Exams › JEE Advanced › Maths

Let z = 1 + ai be a complex number with a > 0, such that z³ is a real number. Then the sum 1 + z + z² + z³ +... + z¹¹ equals:

1. -1365*sqrt(3)*i
2. 1365*sqrt(3)*i
3. -1365*(1 + sqrt(3)*i)
4. 1365*(1 - sqrt(3)*i)

Correct answer: -1365*sqrt(3)*i

Solution

z = 1+ai, z³ is real. z³ = (1+ai)³ = (1-3a²) + i(3a-a³). For imaginary part = 0: a(3-a²)=0, a=sqrt(3) (since a>0). So z = 1+i*sqrt(3), |z|=2, arg(z)=pi/3. z³ = 8*(cos(pi)+i*sin(pi)) = -8. Sum S = (z¹²-1)/(z-1). z¹² = (z³)⁴ = (-8)⁴ = 4096. S = 4095/(i*sqrt(3)) = 4095*(-i)/(sqrt(3)) = (4095/sqrt(3))*(-i) = 1365*sqrt(3)*(-i) = -1365*sqrt(3)*i.

Related JEE Advanced Maths questions

• What is the cubic equation formed when r₁, r₂, and r₃ are the roots?
• Let complex numbers α and 1/α lie on circles (x - x0)² + (y - y0)² = r² and (x - x0)² + (y - y0)² = 4r² respectively. If z0 = x0 + iy0 satisfies the equation 2|z0|² = r² + 2, then |α| =
• If a quadratic equation with real coefficients has roots that are purely imaginary, what can be said about the nature of the roots of the equation formed by substituting the quadratic into itself?
• Consider the set S comprising all non-zero real values of α such that the quadratic equation αx² - x + α = 0 possesses two distinct real roots x₁ and x₂, which satisfy the condition |x₁ - x₂| < 1. Which of the following intervals lies entirely within S?
• Given -π/6 < θ < -π/12, let α₁ and β₁ be the solutions of the quadratic equation x² - 2x secθ + 1 = 0, and α₂ and β₂ be the solutions of the quadratic equation x² + 2x tanθ - 1 = 0. If α₁ > β₁ and α₂ > β₂, what is the value of α₁ + β₂?
• Let a and b represent the distinct real roots of the quadratic equation x² + 20x − 2020, and c and d represent the distinct complex roots of the quadratic equation x² − 20x + 2020. What is the value of ac(a − c) + ad(a − d) + bc(b − c) + bd(b − d)?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →