Let z be a complex number such that |z − 6| = 5 and |z + 2 − 6i| = 5. Then the value of z3 + 3z2 - 15z + 141 is equal to :
- 61
- 37
- 42
- 50
Answer
(4) 50
Solution
|z − 6| = 5 is a circle with centre (6, 0) and radius 5, and |z + 2 − 6i| = 5 is a circle with centre (−2, 6) and radius 5.
The distance between the centres is 82 + 62 = 10, which equals the sum of the radii, so the circles touch externally at the mid-point of the centres, i.e. z = 2 + 3i.
z2 = 4 - 9 + 12i = -5 + 12i
z3 = (2 + 3i)(-5 + 12i) = -46 + 9i
z3 + 3z2 - 15z + 141
= (-46 + 9i) + (-15 + 36i) - (30 + 45i) + 141
= 50 + 0i
value = 50
Answer: (4)