The area (in sq. units) of the region S = \z C : |z - 1| 2; (z + z) + i(z - z) 2, Im(z) 0\ is
- 7 4
- 3 2
- 7 3
- 17 8
Answer
(2) 3 2
Solution
Put z = x + iy:
|z - 1| 2 (x - 1)2 + y2 4 (disc: centre (1, 0), radius 2)
(z + z) + i(z - z) = 2x + i(2iy) = 2x - 2y 2 y x - 1
Im(z) 0 y 0
[figure]
∵ y = x - 1 passes through the centre at 45° ⇒ region = sector of angle 135 = 3 4
Area = 12r2 = 12 4 3 4 = 3 2
Answer: (2)