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)42
(B)37
(C)50✓ Correct
(D)61
Answer: (C)
Solution
The conditions describe circles of radius 5 centred at (6,0) and (−2,6). The distance between the centres is 82+62=10=5+5, so the circles touch externally at the single point dividing the join of the centres equally, namely z=26+(−2)+i20+6=2+3i. Then z2=−5+12i and z3=−46+9i, so z3+3z2−15z+141=(−46+9i)+(−15+36i)−(30+45i)+141=50.
JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option