If z=1/2-2 i is such that |z+1|=α z+β(1+i), i=√ -1 and α, β ∈ R, then α+β is equal to

If $\displaystyle z=\frac{1}{2}-2 i$ is such that $\displaystyle |z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\displaystyle \alpha, \beta \in R$, then $\displaystyle \alpha+\beta$ is equal to