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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.

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

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