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

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.

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 $$x=\frac{2\pi sb}{q}$$.