ML-inequality: Why does this hold $|e^{-3y+3ix}| = e^{-3y}$ during numerator estimation of $f(z) = \frac{e^{3iz}}{z^2 + 1}$

Question

Given the following related to an ML-inequality for $R > 1$:

Estimation of the numerator from the function $f(z)$ is supposed to develop as follows:

I'm wondering why and how exactly the power $3ix$ cancels out (why does it vanish/equal to zero) in the equality $|e^{-3y+3ix}| = e^{-3y}$ ?

Answer 1

Let $z=\alpha+\beta i$ be a complex number. Then \begin{align} |e^z| &= |e^{\alpha +\beta i}|\\ &=|e^{\alpha}e^{\beta i}|\\ &=|e^\alpha| |e^{\beta i}|\\ &=e^\alpha|\cos\beta +i\sin\beta|\\ &=e^\alpha\sqrt{\cos^2\beta +\sin^2\beta}\\ &=e^\alpha\cdot \sqrt{1}\\ &=e^\alpha. \end{align}

Answer 2

Let $x \in \mathbb{R}$. Now, $e^{ix} = \cos(x) + i \sin(x)$, so $|e^{ix}| = \cos^2(x) + \sin^2(x) = 1$.