Complex Number Question 2
Question 2 - 2024 (01 Feb Shift 1)
Let $\mathrm{P}={\mathrm{z} \in \mathbb{C}:|\mathrm{z}+2-3 \mathrm{i}| \leq 1}$ and $Q={z \in \mathbb{C}: z(1+i)+\bar{z}(1-i) \leq-8}$. Let in $\mathrm{P} \cap \mathrm{Q},|\mathrm{z}-3+2 \mathrm{i}|$ be maximum and minimum at $z_{1}$ and $z_{2}$ respectively. If $\left|z_{1}\right|^{2}+2|z|^{2}=\alpha+\beta \sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ equals
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Answer (36)
Solution
Clearly for the shaded region $\mathrm{z}{1}$ is the intersection of the circle and the line passing through $\mathrm{P}\left(\mathrm{L}{1}\right)$ and $\mathrm{z}{2}$ is intersection of line $\mathrm{L}{1} & \mathrm{~L}_{2}$
Circle: $(x+2)^{2}+(y-3)^{2}=1$
$\mathrm{L}_{1}: \mathrm{x}+\mathrm{y}-1=0$
$\mathrm{L}_{2}: \mathrm{x}-\mathrm{y}+4=0$
On solving circle $& \mathrm{~L}_{1}$ we get
$\mathrm{z}_{1}:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right)$
On solving $L_{1}$ and $z_{2}$ is intersection of line $L_{1} & L_{2}$
we get $z_{2}:\left(\frac{-3}{2}, \frac{5}{2}\right)$
$$ \begin{aligned} & \quad\left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}=14+5 \sqrt{2}+17 \ & \quad=31+5 \sqrt{2} \ & \alpha=31 \end{aligned} $$
So $\beta=5$
$\alpha+\beta=36$
