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  • Let z be those complex numbers which satisfy  <img alt="\left |z+5 \right | \leq 4 \text{ and } z (1+i)+\overline{z} (1-i)\geqslant -10, i= \sqrt{-1}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%7Cz&plus;5%20%5Cright%20%7C%20

Let z be those complex numbers which satisfy  \left |z+5 \right | \leq 4 \text{ and } z (1+i)+\overline{z} (1-i)\geqslant -10, i= \sqrt{-1} If the maximum value of \left |z+1 \right |^2 \text{ is } \alpha +\beta \sqrt{2}, then the value of (\alpha +\beta ) is ________
Option: 1 36
Option: 2 42
Option: 3 48
Option: 4 64

Answers (1)

best_answer

Let z = x + iy

\\|z+5| \leq 4 \\ (x+5)^{2}+y^{2} \leq 16 \\ z(1+i)+\bar{z}(1-i) \geq-10

\\(z+\bar{z})+i(z-\bar{z}) \geq-10 \\ (x+i y+x-i y)+i(x+i y-(x-i y)) \geq-10 \\ 2 x+i(x+i y-x+i y) \geq-10 \\ 2 x+2 y i^{2} \geq-10 \\ 2 x-2 y \geq-10 \\ x-y \geq-5 \\x-y+5 \geq 0

\\|z+1|^2=|z-(-1)|^2\\

Let P = (-1,0)

\\|z+1|^2_{\text{max}}=PB^2\\

For the point of intersection, solve 

\\(x+5)^{2}+y^{2}=16 \\ x-y+5=0

we get

\\\mathrm{A}(2 \sqrt{2}-5,2 \sqrt{2}) \quad \mathrm{B}(-2 \sqrt{2}-5,-2 \sqrt{2}) \\ \mathrm{PB}^{2}=(+2 \sqrt{2}+4)^{2}+(2 \sqrt{2})^{2} \\ |\mathrm{z}+1|^{2}=8+16+16 \sqrt{2}+8 \\ \alpha+\beta \sqrt{2}=32+16 \sqrt{2} \\ \alpha=32, \beta=16 \Rightarrow \alpha+\beta=48

 

 

Posted by

himanshu.meshram

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