Question

If 5 < |z| 2 ≤ 12 and z² + z-2 - 2zz + 8z + 8z > 0, then

(A)  1 < Re(z) ≤ 2√3 and |Im(z)| < 2√2

(B)  - 2√3 ≤ Re(z) < -1 and |Im(z)| < 2√2

(C)  1 < Re(z) ≤ 2√3 and |Im(z)| < 2√3

(D)  - 2√3 ≤ Re(z)< -1 and |Im(z)| < 2√3

Answer 1

Correct option (A)  1 < Re(z) ≤ 2√3 and |Im(z)| < 2√2

Let z = x + iy 

 Therefore, according to given inequations, we have

5 < x² + y² ≤ 12

So, it represents the region bounded in between two concentric circles centred at origin of radii √5 and 2√3 units.

and  (z - z)² + 8(z + z) > 0

⇒ (2y/i)² + 8(2x) > 0

⇒ -4y² + 16x > 0 ⇒ y² < 4x

represents the region inside the parabola y² = 4x. The common region bounded is shown in Fig.

The point of intersections are

x² + y² = 5 and y² = 4x

⇒ x² + 4x - 5 = 0

⇒ x = 1, - 5 x² + y² = 12 and y² = 4x

⇒ x² + 4x - 12 = 0