why is abs(e^(-i*ωt))*abs(e^(-i*ωt)) = 1
all complex exponentials have modulus 1think about polar form expressed using a complex exponential; the magnitude is encoded in what multiplies the complex exponential which encodes the phase
>>16784893[math]|e^{ix}|^2 = e^{ix} \cdot e^{-ix} = e^{(ix - ix)} = e^0 = 1[/math]
