$$\frac{[(\omega_0^2-\omega^2)-2i\omega\gamma]^2}{[(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2]^2}=\frac{1}{[(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2]}$$

I don't understand how can I get to that solution. Any hint will be very thankful. The denominator for that fraction it can be write like adjoint but after that I will remain with a quadric equation and it won't be the same as the solution.

• Assuming that you want to solve the equation above. Let $z\in\mathbb{C}$. Suppose that $z^2$ is a positive real number. Show that $z\in\mathbb{R}$. If you want to prove that two sides of the equation are equal, then good luck. They are NOT equal except at very few values of $\gamma$ and $\omega$. – Batominovski Jul 29 '15 at 18:11

Note that on the numerator of the LHS of the equation, the [...]$^2$ operation is actually taking the modulus squared. That is, $|a+bi|^2 = a^2+b^2$.