For $a>0$ and $s=x+iy$, I've got followig equations in my scriptum:
$|a^{s-1}e^{-na}|=|a^{x-1}e^{-na}e^{ilogy}|=a^{x-1}e^{-na}$
I've got no idea how they've calculated the second and third equation. Anyone can give me hint? I know that, $|a^{s-1}e^{-na}|=|a^{(x+iy)-1}e^{-na}|=|a^{x-1}a^{iy}e^{-na}|=..$
Now I don't know how to go on.
Well, we have:
$$\text{D}:=\left|\text{a}^{\text{s}-1}\cdot\exp\left(-\text{a}\text{n}\right)\right|=\frac{\left|\text{a}^{\text{s}-1}\right|}{\left|\exp\left(\text{a}\text{n}\right)\right|}=\frac{\left|\text{a}^{\Re\left(\text{s}\right)-1}\cdot\text{a}^{\Im\left(\text{s}\right)i}\right|}{\exp\left(\text{a}\text{n}\right)}=\frac{\left|\text{a}^{\Re\left(\text{s}\right)-1}\right|\cdot\left|\text{a}^{\Im\left(\text{s}\right)i}\right|}{\exp\left(\text{a}\text{n}\right)}\tag1$$
Assuming $\text{n}\in\mathbb{R}^+_0$
Now, we get:
$$\left|\text{a}^{\Im\left(\text{s}\right)i}\right|=\left|\exp\left(\Im\left(\text{s}\right)\ln\left(\text{a}\right)i\right)\right|=1\tag2$$
So, we get:
$$\text{D}=\frac{\left|\text{a}^{\Re\left(\text{s}\right)-1}\right|}{\exp\left(\text{a}\text{n}\right)}=\frac{\left|\text{a}^{\Re\left(\text{s}\right)}\cdot\text{a}^{-1}\right|}{\exp\left(\text{a}\text{n}\right)}=\frac{1}{\text{a}}\cdot\frac{\left|\text{a}^{\Re\left(\text{s}\right)}\right|}{\exp\left(\text{a}\text{n}\right)}=\frac{1}{\text{a}}\cdot\frac{\text{a}^{\Re\left(\text{s}\right)}}{\exp\left(\text{a}\text{n}\right)}=\frac{\text{a}^{\Re\left(\text{s}\right)-1}}{\exp\left(\text{a}\text{n}\right)}\tag3$$