Suppose we have angular momentum operators $L_1,L_2,L_3$ which satisfy $[L_1,L_2]=iL_3$, $[L_2,L_3]=iL_1$ and $[L_3,L_1]=iL_2$. We can show that the operator $L^2:=L_1^2+L_2^2+L_3^2$ commutes with $L_1,L_2$ and $L_3$. Now define $L_{\pm}=L_1\pm iL_2$ and then we can also show that $$L_+L_-=L^2-L_3(L_3-I)$$ and $$L_-L_+=L^2-L_3(L_3+I)$$ If now we have a vector $v$ in our Hilbert space $\mathcal{H}$ such that $L^2v=\lambda v$ and $L_3 v=\lambda_3v$. Then it is easy to show using the commutation relations that $L_+v$ is an eigenvector of $L^2$ with eigenvalue $\lambda$ and it is also an eigenvector of $L_3$ with eigenvalue $1+\lambda_3$. Now I have to show that $$||L_+v||^2=|\lambda-\lambda_3(\lambda_3+1)|||v||^2$$
I dont see how this can be deduced from the above calculations, we do know that $$||L_-L_+v||^2=|\lambda-\lambda_3(\lambda_3+1)|^2||v||^2$$ This is true because of the second formula above for $L_-L_+$.
I need a hint for this calculation. Thanks.
(promoting my comment to an answer)
You can use the fact that $L_+$ and $L_-$ are each others adjoints, IOW $$\langle L_+x|y\rangle=\langle x|L_-y\rangle$$ for all $x,y$. Applying this to $y=L_+v, x=v$ gives $$ \begin{aligned} \Vert L_+v\Vert^2&=\langle L_+v\mid L_+v\rangle\\ &=\langle v\mid L_-L_+ v\rangle\\ &=\langle v\mid (\lambda-\lambda_3(1+\lambda_3))v\rangle\\ &=(\lambda-\lambda_3(1+\lambda_3))\Vert v\Vert^2. \end{aligned} $$ Consequently the scalar $\lambda-\lambda_3(1+\lambda_3)$ is a non-negative real number (most likely you already knew this by other means), so it is unnecessary to wrap it inside absolute value signs.