Finding the Roots of Unity

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I have the following equation, $$z^5 = -16 + (16\sqrt 3)i$$I am asked to write down the 5th roots of unity and find all the roots for the above equation expressing each root in the form $re^{i\theta}$. I am just wondering if my solutions are correct. Here are my solutions,

5th roots of unity, $$z = 1, e^{\frac{2\pi}{5}i}, e^{\frac{4\pi}{5}i}, e^{-\frac{2\pi}{5}i}, e^{-\frac{4\pi}{5}i}$$

and for all the roots for the above equation, $$z = e^{\frac{2\pi k} {5}i}, 2e^{\frac{(6k+2)\pi}{15}i}$$ where k = 0, 1, 2, 3, 4.

Please correct me if there are any mistakes. I will leave my working at the answer section for future reference.

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If u have used the theta as 2kπ/theta and got it then its right

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For the fifth roots of unity, the roots are $\frac{2\pi}{5}$ radian apart from one another. Since the principal value for arg(z) is (-$\pi$,$\pi$], hence, the fifth roots of unity are $$z = 1, e^{\frac{2\pi}{5}i}, e^{\frac{4\pi}{5}i}, e^{-\frac{2\pi}{5}i}, e^{-\frac{4\pi}{5}i}$$ For the all the roots of the above equation, one of the roots will always be $$z = e^{\frac{2k\pi}{n}i}$$ where n = the power of z and k = 0, 1, 2, ..., n-1

Hence, $$z = e^{\frac{2k\pi}{5}i}$$ where k = 0, 1, 2, 3, 4

For the other root, $$|z^5| = \sqrt {(-16)^2 + (16\sqrt3)^2}$$ $$|z| = \sqrt[5] {32} = 2$$ and since the coordinate lies in the 2nd quadrant, $$arg(z) = \pi - tan^{-1} (\frac{16\sqrt3}{16}) = \frac{2\pi}{3}$$ Therefore, $$\begin{align} z & = re^{i\theta}\\ & = 2e^{(\frac{2k\pi}{5}+\frac{1}{5}\cdot\frac{2\pi}{3})i}\\ & = 2e^{\frac{(6k+2)\pi}{15}i}\\ \end{align}$$