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 & $e^{-y}$
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:
$|e^{-y} \cos x + i e^{-y} \sin x|=|e^{-y}||\cos x + i \sin x|=e^{-y}|e^{ix}|=e^{-y}$
If
$z = x + iy$
then
$|z| = \sqrt{x^2+y^2}$
Letting $z = e^{-y} \cos x + i e^{-y}\sin x$, we have $\Re (z) = e^{-y} \cos x$ and $\Im (z) = e^{-y}\sin x$. Assuming $y \in \mathbb R$:
$|z| = \sqrt{(e^{-y} \cos x)^2 + (e^{-y}\sin x)^2} = \sqrt{e^{-2y} ((\cos x)^2 + (\sin x)^2)} = \sqrt{e^{-2y}} = e^{-y}$
If you take the square root of both sides, you will see your equality is if fact false.
Since ${e^{ix}} = \cos (x) + i\sin (x)$ we have $\left| {{e^{ix}}} \right| = \sqrt {{{\cos }^2}(x) + {{\sin }^2}(x)} = 1$ so ...
What is the formula for $|e^{-y}\cos x+ie^{-y}\sin x|$? Do you see any trig identities you can use? The complex number isn't on the unit circle because it's been scaled by $e^{-y}$.