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.

  • $\begingroup$ 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

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)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.