Finding roots and complex roots of a determinant

[Redwaves]

Homework Statement

Finding roots and complex roots of a determinant

Relevant Equations

$(-\Omega^2 + i\gamma\Omega + \frac{2k}{3m})(-\Omega^2 + i\gamma\Omega + \frac{2k}{3m}) - (-i\gamma\Omega)(-i\gamma\Omega) = 0$

I need to find the values of $\Omega$ where $(-\Omega^2 + i\gamma\Omega + \frac{2k}{3m})(-\Omega^2 + i\gamma\Omega + \frac{2k}{3m}) - (-i\gamma\Omega)(-i\gamma\Omega) = 0$

I get $\Omega^4 -2i\gamma \Omega^3 - \frac{4k}{3m}\Omega^2 + i\frac{4k}{3m}\gamma\Omega + \frac{4k^2}{9m^2} = 0$

I don't think this is the correct way.
I don't find a way to resolve a quadratic.
I have to factor some terms, but I don't see where.

 

[fresh_42]

You have $f(\Omega)^2=g(\Omega)^2$. Why don't you check all possibilities from here?

 

[pasmith]

a^2 - b^2 = (a +b)(a - b) also holds for complex a and b...

 

[Redwaves]

I found the the reals roots which are $\Omega = + -\sqrt{\frac{2k}{3m}}$ using $(-\Omega^2 +i\gamma\Omega + \frac{2k}{3m}))^2 = i^2\gamma^2\Omega^2$

However, I don't see how to get the complex roots.
To find the complex roots, should I replace $i^2$ by -1 ?

 

[fresh_42]

You have $f(\Omega) =\pm g(\Omega)$ or $a=\pm b$ with @pasmith 's notation.​

Now $f(\Omega)$ is quadratic, that gives you $4$ roots, two real and two complex.

 

[Redwaves]

Alright, Thanks!
I found it using Pasmith's notation.
Thanks guys!