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

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.

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  • $\begingroup$ 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 $\omega$. $\endgroup$ – Batominovski Jul 29 '15 at 18:11
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Note that on the numerator of the LHS of the equation, the [...]$^2$ operation is actually taking the modulus squared. That is, $|a+bi|^2 = a^2+b^2$.

Work it out. The conclusion is trivial after you note that.

Complex modulus is the notion of "length" in the complex plane (http://mathworld.wolfram.com/ComplexModulus.html)

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  • $\begingroup$ and this is a PERFECT end for a 'workaholic' day. thank you so much for your answer, this really helped me. $\endgroup$ – dsadfas Jul 29 '15 at 18:30

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