<s>[INST] Expressing complex roots in a trigonometric form 
Given that
$$(z+2)^{12}=z^{12}$$
Show that its roots may be expressed in the form
$$-1-i\cot\left(\frac{1}{12}k\pi\right)$$
where $k=\pm1,\pm2,\pm3,\pm4,\pm5.$

My attempt at solving this:
$$\frac{(z+2)^{12}}{z^{12}}=1 \rightarrow \left({\frac{z+2}{z}}\right)^{12}=1$$
$$\frac{z+2}{z}=e^{i\frac{1}{6}k\pi} \rightarrow\frac{2}{z}=-1+e^{i\frac{1}{6}k\pi} \rightarrow z=\frac{2}{-1+e^{i\frac{1}{6}k\pi}}$$
I'm stuck from this point onwards. Have I made a mistake in my working or am unable to proceed due to my deficiency in problem solving?
 [/INST] Your calculations are correct so far, you can continue with
$$
\frac{2}{-1+e^{i\frac{1}{6}k\pi}} = -1 + \frac{e^{i\frac{1}{6}k\pi}+1}{e^{i\frac{1}{6}k\pi}-1}
= -1 + \frac{e^{i\frac{1}{12}k\pi}+e^{-i\frac{1}{12}k\pi}}{e^{i\frac{1}{12}k\pi}-e^{-i\frac{1}{12}k\pi}}
= -1 +\frac{2\cos(\frac{1}{12}k\pi)}{2i\sin(\frac{1}{12}k\pi)}
$$
Note that $k=6$ (corresponding to $z=-1$) must be included to get all $11$ solutions of the original equation.
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