Question

Let $\quad S=\{z \in C:|z-2| \leq 1, z(1+i)+\bar{z}(1-$ 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 ________.

Answer 1

$| z -2| \leq 1$

$(x-2)^{2}+y^{2} \leq 1 \ldots \ldots(1)$
$\&$
$z(1+i)+\bar{Z}(1-i) \leq 2$
Put $z=x+i y$
$\therefore x - y \leq 1 \ldots(2)$
$ PA =\sqrt{17}, PB =\sqrt{13}$
Maximum is $PA $ & Minimum is $PD$
Let $D (2+\cos \theta, 0+\sin \theta)$
$\therefore m _{ cp }=\tan \theta=-2$
$\cos \theta=-\frac{1}{\sqrt{5}}, \sin \theta=\frac{2}{\sqrt{5}}$
$\therefore D \left(2-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)$
$\Rightarrow z _{1}=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2 i }{\sqrt{5}} $
$\left| z _{1}\right|=\frac{25-4 \sqrt{5}}{5} \& z _{2}=1$
$\therefore\left| z _{2}\right|^{2}=1$
$\therefore 5\left(\left| z _{1}\right|^{2}+\left| z _{2}\right|^{2}\right)=30-4 \sqrt{5} $
$\therefore \alpha=30$
$\beta=-4 $
$\therefore \alpha+\beta=26$