JEE Main 2022 — Mathematics Algebra

Given $|z-2|\leq 1$∣z−2∣≤1

${(x-2)}^{2}+{y}^{2}\leq 1\cdots (i)$(x−2)2+y2≤1⋯(i) represents interior region of a circle with centre $(2,0)$(2,0) and radius $1$1

Now $z(1+i)+\bar{z}(1-i)\leq 2$z(1+i)+zˉ(1−i)≤2

Puting $z=x+iy$z=x+iy, we get

$$(x+iy)(1+i)+(x-iy)(1-i)\leq 2$$(x+iy)(1+i)+(x−iy)(1−i)≤2

$$\Rightarrow x-y+i(x+y)+x-y-i(x+y)\leq 2$$⇒x−y+i(x+y)+x−y−i(x+y)≤2

$$\therefore x-y\leq 1\ldots (i)$$∴x−y≤1…(i)

Let $A(1,0)$A(1,0) and $B(2,1)$B(2,1) be the points on the line $x-y=1$x−y=1 and the circle ${(x-2)}^{2}+{y}^{2}=1$(x−2)2+y2=1

$|z-4i|$∣z−4i∣ represents the distance from $P(0,4)$P(0,4) to any point in the region $S$S

Here $PA=\sqrt{17},PB=\sqrt{13}$PA=17​,PB=13​

So $A(1,0)$A(1,0) is the point representing ${z}_{2}$z2​

Let $D(2+\mathrm{cos}\theta ,0+\mathrm{sin}\theta )$D(2+cosθ,0+sinθ) be the point representing ${z}_{1}$z1​

$$\therefore {m}_{CP}=\mathrm{tan}\theta =-2$$∴mCP​=tanθ=−2

$$\mathrm{cos}\theta =-\frac{1}{\sqrt{5}},\mathrm{sin}\theta =\frac{2}{\sqrt{5}}$$cosθ=−5​1​,sinθ=5​2​

$$\therefore D\equiv (2-\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$$∴D≡(2−5​1​,5​2​)

$$\Rightarrow {z}_{1}=(2-\frac{1}{\sqrt{5}})+\frac{2i}{\sqrt{5}}$$⇒z1​=(2−5​1​)+5​2i​

So |{z}_{1}|=\frac{25-4\sqrt{5}}{5}&{z}_{2}=1

$$\therefore {|{z}_{2}|}^{2}=1$$∴∣z2​∣2=1

$$\therefore 5({|{z}_{1}|}^{2}+{|{z}_{2}|}^{2})=30-4\sqrt{5}$$∴5(∣z1​∣2+∣z2​∣2)=30−45​

$$\therefore \alpha =30$$∴α=30

$$\beta =-4$$β=−4

Hence $\alpha +\beta =26$α+β=26