Let $L_m$ be the $m+1$ dimensional simple module over $\mathfrak{sl}_2(\Bbb{C})$, and $M_\lambda,\; \lambda\in \Bbb C$, the Verma module with highest weight $\lambda$ over $\frak{sl}_2(\Bbb C)$. Show the following :
For any $\lambda$, $L_m \otimes M_\lambda$ carries a filtration by submodules $$\{0\}=N_{m+1}\subset N_m\subset \cdots \subset N_1\subset N_0=L_m\otimes M_\lambda$$ such that $N_i/N_{i+1}\cong M_{\lambda+m-2i}$.
Prove that if $\lambda\notin \{0,1,2,\dots, m-1\}$, then $L_m \otimes M_\lambda$ is completely reducible and, furthermore, $$L_m\otimes M_\lambda \cong M_{m+\lambda}\otimes M_{m+\lambda-2}\oplus \cdots \oplus M_{\lambda-m}$$
I am stumped by both these problems. For the first I tried to proceed in the following way : first I wanted to find a singular vector for the copy of $M_{\lambda-m}\cong N_m/N_{m+1}\cong N_m$ inside $M_\lambda$. It must lie inside the eigenspace with eigenvalue $\lambda-m$, which is generated by $f^iv_m\otimes f^{m-i}v_\lambda, 0\le i\le m$ where $v_m$ amd $v_\lambda$ are highest weight vectors in $L_m$ and $M_\lambda$ respectively. Then I tried to solve $$e\cdot(f^{m}v_m\otimes v_\lambda +\sum_{i=1}^m x_if^{m-i}v_m\otimes f^{i}v_\lambda)=0 $$ but that became too complicated. Even if I could solve this, I don't know how to choose the next submodule in the filtration.
For the second part, I have no idea, what I'm mostly confused about is why is there a restriction about $\lambda\notin\{0, \dots, m-1\}$? What happens if $\lambda$ is one of those integers? What fails? I feel to answer that I would need to solve the first part but I am not able to do either. Any help would be appreciated, thanks!
BTW : I am using the decomposition of $\mathfrak{sl}_2(\Bbb c)$ with basis $e, f, h$ with $[e,f]=h, [h,e]=2e, [h, f]=-2f$.