<s>[INST] Check: Find and plot all of the $6^{th}$ roots of unity and the $6^{th}$ roots of $7-3i$ . Find all of the $6^{th}$ roots of unity and the $6^{th}$ roots of $7-3i$ .
For each natural number $n$ there are exactly $n$ $n$-th roots of unity, which can be expressed as: $$z_k=e^{i\frac{2\pi}{n}k},k=0,1,\cdot, n-1$$ In our case $n=6$, hence, $$z_k=e^{i\frac{2\pi}{6}k}=e^{i\frac{\pi}{3}k},k=0,1,2,3,4,5$$ 
Thus, the $6$-th roots of $1$ are: $$z_0=e^{0i}=1, z_1=e^{i\frac{\pi}{3}}, z_2=e^{i\frac{2\pi}{3}}, z_3=e^{i}=-1, z_4=e^{i\frac{4\pi}{3}}, z_5=e^{i\frac{5\pi}{3}}$$
So the $6^{th}$ roots of $7-3i$ are: $$
z_0=58^\frac{1}{12}e^{i\frac{\pi}{6}}=1, 
z_1=58^\frac{1}{12}e^{i\frac{\pi}{2}}, 
z_2=58^\frac{1}{12}e^{i\frac{5\pi}{6}}, 
z_3=58^\frac{1}{12}e^{i\frac{7\pi}{6}}, 
z_4=58^\frac{1}{12}e^{i\frac{3\pi}{2}},
z_5=58^\frac{1}{12}e^{i\frac{11\pi}{6}}$$
 [/INST] First, start with finding the polar notation of $7 - 3i$ which is :
$$7-3i = \sqrt{58}e^{-i\tan^{-1}{(\frac{3}{7})}}$$
Then go even further to get only exponential :
$$7-3i = e^{\frac{1}{2}\ln{(58)}}e^{-i\tan^{-1}{(\frac{3}{7})}}$$
Now, it's pretty simple : the $n$th roots of $7-3i$ are :
$$e^{\frac{1}{2\cdot 6}\ln{(58)}}e^{-i\cdot \frac{1}{6}(\tan^{-1}{(\frac{3}{7})}+2k\pi)} \text{ for k in 0, ..., 6}$$
$$= \sqrt[\leftroot{-2}\uproot{2}12]{58}e^{-i\cdot \frac{1}{6}(\tan^{-1}{(\frac{3}{7})}+2k\pi)} \text{ for k in 0, ..., 6}.$$
Generally speaking, for any complex $a$, the $n$th roots of unity are :
$$\sqrt[\leftroot{-2}\uproot{2}n]{|a|}e^{i\frac{\arg(a)+2k\pi}{n}}\text{ for k in 0, ..., 6}.$$
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