Subject: Mathematics — Complex Numbers
Question: Let
$S = \{ z \in C:|z = 2| \leqslant 1,z(1 + i) + \overline{z}(1 - i) \leq 2\}$.
Let $|z - 4i|$ 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
Answer: 26
Solution: Let $z = x + iy \Rightarrow |z - 2| \leq 1$
[IMAGE] ${(x - 2)}^{2} + y^{2} \leq 1$ ......(i)
and $z(1 + i) + \bar{z}(1 - i) \leq 2$\
$\therefore x - y \leq 1$ ......(ii)\
$$PA = \sqrt{17},PB = \sqrt{13}
$$∴ Maximum is PA and Minimum is PD\
Let $D(2 + cos\theta,0 + sin\theta)$\
∴ slope of
$CP = tan\theta = - 2 \Rightarrow cos\theta = - \frac{1}{\sqrt{5}},sin\theta = \frac{2}{\sqrt{5}}$\
$$\therefore D\left( 2 - \frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}} \right) \Rightarrow z_{1} = \left( 2 - \frac{1}{\sqrt{5}} \right) + \frac{2i}{\sqrt{5}}
$$$\left| z_{1} \right| = \frac{25 - 4\sqrt{5}}{5}$ and $z_{2} = 1$\
$${\therefore\left| z_{2} \right|^{2} = 1
}{\therefore 5\left( \left| z_{1} \right|^{2} + \left| z_{2} \right|^{2} \right) = 30 - 4\sqrt{5}\ \therefore\alpha = 30
}{\beta = - 4
}{\therefore\alpha + \beta = 26}$$