I would like to show that
The image of any group homomorphism $\phi: SL_2(\Bbb Z) \to \Bbb C^\times$ is contained in the finite subgroup $\{1, \omega, \ldots, \omega^{11}\}$ of $\Bbb C^\times$, where $\omega = e^{\frac{i\pi}{6}}$.
We can find $S$ and $T$ such that $|S| = 4$, $|T| = 6$, and $SL_2(\Bbb Z) = \langle S,T\rangle$. Here, $|\cdot|$ denotes the order of an element in a group. Now, the image of $\phi$ is a subgroup of $\Bbb C^\times$, and it is generated by $\phi(S)$ and $\phi(T)$. That is, $\phi(SL_2(\Bbb Z)) = \langle \phi(S), \phi(T)\rangle$. As $S^4 = T^6 = 1$, we have $(\phi(S))^4 = (\phi(S))^6 = 1$. $\phi(S)$ is a fourth root of unity, and $\phi(T)$ is a sixth root of unity. With this information, how can we show that $$\phi(SL_2(\Bbb Z)) = \langle \phi(S), \phi(T)\rangle \subset \{1, \omega, \ldots, \omega^{11}\}$$ where $\omega = e^{\frac{i\pi}{6}}$?
You already know that $\phi(S)= e^{\frac{2\pi ia} 4}$ and $\phi(T)= e^{\frac{2\pi i b}{6}} $ for some $a,b \in \mathbb Z$.
Since $\mathbb C^*$ is abelian, for any $\gamma \in SL_2(\mathbb Z)$ it holds $\phi(\gamma)=\phi(S)^u\cdot \phi(T)^v$ for some $u,v \in \mathbb Z$. Therefore, $$\phi(\gamma)= e^{\dfrac{2\pi i au}4}\cdot e^{\dfrac{2\pi i vb}{6}} =e^{\dfrac{2\pi i (3au+2bv)}{12}} = \omega^{ 3au+2bv}\in \{1,\omega,\ldots, \omega ^{11}\}$$