Why is the following exponential ignored (or equals 1) in the probability amplitude?

Question

I'm reading Ronald de Wolf's lecture notes and when explaining Shor's algorithm on page 40 after applying a QFT to

$$ \frac{1}{\sqrt{m}} \sum_{j=0}^{m-1} |jr+s\rangle $$

the following expression is shown to describe the state of the first register: $$ \frac{1}{\sqrt{m}} \sum_{j=0}^{m-1} \frac{1}{\sqrt{q}} \sum_{b=0}^{q-1} e^{\frac{2\pi i(jr+s)b}{q}} |b\rangle = \frac{1}{\sqrt{mq}} \sum_{b=0}^{q-1} e^{\frac{2\pi isb}{q}} \left( \sum_{j=0}^{m-1} e^{\frac{2\pi ijr b}{q}} \right) |b\rangle $$

Soon after when looking at the case where $r$ divides $q$ it is noted that every $b$ for which $e^{\frac{2\pi ijr b}{q}}$ is $1$ the inner sum will equal $m$, resulting in these $b$ outcomes having a squared amplitude of $(m/\sqrt{mq})^2=1/r$. I don't understand at what point the $e^{\frac{2\pi isb}{q}}$ exponential got lost, I would assume that it must be 1 for all $b$ and $q$ but I don't see how.

Answer 1

The amplitude for these $|b\rangle$ will be $$\left|\frac{me^{\frac{2\pi i s b}{q}}}{\sqrt{mq}} \right|^2 = \frac{1}{r}\left|e^{\frac{2\pi i s b}{q}}\right|^2 = \frac{1}{r}$$

since $|e^{ix}| = |\cos(x)+i\sin(x)| = \sqrt{\cos^2(x) + \sin^2(x)}=1$ for every $x\in \mathbb{R}$. Here $x=\frac{2\pi sb}{q}$.