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(B) Let $z = x + iy$. Substituting this into the equation:$x + iy + \alpha|x + iy - 1| + 2i = 0$$x + \alpha\sqrt{(x-1)^2 + y^2} + i(y + 2) = 0$Equatin

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$10$

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(B) Let $z = x + iy$. Substituting this into the equation:$x + iy + \alpha|x + iy - 1| + 2i = 0$$x + \alpha\sqrt{(x-1)^2 + y^2} + i(y + 2) = 0$Equating real and imaginary parts to zero:$y + 2 = 0 \implies y = -2$$x + \alpha\sqrt{(x-1)^2 + (-2)^2} = 0 \implies \alpha = -\frac{x}{\sqrt{x^2 - 2x + 5}}$Squaring both sides:$\alpha^2 = \frac{x^2}{x^2 - 2x + 5}$Let $f(x) = \frac{x^2}{x^2 - 2x + 5}$. To find the range,we solve $f'(x) = 0$ or analyze the function.The range of $f(x)$ is $[0, \frac{5}{4}]$.Thus,$\alpha^2 \in [0, \frac{5}{4}]$,which implies $\alpha \in [-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}]$.Here,$p = -\frac{\sqrt{5}}{2}$ and $q = \frac{\sqrt{5}}{2}$.Then $4(p^2 + q^2) = 4(\frac{5}{4} + \frac{5}{4}) = 4(\frac{10}{4}) = 10$.

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

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