Let $\alpha ,\beta$ be real and $z$ be a complex number. If $z^2+\alpha z+\beta =0$ has two distinct roots on the line $\text{Re}(z)=1$, then it is necessary that:

Let roots be $1+iy$ and $1-iy$ (since coefficients are real, roots must be conjugates for the real part to be same).Sum of roots $=(1+iy)+(1-iy)=2=-\alpha ⟹\alpha =-2$.Product of roots $=(1+iy)(1-iy)=1+y^2=\beta$.Since roots are distinct, $y\ne 0$, so $y^2>0$.Therefore $\beta =1+y^2>1$.$\beta \in (1,\infty )$.