Complex Numbers and Quadratic Equations · Question 5

JEE Advanced 2025 Paper 2NumericalEasy📈 Topic in 33% papers+4

For a non-zero complex number z, let (z) denote the principal argument of z, with - < (z) . Let be the cube root of unity for which 0< ( )< . Let = ( n=12025(- )n ). Then the value of 3 is .

Answer

-2

Solution

=ei2 /3 - =e-i /3=r, with r6=1 and r+r2+ +r6=0.

2025=6 337+3, so the sum =r+r2+r3.

r= 12- 32i, r2=- 12- 32i, r3=-1; sum =-1- 3 i.

=- 2 3 3 =-2

Answer: -2

JEE Advanced 2025 Paper 2

Keep practising — Complex Numbers and Quadratic Equations

Let f(x)=x4+ax3+bx2+c be a polynomial with real coefficients such that f(1)=-9. Suppose that i 3 is a root of the equation 4x3+3ax2+2bx=0, where i= -1. If 1, 2, 3, and 4 are all the roots of the equation f(x)=0, then | 1|2+| 2|2+| 3|2+| 4|2 is equal to .

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