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

Let z be a complex number such that |z − 6| = 5 and |z + 2 − 6i| = 5. Find the value of z³ + 3z² − 15z + 141

Q. 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

(A) 61

(B) 37

(D) 50

Correct Answer: 50

Explanation

The given conditions represent two circles in the Argand plane.

|z − 6| = 5 ⇒ Centre (6, 0), Radius = 5

|z + 2 − 6i| = 5 ⇒ Centre (−2, 6), Radius = 5

Distance between the centres:

√[(6 + 2)² + (0 − 6)²] = √(64 + 36) = √100 = 10

Since the distance between the centres is equal to the sum of radii (5 + 5), the circles touch each other externally at exactly one point.

That point lies on the line joining the centres and divides it internally in the ratio 1 : 1.

So, z = (6 − 2)/2 + (0 + 6)/2 i = 2 + 3i

Now substitute z = 2 + 3i in the given expression:

z³ + 3z² − 15z + 141

After simplification, the value comes out to be:

Hence, the required value is 50.