Problem
Let $\mathbb D_3$ be the symmetry group of the equilateral triangle. Give a complete list of the representatives of the simple left $\mathbb C[\mathbb D_3]$-modules.
My attempt at a solution
I know that a $K[G]$-module is in correspondence with a representation $(V,\rho)$ of the group $G$.
So maybe the idea is to define morphisms $$\phi: \mathbb D_3 \to \operatorname{Aut}_{\mathbb C}(\mathbb C)$$
It occurred to me to define two:
$$\phi_1:\mathbb D_3 \to \operatorname{Aut}_{\mathbb C}(\mathbb C)$$$$r \to 1.$$$$s \to 1.$$ $$\phi_2:\mathbb D_3 \to \operatorname{Aut}_{\mathbb C}(\mathbb C)$$$$r \to 1.$$$$s \to -1.$$
My big question is, how do I know there aren't any more morphisms?
Simple modules are irreducible representations of $D_3$. It is a well-known fact that there is a bijection between the set of isomorphism classes of irreducible representations of a finite group $G$ and the set of conjugacy classes in $G$. There are $3$ conjugacy classes in $D_3$, so $3$ irreducible reps (up to iso). Another well-known fact is that if $\{d_i\}$ is the set of the dimensions of the irreducible reps of $G$ then $\sum d_i^2=|G|$. Therefore in your case there are $3$ integers such thath $d_1^2+d_2^2+d_3^2=6$. It is easy to see that the only possibility is $d_1=d_2=1$, $d_3=2$ (up to order of course). So the answer to your question is: if the two reps that you found are non-isomorphic, and they are as you can easily check, then they are the only one up to isomorphism. Since isomorphisms of reps are given by conjugation, and $\text{Aut}(\mathbb{C})=\mathbb C$ is abelian, the two homomorphisms that you found are the only ones.
This of course you could also see it in an elementary way, since homomorphisms are determined by normal subgroups. You are still lacking an irreducible $2$-dimensional representation, though, namely a simple $2$-dimensional $\mathbb C[D_3]$-module.