Let C be the set of all complex numbers.

Question

Let C be the set of all complex numbers. Let

S1 = {z$$\in$$C : |z - 2| $$\le$$ 1} and

S2 = {z$$\in$$C : z(1 + i) + $$\overline z (1 - i) \[\ge$$ 4}.

Then, the maximum value of $${\left| {z - {5 \over 2}} \right|^2}$$ for z$$\in$$S1 $$\cap$$ S2 is equal to :

Accepted Answer

|t - 2| $$\le$$ 1

Put t = x + iy

(x - 2)2 + y2 $$\le$$ 1

Also, t(1 + i) + $$\overline t(1 - i) \[\ge$$ 4

x - y $$\ge$$ 2

Let point on circle be A(2 + cos$$\theta$$, sin$$\theta$$)

$$\theta  \in \left[ { - {{3\pi } \over 4},{\pi  \over 4}} \right]$$

$${(AP)^2} = {\left( {2 + \cos \theta  - {5 \over 2}} \right)^2} + \sin \theta $$

$$ = {\cos ^2}\theta  - \cos \theta  + {1 \over 4} + {\sin ^2}\theta $$

$$ = {5 \over 4} - \cos \theta $$

For (AP)2 maximum $$\theta  =  - {{3\pi } \over 4}$$

$${(AP)^2} = {5 \over 4} + {1 \over {\sqrt 2 }} = {{5\sqrt 2  + 4} \over {4\sqrt 2 }}$$

Solutions · 1