Question

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

Answer 1

$x+iy+\alpha \sqrt{(x-1{)}^2+y^2}+2i=0$
$\Rightarrow y+2=0$ and $x+\alpha \sqrt{(x-1{)}^2+y^2}=0$
$y=-2\&x^2={\alpha }^2(x^2-2x+1+4)$
$\Rightarrow x^2({\alpha }^2-1)-2x{\alpha }^2+5{\alpha }^2=0$
$\because x\in R\Rightarrow D\ge 0$
$\Rightarrow 4{\alpha }^4-4({\alpha }^2-1)5{\alpha }^2\ge 0$
$\Rightarrow {\alpha }^2[4{\alpha }^2-20{\alpha }^2+20]\ge 0$
$\Rightarrow {\alpha }^2[-16{\alpha }^2+20]\ge 0$
$\Rightarrow {\alpha }^2[{\alpha }^2-\frac{5}{4}]\ge 0$
$\Rightarrow {\alpha }^2\in [0,\frac{5}{4}]$
$\Rightarrow \alpha \in [-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}]$
then $4[p^2+q^2]=4[\frac{5}{4}+\frac{5}{4}]=10$