Let C be the set of all complex numbers. Let S₁ = {z∈ C:|z-2|≤1} and S₂ = {z∈ C: z(1+i)+z̄(1-i) ≥ 4}. Then, the maximum value of |z - 5/2|² for z ∈ S₁ ∩ S₂ is equal to:
Let C be the set of all complex numbers. Let S₁ = {z∈ C:|z-2|≤1} and S₂ = {z∈ C: z(1+i)+z̄(1-i) ≥ 4}. Then, the maximum value of |z - 5/2|² for z ∈ S₁ ∩ S₂ is equal to:
Option 1 -
(5+2√2)/4
Option 2 -
(5+2√2)/2
Option 3 -
(3+2√2)/4
Option 4 -
(3+2√2)/2
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1 Answer
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Correct Option - 1
Detailed Solution:S? : |z−2|≤1 is circle with centre (2,0) and radius less than equal to 1.
S? : z (1+i)+z? (1−i)≥4
Put z=x+iy
y≤x−2
Solving with S1
⇒x=2−1/√2, y=-1/√2
Point of intersection P= (2−1/√2, −i/√2)
|z−5/2|² = | (2−1/√2)−i (1/√2)−5/2|² = (5√2+4)/4√2 = (5+2√2)/4
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