<s>[INST] Complex Number - Find all roots of the equation $$e^{i \frac{{\pi}}{3}}z^5+4e^{i\frac{(2+3){\pi}}{6}}z^3 + z^2 + 4i = 0.$$
By using Euler's formula, I got:
$$e^{i \frac{{\pi}}{3}} = \cos{\frac {\pi}{3}} + i\sin{\frac {\pi}{3}} = (\frac{1}{2} + i{\frac {\sqrt{3}}{2}}) $$
$$4e^{i  \frac{(2+3){\pi}}{6}} = 4(\cos{\frac {5\pi}{6}} + i\sin{\frac {5\pi}{6}}) = 4 ({\frac {-\sqrt{3}}{2}} + i \frac{1}{2}) $$
but how do i proceed from here? Any help is much appreciated!
Thank you.
 [/INST] Hint: The polynomial can be factored as
$$
e^{i\pi/3} (z^3+a)(z^2+b).
$$
More: Multiply it out:
$$
e^{i\pi/3} (z^3+a)(z^2+b) = e^{i\pi/3}z^5+be^{i\pi/3}z^3 + ae^{i\pi/3}z^2 + abe^{i\pi/3}.
$$
To make this match the original polynomial, $e^{i\pi/3}z^5 + 4e^{i5\pi/6}z^3 + z^2 + 4i$, we need
$$
be^{i\pi/3} = 4e^{i5\pi/6}, \qquad ae^{i\pi/3} = 1, \qquad abe^{i\pi/3} = 4i.
$$
From the first equation we get
$$
b = 4e^{i3\pi/6} = 4i
$$
and from the second we get
$$
a = e^{i5\pi/3}.
$$
Thus
$$
e^{i\pi/3}z^5 + 4e^{i5\pi/6}z^3 + z^2 + 4i = e^{i\pi/3}(z^3 + e^{i\pi/3})(z^2+4i).
$$
Now just solve
$$
z^3 + e^{i\pi/3} = 0
$$
and
$$
z^2+4i = 0
$$
to find the roots of the equation.
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