Solving $z^5=-16+16\sqrt3i$ using the fifth roots of unity

Question

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?

Answer 1

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.

Answer 2

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 $u^5 = 1$

Answer 3

$$ z^5=-16+16\sqrt3 i=32\;e^{i\big(2n\pi+\frac{2\pi}3\big)}\\ z=2\ e^{i\big(\frac{2n}5+\frac 2{15}\big)\pi}$$