I want to show the following: if for $t_0\ne 0$ we have $\lvert\varphi(t_0)\rvert = 1$, then there exists a constant $a$ such that $P(X\in a+\frac{2\pi}{t_0}\mathbb{Z}) = 1$.
I'm stuck on how to actually use the premise. Rewriting directly led me to $$1=\lvert \varphi(t_0)\rvert = \left\lvert\int e^{it_0 x}P_X(dx)\right\rvert \implies \int e^{it_0x}P_X(dx) = \cos(r) + i\sin(r)$$
But I don't see how this might help. I think I need to approach it from a different angle and then use the premise, but I don't know what that angle might be. Does anyone have any ideas?
You have that $\mathbb{E}[e^{it_0 X}] = e^{it_0 a}$ for some $a \in \mathbb{R}$. It follows that $\mathbb{E}[e^{it_0 (X - a)}] = 1.$ In particular, we get for the real part $\Re$ $$1 = \Re(\mathbb{E}[e^{it_0( X - a)}]) = \mathbb{E}[\cos(t_0(X -a))].$$ Note that $\cos(t_0(X -a)) < 1$ if $t_0(X -a) \notin 2\pi\mathbb{Z}$. It follows that $$\mathbb{P}\left(X \in a + \frac{2\pi}{t_0}\mathbb{Z}\right) = 1.$$