Question

Let $C$ be the set of all complex numbers. Let
$S_1=\{z\in C:|z-2|\le 1\}$ and
$S_2=\{z\in C:z(1+i)+\overline{z}(1-i)\ge 4\}$. Then the maximum value of ${|z-\frac{5}{2}|}^2$ for $z\in S_1\cap S_2$ is equal to

• A. $\frac{5+2\sqrt{2}}{2}$
• B. $\frac{5+2\sqrt{2}}{4}$
• C. $\frac{3+2\sqrt{2}}{4}$
• D. $\frac{3+2\sqrt{2}}{2}$

Answer 1

The correct option is B $\frac{5+2\sqrt{2}}{4}$

$S_1\equiv |z-2|\le 1\Rightarrow (x-2{)}^2+y^2\le 1$
$S_2\equiv x-y\ge 2$
$S_1\cap S_2$

Solving both the above equations, we get
$y^2=\frac{1}{2}\Rightarrow y=\pm \frac{1}{\sqrt{2}},x=2\pm \frac{1}{\sqrt{2}}$
Points of intersection are $(2\pm \frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{2}})$

Maximum value of $|z-\frac{5}{2}|$ is the distance between $(2-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ and $(\frac{5}{2},0)$

$max\left({|z-\frac{5}{2}|}^2\right)={(2-\frac{1}{\sqrt{2}}-\frac{5}{2})}^2+{(-\frac{1}{\sqrt{2}})}^2={(\frac{1}{2}+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}$
$=\frac{3+2\sqrt{2}}{4}+\frac{2}{4}=\frac{5+2\sqrt{2}}{4}$