Let mathbbC be the set of all complex numbers. Let S_1=z∈ mathbbC:|z-2|≤ 1 and S_2=z∈ mathbbC:z(1+i)+ z(1-i)≥ 4. Then, the maximum value of |z- 5…

Here, $|z-2|\leq 1$∣z−2∣≤1 Put $z=x+iy$z=x+iy

$${(x-2)}^{2}+{y}^{2}\leq 1$$(x−2)2+y2≤1

Also, $z(1+i)+\bar{z}(1-i)\geq 4$z(1+i)+zˉ(1−i)≥4

Gives $x-y\geq 2$x−y≥2

Let point on circle be $A(2+\mathrm{cos}\theta ,\mathrm{sin}\theta )$A(2+cosθ,sinθ)

$$\theta \in [-\frac{3\pi }{4},\frac{\pi }{4}]$$θ∈[−43π​,4π​]

Let $P$P be $(\frac{5}{2},0)$(25​,0)

So,

$${|z-\frac{5}{2}|}^{2}={(\mathrm{AP})}^{2}={(2+\mathrm{cos}\theta -\frac{5}{2})}^{2}+{\mathrm{sin}}^{2}\theta$$∣z−25​∣2=(AP)2=(2+cosθ−25​)2+sin2θ

$$={\mathrm{cos}}^{2}\theta -\mathrm{cos}\theta +\frac{1}{4}+{\mathrm{sin}}^{2}\theta$$=cos2θ−cosθ+41​+sin2θ

$$=\frac{5}{4}-\mathrm{cos}\theta$$=45​−cosθ

For ${(\mathrm{AP})}^{2}$(AP)2 to be maximum $\theta =-\frac{3\pi }{4}$θ=−43π​

$${(\mathrm{AP})}^{2}=\frac{5}{4}+\frac{1}{\sqrt{2}}=\frac{5\sqrt{2}+4}{4\sqrt{2}}=\frac{5+2\sqrt{2}}{4}$$(AP)2=45​+2​1​=42​52​+4​=45+22​​