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1 Write down the fifth roots of unity. Hence, or otherwise, find all the roots of the equation $$z^5=-16+16\sqrt3i,$$ giving each root in the form $re^{i\theta}$. [4]

In this question, I can work out the answer using de Moivre's theorem, but the question implies that I have to use the 5 roots of unity somehow. Does it just mean that I have to solve the equation in the same way as I solved it for the 5 roots of unity, or whether I have to actually use the 5 roots of unity?

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  • $\begingroup$ After more than one month being a member and after asking 8 questions, is about time you learn to use MathJaX $\endgroup$ – DonAntonio May 27 '17 at 7:36
  • $\begingroup$ @DonAntonio But I did it for him! $\endgroup$ – Parcly Taxel May 27 '17 at 7:36
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    $\begingroup$ @ParclyTaxel So nice of you, yet you are not going to be all the time by his side to fix his posts...are you? $\endgroup$ – DonAntonio May 27 '17 at 7:37
  • $\begingroup$ @DonAntonio As long as I'm on here actually and I find an "interesting" question, then I might fix. Nevertheless, MathJax is something everyone should learn. $\endgroup$ – Parcly Taxel May 27 '17 at 7:38
  • $\begingroup$ Sorry, I've had my exams for the past month. I shall learn it by next week. $\endgroup$ – user440261 May 27 '17 at 8:16
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The modulus of $-16+16\sqrt{3}\,i$ is $32=2^5$. Setting $z=2w$, the equation becomes $$ w^5=-\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{2\pi i/3} $$ If $\zeta$ is a fifth root of unity, then $$ w=\zeta e^{2\pi i/15} $$ is a solution of the equation. So, if $\zeta_0=1$, $\zeta_1=e^{2\pi i/5}$, $\zeta_2=e^{4\pi i/5}$, $\zeta_3=e^{6\pi i/5}$ and $\zeta_4=e^{8\pi i/5}$ are all the fifth roots of unity, the solutions of your equation are $$ 2\zeta_0 e^{2\pi i/15} \quad 2\zeta_1 e^{2\pi i/15} \quad 2\zeta_2 e^{2\pi i/15} \quad 2\zeta_3 e^{2\pi i/15} \quad 2\zeta_4 e^{2\pi i/15} $$ and you can do the final computations.

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  • $\begingroup$ Sorry, I'm confused about your notation.. what does the symbol ζ mean? And how do you go from the first step to the second? $\endgroup$ – user440261 May 27 '17 at 8:58
  • $\begingroup$ @Saad Just what it says: if $\zeta$ is one of the fifth roots of unity, then $\zeta e^{2\pi i/15}$ is a solution of $w^5=e^{2\pi i/3}$. Thus, using all five of them you can express the full set of solutions. $\endgroup$ – egreg May 27 '17 at 9:03
  • $\begingroup$ Oh, so you do w^5 / (e^2i pi/15)^5 = 1 Then, you let the L.H.S equal u^5, for example, and then we know the roots of u^5 = 1, so we can find the final solutions.. Thank you. $\endgroup$ – user440261 May 27 '17 at 9:29
  • $\begingroup$ @Saad Yes, I find five different solutions, so all of them. $\endgroup$ – egreg May 27 '17 at 9:42
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From $z^5=-16+16\sqrt3i$ we have $(\frac z {\sqrt[5]32})^5=\frac {-1} 2+ \frac {\sqrt3i} 2$ and $(\frac z 2)^5=\frac {-1} 2+ \frac {\sqrt3i} 2= \cos \frac {\pi} 3 + i \sin \frac {\pi} 3= (\cos \frac {\pi} {15} + i \sin \frac {\pi} {15})^5$

Now making a notation $u=\frac {z}{2(\cos \frac {\pi} {15} + i \sin \frac {\pi} {15})}$ we get $u^5 = 1$

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  • $\begingroup$ Thank you for the answer. It does use the 5 roots of unity, but this is a very roundabout method.. Is there a more direct approach? $\endgroup$ – user440261 May 27 '17 at 8:17
  • $\begingroup$ @Saad In fact, the above only shows one $\;5\,-$ th root of unity...and thus shows only one single root of the equation. $\endgroup$ – DonAntonio May 27 '17 at 8:45
  • $\begingroup$ @DonAntonio That's wrong. If $u_1,u_2,u_3,u_4,u_5,$ are the solutions of $u^5 = 1$ then $z_1=2(\cos \frac {\pi} {15} + i \sin \frac {\pi} {15})u_1,z_2=2(\cos \frac {\pi} {15} + i \sin \frac {\pi} {15})u_2, ...$ are the 5 solutions of $z$ equation $\endgroup$ – user261263 May 27 '17 at 12:38
  • $\begingroup$ @EugenCovaci How do you count the roots?I see one single $\;u\;$, as opposed to, for example, the accepted answer where the five distinct roots appear. Here there is neither one root times other one, nor powers of one single roots, or moving the argument... $\endgroup$ – DonAntonio May 27 '17 at 13:20
  • $\begingroup$ @DonAntonio It's becoming quite funny ... $\endgroup$ – user261263 May 27 '17 at 15:14
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$$ z^5=-16+16\sqrt3 i=32\;e^{i\big(2n\pi+\frac{2\pi}3\big)}\\ z=2\ e^{i\big(\frac{2n}5+\frac 2{15}\big)\pi}$$

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