If the least and the largest real values of $\alpha$, for which the equation $z+\alpha|z-1|+2 i=0 \q

Question

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

Professor PC Thomas Classes · Answer

(10) : The given equation is\ $z+\alpha|z-1|+2 i=0$\ Putting $z=x+i y$, we get\ $x+i y+\alpha|x+i y-1|+2 i=0$\ $\Rightarrow \quad x+\alpha \sqrt{(x-1)^{2}+y^{2}}+i(y+2)=0+0 i$\ $\Rightarrow y+2=0$ and $x+\alpha \sqrt{(x-1)^{2}+y^{2}}=0$\ $\Rightarrow y=-2$ and $\alpha^{2}=\frac{x^{2}}{x^{2}-2 x+5}$\ Now, $\frac{x^{2}}{x^{2}-2 x+5} \in\left[0, \frac{5}{4}ight]$\ $	herefore \alpha^{2} \in\left[0, \frac{5}{4}ight] \Rightarrow \alpha \in\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}ight]$\ $	herefore p=-\frac{\sqrt{5}}{2} ; q=\frac{\sqrt{5}}{2}$\ $\Rightarrow \quad 4\left(p^{2}+q^{2}ight)=4\left(\frac{5}{4}+\frac{5}{4}ight)=10$\