How to solve the complex equation $ω^2=-11/4+15i$ The question is stated as following:
"First, solve the equation: 
$ω^2=-11/4+15i$
and after, with the help of that, solve:
$z^2-(3-2i)z+(4-18i)=0$"
The problem for me lies in solving the system of equations for ω; 
$Re:a^2-b^2=-11/4$ and $Im:2ab=15$
Where I eventually end up with the fourth degree equation: $4b^4+11b^2+15^2=0$ which I don't know how to solve, is this solveable or am I on the wrong track? I would very much appreciate some help, thank you for your time.

As you have noted, equating the real parts gives $a^2 - b^2 = -\frac{11}{4}$. Now since the modulus of $\omega^2$ is the squared modulus of $\omega$ (think in polar form), $|\omega|^2 = a^2 + b^2 = \sqrt{(-11/4)^2 + 15^2} = \frac{61}{4}$.
Thus adding and subtracting, $a^2 = \frac{1}{2} \left(\frac{61}{4} + -\frac{11}{4} \right) = \frac{25}{4}$ and $b^2 = \frac{1}{2} \left(\frac{61}{4} - \frac{11}{4}\right) = 9$. But since the argument of $\omega^2$ is either in the second quadrant or the second quadrant + $360º$, the argument of $\omega$ is half that of $\omega^2$, which results in the first quadrant or the third quadrant. Hence $a+bi = \frac{5}{2} + 3i, -\frac{5}{2} - 3i$.
