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

Question

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

Answer: 10

Accepted Answer

Let $z=x+iy$. Then $x+\alpha\sqrt{(x-1)^2+y^2}+i(y+2)=0$ gives $y=-2$ and $x=-\alpha\sqrt{(x-1)^2+4}$.

Squaring: $x^2=\alpha^2(x^2-2x+5) \implies \alpha^2=\dfrac{x^2}{x^2-2x+5}$, whose range is $\left[0,\dfrac54\right]$, so $\alpha\in\left[-\dfrac{\sqrt5}2,\dfrac{\sqrt5}2\right]$.

$$4(p^2+q^2)=4\left(\frac54+\frac54\right)=10.$$

Solutions · 1