If $\alpha ,\beta \in R$  are such that $1-2i(here{i}^2=-1)$is a root of $z^2+\alpha z+\beta =0,then(\alpha -\beta )$ is equal to

Given $z^2+\alpha z+\beta =0$ has one root 1 – 2i

            Hence, the other root is conjugate of the other   ( since the coefficients $\alpha ,\beta$ are real ), then the other root is 1 + 2i

            Therefore, $$\begin{gathered} sum of the roots =-\alpha =2; and product of the roots =\beta =\left(1+2i\right)\left(1-2i\right)=5 \\ \Rightarrow \alpha -\beta =-2-5=\boxed{-7} \end{gathered}$$