-\sqrt{5}i}{2}}$, $-\sqrt{\frac{-1-\sqrt{5}i}{2}}$, $\sqrt{\frac{-1+\sqrt{5}i}{2}}$, $-\sqrt{\frac{-1+\sqrt{5}i}{2}}$.

Another way is to multiply equation by $z-1$ as @RobertShore has advised.

In this case $(z^5+z^4+z^3+z^2+z+1)(z-1)=0$ or $z^6-1=0$.

Can you proceed from here?

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    $\begingroup$ Not a good idea to answer a question that shows no signs of any independent effort to reach a solution. Also, it's not a good idea to leave the $i$ beneath the radical. $\endgroup$ Sep 6, 2022 at 19:58
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    $\begingroup$ @RobertShore, by downvoting the correct answer you mislead the people, don't you? $\endgroup$
    – eMathHelp
    Sep 6, 2022 at 20:01
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    $\begingroup$ I don't consider fair at all to downvote your answer, but you should advise grunger95 in a different way (even if it doesn't go in the logic of this exercise) : based on the recognition a geometric series, then ask if he/she knows that it is $(z^6-1)/(z-1)$ then ask him/her if a knows $n$th roots of unity.... $\endgroup$
    – Jean Marie
    Sep 6, 2022 at 20:11
  • $\begingroup$ Agree with you... $\endgroup$
    – eMathHelp
    Sep 6, 2022 at 20:12
  • $\begingroup$ I'm not the one who downvoted your answer. $\endgroup$ Sep 6, 2022 at 21:25

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