JEE PYQ: Complex Numbers Question 9
Question 9 - 2021 (24 Feb Shift 1)
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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Answer: 10
Solution
Let $z = x + iy$. Then $y = -2$ and $x + \alpha\sqrt{(x-1)^2 + 4} = 0$. So $\alpha = \frac{-x}{\sqrt{(x-1)^2+4}}$. Range of $\alpha$: $\alpha^2((x-1)^2+4) = x^2$. Extrema give $p^2 + q^2 = \frac{5}{2}$, so $4(p^2+q^2) = 10$.
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