Finding the solution of a complex number in polar form?

Question

Which of the following is a solution to the equation $z^3 = (−1 − \sqrt{3}i)$?

a) $2^{1/3}[cos(\frac{19\pi}{18})+isin(\frac{19\pi}{18})]$

b) $2^{1/3}[cos(\frac{10\pi}{9})+isin(\frac{10\pi}{9})]$

c) $2^{1/3}[cos(\frac{8\pi}{9})+isin(\frac{8\pi}{9})]$

d) $2^{1/3}[cos(\frac{17\pi}{18})+isin(\frac{17\pi}{18})]$

e) $2^{1/3}[cos(\frac{13\pi}{18})+isin(\frac{13\pi}{18})]$

f) $2^{1/3}[cos(\frac{23\pi}{18})+isin(\frac{23\pi}{18})]$

g) $2^{1/3}[cos(\frac{11\pi}{9})+isin(\frac{1\pi}{9})]$

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My solution:

1) Converting into polar form:

$r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = 2$

$\theta = \tan^{-1}{\frac{-\sqrt{3}}{-1}} = \pi/3$

This angle falls in the wrong quadrant. We need quadrant three, so $\pi/3 + \pi = 4\pi/3$

$z^3 = 2[cos(4\pi/3)+isin(4\pi/3)]$

2) Apply De Moivre's formula:

$z = 2^{1/3}[cos(4\pi/9)+isin(4\pi/9)]$

However, none of the answers match my solution. Please help! Thanks.

Answer 1

There is more than one root.

$\theta=\frac{4\pi}{3}+2k\pi$

$z=2^{1/3}[cos(\frac{4\pi}{9}+\frac{2k\pi}{3})+isin(\frac{4\pi}{9}+\frac{2k\pi}{3})]$

$0<\frac{4\pi}{9}+\frac{2k\pi}{3}<2\pi$

$-\frac{2}{3}<k<\frac{7}{3}$

$k=0,1,2$

$k=0,z=2^{1/3}[cos(\frac{4\pi}{9})+isin(\frac{4\pi}{9})]$(answer)

$k=1,z=2^{1/3}[cos(\frac{10\pi}{9})+isin(\frac{10\pi}{9})]$(answer)

$k=2,z=2^{1/3}[cos(\frac{16\pi}{9})+isin(\frac{16\pi}{9})]$(another root)

Answer 2

Multiplying by a $3$rd root of unity will give another cube root. As mentioned in the comments there are $2$ more.

Multiplying by $e^{\frac{2\pi i}3}$ gives (b)...