If the least and the largest real values of $\alpha$, for which the equation $\mathrm{z}+\alpha|\mathrm{z}-1|+2 i=0$
( $\mathrm{z} \in \mathrm{C}$ and $i=\sqrt{-1}$ ) has a solution, are $\mathrm{p}$ and $\mathrm{q}$ respectively; then $4\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)$ is equal to
Put $z=x+i y$
$x+i y+\alpha|x+i y-1|+2 i=0$
$\Rightarrow \quad x+\alpha \sqrt{(x-1)^{2}+y^{2}}+i(y+2)=0+0 i$
$\Rightarrow \quad y+2=0$ 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 $\frac{x^{2}}{x^{2}-2 x+5} \in\left[0, \frac{5}{4}\right]$
$\therefore \quad \alpha^{2} \in\left[0, \frac{5}{4}\right] \quad \Rightarrow \alpha \in\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}\right]$
$\therefore \quad \mathrm{p}=-\frac{\sqrt{5}}{2} ; \mathrm{q}=\frac{\sqrt{5}}{2}$
$\Rightarrow \quad 4\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$
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