Complex Number Question 2
Question 2 - 2024 (01 Feb Shift 1)
Let $P={z \in \mathbb{C}:|z+2-3 i| \leq 1}$ and $Q={z \in \mathbb{C}: z(1+i)+\bar{z}(1-i) \leq-8}$. Let in $P \cap Q,|z-3+2 i|$ 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 $z _1$ is the intersection of the circle and the line passing through $P\left(L _1\right)$ and $z _2$ is intersection of line $L _1 \& L _2$
Circle: $(x+2)^{2}+(y-3)^{2}=1$
$L _1: x+y-1=0$
$L _2: x-y+4=0$
On solving circle $\& L _1$ we get
$z _1:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right)$
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 $\beta=5$
$\alpha+\beta=36$
