[proofplan] We use the finite-dimensional classification of irreducible $\mathfrak{sl}_2(\mathbb C)$-modules to choose a basis of $L(m)$ consisting of weight vectors. This identifies all weights of $L(m)$ as $m,m-2,\dots,-m$, each with multiplicity one, so the character is the corresponding finite sum. The quotient formula is then only the closed form of this finite Laurent geometric progression. [/proofplan] [step:Read the weights from the standard basis of $L(m)$] By [citetheorem:9363], the module $L(m)$ has a basis \begin{align*} \{v_i:0\le i\le m\} \end{align*} such that \begin{align*} h v_i=(m-2i)v_i \end{align*} for every integer $i$ with $0\le i\le m$. For each such $i$, the vector $v_i$ is therefore a weight vector of weight $m-2i$. The numbers $m-2i$ are pairwise distinct for $0\le i\le m$: if $m-2i=m-2j$, then $i=j$. Since the vectors $v_i$ form a basis of $L(m)$, it follows that \begin{align*} L(m)_{m-2i}=\mathbb C v_i \end{align*} for $0\le i\le m$, and every other weight space is zero. [guided] The classification theorem gives more than the existence of $L(m)$: it gives an explicit weight basis. By [citetheorem:9363], there are vectors \begin{align*} v_i\in L(m) \end{align*} indexed by the integers $i$ satisfying $0\le i\le m$, and these vectors form a basis of $L(m)$. The action of the Cartan element $h\in\mathfrak{sl}_2(\mathbb C)$ on this basis is \begin{align*} h v_i=(m-2i)v_i. \end{align*} This equation says exactly that $v_i$ lies in the weight space of weight $m-2i$. Thus \begin{align*} v_i\in L(m)_{m-2i} \end{align*} for every $0\le i\le m$. We must also check the multiplicities. The weights listed above are pairwise distinct: if \begin{align*} m-2i=m-2j, \end{align*} then subtracting $m$ from both sides gives $-2i=-2j$, hence $i=j$. Therefore no two basis vectors in this list contribute to the same weight space. Since the vectors $v_i$ form a basis of the entire module, there are no additional weight vectors outside their span. Hence \begin{align*} L(m)_{m-2i}=\mathbb C v_i \end{align*} for every $0\le i\le m$, and all other weight spaces are zero. [/guided] [/step] [step:Sum the one-dimensional weight spaces to compute the character] By the definition of the character of a finite-dimensional weight module, \begin{align*} \operatorname{ch} L(m)=\sum_{\lambda\in\mathbb C}\dim L(m)_\lambda q^\lambda. \end{align*} The preceding step shows that the only nonzero weight spaces are $L(m)_{m-2i}$ for $0\le i\le m$, and each has dimension $1$. Therefore \begin{align*} \operatorname{ch} L(m)=\sum_{i=0}^{m} q^{m-2i}. \end{align*} [/step] [step:Rewrite the finite character as a Laurent geometric quotient] Let \begin{align*} S_m:=\sum_{i=0}^{m}q^{m-2i}. \end{align*} Factoring out $q^m$ gives \begin{align*} S_m=q^m\sum_{i=0}^{m}q^{-2i}. \end{align*} Using the finite geometric series identity with ratio $q^{-2}$, \begin{align*} \sum_{i=0}^{m}q^{-2i}=\frac{1-q^{-2m-2}}{1-q^{-2}}. \end{align*} Thus \begin{align*} S_m=q^m\frac{1-q^{-2m-2}}{1-q^{-2}}. \end{align*} Multiplying numerator and denominator by $q$ after simplifying the powers gives \begin{align*} S_m=\frac{q^{m+1}-q^{-(m+1)}}{q-q^{-1}}. \end{align*} Since $S_m=\operatorname{ch}L(m)$ by the previous step, this proves the quotient formula and completes the proof. [/step]