I have the following number that I have to calculate the square but I have doubts on the complex part... Can someone help me solving my problem?
$$\left(\frac{2-4(\cos(2 \pi \nu ) - j \sin(2 \pi \nu))}{2-(\cos(2 \pi \nu ) - j \sin(2 \pi \nu))}\right)^2$$
Thank you
If you square the numerator and denominator separately and then do the division, you get the same result as if you divide first and then square, so which is better may be just a matter of which costs you the least effort. I'm not sure which one that will be. \begin{align} & \frac{2 - 4\exp(-2\pi\nu j)}{2 - \exp(-2\pi\nu j)} \\[8pt] = {} & \frac{2 - 4\exp(-2\pi\nu j)}{2 - \exp(-2\pi\nu j)} \cdot \frac{2-\exp(2\pi\nu j)}{2-\exp(2\pi\nu j)} \\[8pt] = {} & \frac{4 - 2\exp(2\pi\nu j)-8\exp(-2\pi\nu j) + 4}{2 - 2\exp(2\pi\nu j) - 2\exp(-2\pi\nu j) + 1} \\[8pt] = {} & \frac{8 - 2(\cos(2\pi\nu ) + j\sin(2\pi \nu)) -8(\cos(2\pi\nu) - j\sin(2\pi\nu))}{3- 4\cos(2\pi\nu)} \\[8pt] = {} & \frac{8 - 10\cos(2\pi\nu)}{3-4\cos(2\pi\nu)} + j \frac{6\sin(2\pi\nu)}{3-4\cos(2\pi\nu)} \\[10pt] = {} & a+bj \end{align} Next use the identity $(a+bj)^2 = (a^2-b^2) + 2abj.$
Possibly some simplifications based on trigonometric identities may follow.