Let $\mathrm{P}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}+2-3 \mathrm{i}| \leq 1\}$ and $\mathrm{Q}=\{\mathrm{z} \in \mathrm{C}: \mathrm{z}(1+\mathrm{i})+\overline{\mathrm{z}}(1-\mathrm{i}) \leq-8\}$. 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 $\left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}=\alpha+\beta \sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ equals
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Solution
Shaded region is $\mathrm{P} \cap \mathrm{Q}$. where z lies
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 ofline $\mathrm{L}_{1}$ \& $\mathrm{L}_{2}$\[ \begin{aligned} & \mathrm{P}:|\mathrm{z}+2-3 \mathrm{i}| \leq 1 \text { Circle : }(\mathrm{x}+2)^{2}+(\mathrm{y}-3)^{2} \leq 1, \mathrm{C}_{0}(-2,3) \\ & \mathrm{Q}: \mathrm{z}(1+\mathrm{i})+\bar{z}(1-\mathrm{i}) \leq-8 \\ & \Rightarrow(\mathrm{z}+\bar{z})+\mathrm{i}(\mathrm{z}-\bar{z}) \leq-8 \\ & \Rightarrow 2 \mathrm{x}+\mathrm{i}(2 \mathrm{yi}) \leq-8 \Rightarrow \mathrm{x}-\mathrm{y} \leq-4 \end{aligned} \]$|\mathrm{z}-(3-2 \mathrm{i})|=$ distance between z and point $\mathrm{D}(3,-2)$
For minimum \& maximum required distance lies along common normal of line L and circle.
So, common normal through D is $\mathrm{x}+\mathrm{y}-1=0$, which cuts circle at B and passing through centre of circle
line $\mathrm{AB} \perp$ line L
For position $\mathrm{A},|\mathrm{z}-3+2 \mathrm{i}|_{\text {min }}=\mathrm{AD}$ where A is point of intersection of line $\mathrm{L}: \mathrm{x}-\mathrm{y}+4=0$ and $\mathrm{x}+\mathrm{y}-1=0$\[ \mathrm{A}\left(-\frac{3}{2}, \frac{5}{2}\right),\left|\mathrm{z}_{2}\right|=\mathrm{OA}=\sqrt{\frac{9+25}{4}}=\frac{\sqrt{17}}{\sqrt{2}} \text { For } \mathrm{B},|\mathrm{z}-3+2 \mathrm{i}|_{\max }=\mathrm{BD}=\mathrm{C}_{0} \mathrm{D}+\mathrm{r}=1+5 \sqrt{2} \Rightarrow \frac{\mathrm{x}-3}{-1 / \sqrt{2}}=\frac{\mathrm{y}+2}{1 / \sqrt{2}}=1+5 \sqrt{2} \]\[ \begin{aligned} & B: z_{1}\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right) \Rightarrow\left|z_{1}\right|=\sqrt{\left(-2-\frac{1}{\sqrt{2}}\right)^{2}+\left(3+\frac{1}{\sqrt{2}}\right)^{2}}=\sqrt{4+9+\frac{4}{\sqrt{2}}+\frac{6}{\sqrt{2}}+1}=\sqrt{14+5 \sqrt{2}} \\ & \left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}=14+5 \sqrt{2}+2\left(\frac{17}{2}\right)=31+5 \sqrt{2}=\alpha+\beta \sqrt{2} \Rightarrow \alpha=31, \beta=5 \text { we get } \alpha+\beta=36 \end{aligned} \]
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