I am working through the exercises on Deitmar's Automorphic Forms. For the following question
(Exercise 4.4) Let $ \mathbb{T} := \{z \in \mathbb{C} : |z| = 1\}$ denote the circle group and $\chi: \mathbb{Z}_p \to \mathbb{T}$ be a continuous group homomorphism. Show that there exists $k \in \mathbb{N}$ with $\chi(p^k \mathbb{Z}_p) = 1$. It follows that $\chi$ factors through the finite group $\mathbb{Z}_p/p^k \mathbb{Z}_p \cong \mathbb{Z}/p^k \mathbb{Z}$, so the image of $\chi$ is finite. (Hint: Let $U := \{z \in \mathbb{T}: \Re(z) >0\}$. Then $U$ is open neighbourhood of the unit, so $\chi^{-1}(U)$ is an open neighbourhood of zero.)
My attempt: Following the hint in the question, I am show that there exists a $k \in \mathbb{N}$ so that $p^k \mathbb{Z}_p$ so that $\chi(p^k \mathbb{Z}_p) \subset U$. However, I am not sure how to proceed with this.
I have also considered the closed set $\chi^{-1}(\{1\})$ in $\mathbb{Z}_p$, where $\{1\}$ is compact set in $\mathbb{T}$ and as $\mathbb{Z}_p$ is Hausdorff then $\chi^{-1}(\{1\})$ is compact in $\mathbb{Z}_p$. Again, I am not sure how to proceed with this.
Any hints are appreciated.
From the hint provided by @user10354138:
We have $\chi(p^k \mathbb{Z}_p)$ is subgroup in $\mathbb{T}$ and suppose for a contradiction there exists $$e^{i \theta_0} =z_0 := \chi(p^k m_0) \in \chi(p^k \mathbb{Z}_p)$$ so that $|z_0| = 1$ but $z_0 \neq 1$ and $\theta \in (-\pi/2, 0)\sqcup (0, \pi/2), m_0 \in \mathbb{Z}_p$.
Without loss of generality assume $\theta_0 \in (0, \pi/2)$ and let $r \in \mathbb{N}$ be so that $r\theta_0 \in (\pi/2, \pi)$ we get $$e^{i r \theta_0} = z_0^k = \chi(p^k m_0)\cdots \chi(p^k m_0) = \chi(p^k m_0 + \dots + p^k m_0) \in \chi(p^k \mathbb{Z}_p) \subset U $$ is a contradiction as $\Re(z_0^k) < 0$. With this, we can conclude that $\chi(p^k \mathbb{Z}_p) =1$.
Alternatively, I think this should work as a one-line proof: $\chi(p^k \mathbb{Z}_p)$ is subgroup in $\mathbb{T}$ contained in $U$, and the only subgroup of $\mathbb{T}$ contained in $U$ is precisely $\{1\}$.