Exams › JEE Main › Maths

Let alpha be a primitive (non-real) fifth root of unity and let beta be a primitive (non-real) fourth root of unity. Evaluate (1 + alpha)(1 + beta)(1 + alpha²)(1 + beta²)(1 + beta³)(1 + alpha⁴).

1. 0
2. (1 + alpha + alpha²)(1 - beta²)
3. (1 + alpha)(1 + beta + beta²)
4. 1

Correct answer: 0

Solution

Since beta is a primitive 4th root of unity, beta² = -1, so the factor (1 + beta²) = 0 makes the entire product 0.

Related JEE Main Maths questions

• Given z1 = √3 + i√3 and z2 = √3 + i, determine the quadrant in which the complex number z1/z2 lies.
• For the quadratic equation 2(1+i)x² − 4(2−i)x − (5+3i) = 0, the root with the larger absolute value is
• Find the value of ((cosθ+isinθ)⁴)/((cosθ-isinθ)³).
• How many complex numbers z satisfy the equation z³ + (3|z|²)/(z) = 0, given that |z| = √(3)?
• Let z = x + iy be a variable complex number. If arg((z - 1)/(z + 1)) = π/4, then which relation is satisfied?
• For positive values of a, b and c, consider the quadratic equation ax² + bx + c = 0. What can be said about its two roots?

⚔️ Practice JEE Main Maths free + battle 1v1 →