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Let $\mathrm{S}=\{z \in \mathrm{C}:|z-2| \leq 1, z(1+i)+\bar{z}(1-i) \leq 2\}$ . Let $|z-4 i|$ attains minimum and maximum values, respectively, at $z_{1} \in S$ and $z_{2} \in$ S. If $5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=\alpha+\beta \sqrt{5}$ , where $\alpha$ and $\beta$ are integers, then the value of $\alpha+\beta$ is equal to_____________.
Option: 1
26

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{|z-2| \leq 1 \Rightarrow}$ inside region of circle with centre $\mathrm{(2,0)}$ and radius $\mathrm{1}$

$\mathrm{z(1+i)+\bar{z}(1-i) \leq 2 } \\$

$\mathrm{ \text { Put } z=x+i y} \\$

$\mathrm{(x+i y)(1+i)+(x-i y)(1-i) \leq 2 }\\$

$\mathrm{\Rightarrow x+i x+i y-y+x-i x-i y-y \leq 2} \\$

$\mathrm{\Rightarrow x-y \leq 1}$

Line joining $(0,4)$ and $(2,0)$

$\mathrm{\frac{x}{2}+\frac{y}{4}=1 \Rightarrow 2 x+y=4}$

For $\mathrm{z_{1} \text { solve, }(x-2)^{2}+y^{2}=1 \text { and } 2 x+y=4}$

$\mathrm{(x-2)^{2}+(4-2 x)^{2}=1} \\$

$\mathrm{x^{2}-4 x+4+16+4 x^{2}-16 x=1} \\$

$\mathrm{5 x^{2}-20 x+19=0} \\$

$\mathrm{x=\frac{20 \pm \sqrt{20}}{10}=\frac{20 \pm 2 \sqrt{5}}{10}=2 \pm \frac{1}{\sqrt{5}}} \\$

$\mathrm{x=2-\frac{1}{\sqrt{5}}}$

$\mathrm{y =4-2 x} \\$

$\mathrm{=4-2\left(2-\frac{1}{\sqrt{5}}\right)} \\$

$\mathrm{=4-4+\frac{2}{\sqrt{5}}=\frac{2}{\sqrt{5}} \Rightarrow z_{1}=\left(2-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)}$

$\mathrm{\left|z_{1}\right|^{2} =\left(2-\frac{1}{\sqrt{5}}\right)^{2}+\frac{4}{5}=4+\frac{1}{5}-\frac{4}{\sqrt{5}}+\frac{4}{5}=5-\frac{4}{\sqrt{5}} }\\$

$\mathrm{z_{2} =1 \Rightarrow\left|z_{2}\right|^{2}=1 }\\$

$\mathrm{\therefore 5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=5\left(5-\frac{4}{\sqrt{5}}+1\right)} \\$

$\mathrm{=5\left(6-\frac{4}{\sqrt{5}}\right)} \\$

$\mathrm{=30-4 \sqrt{5}} \\$

$\mathrm{\Rightarrow \alpha=30, \beta=-4 }\\$

$\mathrm{\Rightarrow \alpha+\beta=26}$

Hence answer is $\mathrm{26}$

Posted by

manish painkra

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