If the least and the largest real values of α , for which the equation z + α z - 1 + 2 i = 0 z…

Put $z=x+iy$

$x+iy+\alpha \left|x+iy-1\right|+2i=0$

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

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

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

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

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

$\therefore p=-\frac{\sqrt{5}}{2};q=\frac{\sqrt{5}}{2}$

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