If the least and the largest real values of a, for which the
equation z + $\alpha$|z – 1| + 2i = 0 (z $\in$ C and i = $\sqrt { - 1}$) has a solution, are p and q respectively; then 4(p² + q²) is equal to __________.

$x + iy + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0$<br><br>$\therefore$ y + 2 = 0 and $x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0$<br><br>y = $-$2 &amp; ${x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4)$<br><br>$${\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0$$<br><br>$x \in R \Rightarrow D \ge 0$<br><br>$4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0$<br><br>${\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0$<br><br>${\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0$<br><br>${\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0$<br><br>$0 \le {\alpha ^2} \le {5 \over 4}$<br><br>$\therefore$ ${\alpha ^2} \in \left[ {0,{5 \over 4}} \right]$<br><br>$\therefore$ $\alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right]$<br><br>then $4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10$