I’d like to solve for $x$ in the equation:
$y = \lvert e^{-icx}\rvert$
where $x$ and $y$ are real constants and $c$ is a complex number of the form $a+ib$. Recognizing that the modulus of a complex number is the square root of the sum of the squares of the real and imaginary parts, I’ve tried to expand the equation in terms of sines and cosines but got stuck. Any ideas?
$|e^{-ixc}|= |e^{-ix(a+ib)}|= |e^{-ixa} \cdot e^{xb}|= |e^{-ixa}| \cdot |e^{xb}|= e^{xb}$, since $|e^{iu}|=1 $ for all $u\in \mathbb R$ (Euler's formula). Therefore, $x= \dfrac{\ln y}b$.