How does $\sqrt{|e^{-y}\cos x + ie^{-y}\sin x|}= e^{-y}$ How does $\sqrt{|e^{-y}\cos x + ie^{-y}\sin x|} = e^{-y}$ which is less than $1$? 
This is a step from a question I am doing but I am not sure how the square root equaled & $e^{-y}$ 

I'd start by using properties of the modulus:
$$|e^{-y} \cos x + i e^{-y} \sin x|=|e^{-y}||\cos x + i \sin x|=e^{-y}|e^{ix}|=e^{-y}$$
