Aspire Faculty ID #18090 · Topic: JEE Main 2026 (28 January Morning Shift) · Just now
JEE Main 2026 (28 January Morning Shift)

Let $ z $ be a complex number such that $ |z - 6| = 5 $ and $ |z + 2 - 6i| = 5 $. Then the value of $ z^3 + 3z^2 - 15z + 141 $ is equal to

Solution

Center of first circle $ C_1(6,0),; r_1 = 5 $


Center of second circle $ C_2(-2,6),; r_2 = 5 $


$ \therefore C_1C_2 = r_1 + r_2 $


$ \therefore $ common point $ Z $ is mid point of $ C_1 $ & $ C_2 $


[Image used in solution — two circles diagram]


$ z = 2 + 3i $


$ z^2 = 4z - 13 $


$ z^3 = 3z - 52 $


$ z^3 + 3z^2 - 15z + 141 = 50 $

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