Let P=z∈ mathbbC:|z+2-3i|≤ 1 and Q=z∈ mathbbC:z(1+i)+ z(1-i)≤ -8. Let in P∩ Q,|z-3+2i| be maximum and minimum at z_1 and z_2 respectively. If…

Given: $|z+2-3i|\leq 1$∣z+2−3i∣≤1

Putting, $z=x+iy$z=x+iy.

$$\Rightarrow {(x+2)}^{2}+{(y-3)}^{2}\leq 1...(i)$$⇒(x+2)2+(y−3)2≤1...(i)

For the circle represented in equation $(i)$(i), centre is $(-2,3)$(−2,3) and radius, $r=1$r=1.

It is given that, $z(1+i)+z(1-i)\leq -8$z(1+i)+z(1−i)≤−8

$$\Rightarrow (x+iy)(1+i)+(x-iy)(1-i)\leq -8$$⇒(x+iy)(1+i)+(x−iy)(1−i)≤−8

$$\Rightarrow x+ix-y+iy+x-ix-iy-y\leq -8$$⇒x+ix−y+iy+x−ix−iy−y≤−8

$$\Rightarrow 2x-2y\leq -8$$⇒2x−2y≤−8

$$\Rightarrow x-y+4=0$$⇒x−y+4=0

So, the line passing through $(3,-2)$(3,−2) and perpendicular to $x-y+4=0$x−y+4=0 is given by, $x+y-1=0$x+y−1=0.

So, the distance of line $x-y+4=0$x−y+4=0 from the centre $(-2,3)$(−2,3) is given by,

$$d=|\frac{-2-3+4}{\sqrt{2}}|$$d=∣2​−2−3+4​∣

$$\Rightarrow d=\frac{1}{\sqrt{2}}$$⇒d=2​1​

Now, $x+y-1=0$x+y−1=0 can be rewritten as,

$$\frac{x+2}{\frac{-1}{\sqrt{2}}}=\frac{y-3}{\frac{1}{\sqrt{2}}}=CA\text{or}CB$$2​−1​x+2​=2​1​y−3​=CAorCB

For $CA=\frac{-1}{\sqrt{2}}$CA=2​−1​ and $CB=1$CB=1, $A\equiv (\frac{-3}{2},\frac{5}{2})$A≡(2−3​,25​) and $B\equiv (\frac{-1}{\sqrt{2}}-2,\frac{1}{\sqrt{2}}+3)$B≡(2​−1​−2,2​1​+3).

$$\Rightarrow {|{z}_{1}|}^{2}={(2+\frac{1}{\sqrt{2}})}^{2}+{(3+\frac{1}{\sqrt{2}})}^{2}$$⇒∣z1​∣2=(2+2​1​)2+(3+2​1​)2

$$\Rightarrow {|{z}_{1}|}^{2}=4+\frac{1}{2}+2\sqrt{2}+9+\frac{1}{2}+3\sqrt{2}$$⇒∣z1​∣2=4+21​+22​+9+21​+32​

$$\Rightarrow {|{z}_{1}|}^{2}=14+5\sqrt{2}$$⇒∣z1​∣2=14+52​

$$\Rightarrow {|{z}_{2}|}^{2}=2(\frac{9}{4}+\frac{25}{4})$$⇒∣z2​∣2=2(49​+425​)

$$\Rightarrow {|{z}_{2}|}^{2}=17$$⇒∣z2​∣2=17

$$\Rightarrow {|{z}_{1}|}^{2}+2{|{z}_{2}|}^{2}=31+5\sqrt{2}$$⇒∣z1​∣2+2∣z2​∣2=31+52​

$$\Rightarrow \alpha =31,\beta =5$$⇒α=31,β=5

$$\Rightarrow \alpha +\beta =36$$⇒α+β=36