<s>[INST] Confusion about nth root of a number Let's say we want to find the 6th roots of 64. Then according to the method from my textbook:
$z^6=64$
$z^6=64+0i=64[\cos(0)+i\sin(0)]=64cis(0)=64cis(0+2k\pi)$
Then by De Moivre's theorem:
$z=\color{red}{64^{\frac{1}{6}}}cis(\frac{2k\pi}{6})$  $,k=0, 1, 2, 3, 4, 5$
$z=\color{red}{2}cis(\frac{k\pi}{3})$
So that $z=cis(0), 2cis(\frac{\pi}{3}), 2cis(\frac{2\pi}{3}), 2cis(\frac{3\pi}{3}), 2cis(\frac{4\pi}{3}), 2cis(\frac{5\pi}{3})$
So the way I understand that is that we want to prove that the equation $z^6=64$ has six different solutions. In other words that there are six different values of $z$ that satisfy $z=64^{\frac{1}{6}}$. If so, then why do we simplify $64^{1/6}$ to just $2$ in the process, what about the other five values?
 [/INST] Notice:
$$z^6=64\Longleftrightarrow z^6=|64|e^{\arg(64)i}\Longleftrightarrow z^6=64e^{0i}\Longleftrightarrow$$
$$z=\left(64e^{2\pi ki}\right)^{\frac{1}{6}}\Longleftrightarrow z=2e^{\frac{\pi ki}{3}}$$
With $k\in\mathbb{Z}$ and $k:0-5$
So, the solutions are:
$$z_0=2e^{\frac{\pi\cdot0i}{3}}=2$$
$$z_1=2e^{\frac{\pi\cdot1i}{3}}=2e^{\frac{\pi i}{3}}$$
$$z_2=2e^{\frac{\pi\cdot2i}{3}}=2e^{\frac{2\pi i}{3}}$$
$$z_3=2e^{\frac{\pi\cdot3i}{3}}=-2$$
$$z_4=2e^{\frac{\pi\cdot4i}{3}}=2e^{-\frac{2\pi i}{3}}$$
$$z_5=2e^{\frac{\pi\cdot5i}{3}}=2e^{-\frac{\pi i}{3}}$$
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