Question

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

Answer: 26

$|z-2|\le 1$

$(x-2{)}^2+y^2\le 1\dots \dots (1)$
$\&$
$z(1+i)+\bar{Z}(1-i)\le 2$
Put $z=x+iy$
$\therefore x-y\le 1\dots (2)$
$PA=\sqrt{17},PB=\sqrt{13}$
Maximum is $PA$ & Minimum is $PD$
Let $D(2+\cos \theta ,0+\sin \theta )$
$\therefore m_{cp}=\tan \theta =-2$
$\cos \theta =-\frac{1}{\sqrt{5}},\sin \theta =\frac{2}{\sqrt{5}}$
$\therefore D\left(2-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)$
$\Rightarrow z_1=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2i}{\sqrt{5}}$
$\left|z_1\right|=\frac{25-4\sqrt{5}}{5}\&z_2=1$
$\therefore {\left|z_2\right|}^2=1$
$\therefore 5\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right)=30-4\sqrt{5}$
$\therefore \alpha =30$
$\beta =-4$
$\therefore \alpha +\beta =26$