$$e^{i \frac{{\pi}}{3}}z^5+4e^{i\frac{(2+3){\pi}}{6}}z^3 + z^2 + 4i = 0.$$
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.
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.