Question

Let $C$ be the set of all complex numbers.
Let $S_{1}=\{z \in C:|z-2| \leq 1\}$ and
$S_{2}=\{z \in C: z(1+i)+\bar{z}(1-i) \geq 4\}$
Then, the maximum value of $\left|z-\frac{5}{2}\right|^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to :

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

Answer 1

Answer: (D)

$|t-2| \leq 1$ Put $t=x+i y$

$(x-2)^{2}+y^{2} \leq 1$
Also, $t(1+i)+\bar{t}(1-i) \geq 4$
Gives $x-y \geq 2$
Let point on circle be $A(2+\cos \theta, \sin \theta)$
$\theta \in\left[-\frac{3 \pi}{4}, \frac{\pi}{4}\right] $
$(A P)^{2}=\left(2+\cos \theta-\frac{5}{2}\right)^{2}+\sin ^{2} \theta$
$=\cos ^{2} \theta-\cos \theta+\frac{1}{4}+\sin ^{2} \theta$
$=\frac{5}{4}-\cos \theta$
For $(A P)^{2}$ maximum $\theta=-\frac{3 \pi}{4}$
$(A P)^{2}=\frac{5}{4}+\frac{1}{\sqrt{2}}=\frac{5 \sqrt{2}+4}{4 \sqrt{2}}$