I'm learning about phasors and light intensity and there are two conflicting things I've been told that I can't reconcile.
Firstly I've been told that a phasor $e^{-ikz+i\phi}$ is a mathematical way to represent an electromagnetic plane wave where the physical electric field is given by the real part of the phasor times a time factor $e^{i\omega t}$ (so $\cos(-kz+\omega t+\phi)$ as expected).
I've also been told that the intensity of a beam of light is given by (something proportional to) the square of the amplitude of the whole phasor and not just the real part. So for instance if I had a combined beam made of two waves (say as in a Michelson Interferometer) the intensity at $z = 0, t = 0$ would be given by $$I \propto |A_re^{i\phi_r}+A_se^{i\phi_s}|^2$$
where $A_re^{i\phi_r}$ is the phasor for the first wave and $A_se^{i\phi_s}$ is the phasor for the second wave (at $z=0$).
However this confuses me because I also know that the intensity of a wave is proportional to its amplitude squared and the amplitude of the combined wave equals the real part of the phasor, so $A_r\cos{\phi_r}+A_s\cos{\phi_s}$ at $z=0,t=0$. Therefore the intensity of the combined wave should be proportional to: $$I \propto |A_r\cos(\phi_r)+A_s\cos(\phi_s)|^2$$
These two formulas are very different though. Searching online it looks like the correct formula is the one with the full phasor and not just the real part of it, so my question is then why do we use the full phasor when computing the intensity of a beam of light instead of just the real part of it which corresponds to the physical electric field?
I know it can't be because of the contribution of the magnetic field since that's in phase with the electric field in a plane wave so it adding it in should just multiply the cosine formula for intensity by 2, not replace it with the full phasor. So what then is the reason the intensity is given by the full phasor and where does the $iA\sin(\phi)$ factor come from?
