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: circle, centre (6, 0), radius 5
|z + 2 − 6i| = 5: circle, centre (−2, 6), radius 5
Distance between centres = 82 + 62 = 10 = sum of radii
the circles touch externally at the mid-point of the centres
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)