Put $z=x+$ iy

$x+i y+\alpha \mid x+$ iy $-1 \mid+2 i=0$

$\Rightarrow \quad x+\alpha \sqrt{(x-1)^{2}+y^{2}}+i(y+2)=0+0 i$

$\Rightarrow \quad y+2=0$ and $x+\alpha \sqrt{(x-1)^{2}+y^{2}}=0$

$\Rightarrow \quad y=-2$ and $\alpha^{2}=\frac{x^{2}}{x^{2}-2 x+5}$

Now $\frac{x^{2}}{x^{2}-2 x+5} \in\left[0, \frac{5}{4}\right]$

$\therefore \quad \alpha^{2} \in\left[0, \frac{5}{4}\right] \Rightarrow \alpha \in\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}\right]$

$\Rightarrow \quad 4\left(p^{2}+q^{2}\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$