Can somebody please help me to prove the formula below
$\frac{1}{T} \int_{0}^{T} \operatorname{Re}\left[A e^{i \omega t}\right] \operatorname{Re}\left[B e^{i \omega t}\right] d t=\frac{1}{2} \operatorname{Re}\left[A B^{*}\right]$
Where T is the period, A and B are some complex numbers and B* denotes complex conjugate.
On
It is important to make it clear that $T$ is such that $\omega T=2\pi$. Then it's just a matter of $A=A_r+j A_i$, $B=B_r+jB_i$, $e^{j\omega t}=\cos(\omega t)+j\sin(\omega t)$ so that your integral becomes \begin{align} &\frac{1}{T}\int_0^T dt (A_r\cos\omega t-A_i\sin(\omega t)) (B_r\cos\omega t-B_i\sin(\omega t))\\ &=\frac{1}{T}\int_0^T\,dt (A_rB_r\cos^2(\omega t)+ A_iB_i\sin^2(\omega t))\\ &\qquad\qquad -\frac{1}{T}\int_0^T\,dt\,(A_rB_i+A_iB_r)\cos(\omega t)\sin(\omega t) \end{align} After a few elementary integration using the proper limits, the cross-terms cancel out and you get $$ \frac{1}{T}\frac{\pi}{\omega}(A_rB_r+A_iB_i)=\frac{1}{2} (A_rB_r+A_iB_i)=\frac{1}{2}\hbox{Re}(AB^*) $$
Just write $A$ as $x+iy$, $B$ as $u+iv$, and $e^{i\omega t}$ by Euler's formula, take their real parts, expand and integrate directly. The result will be equal to the RHS given.