JEE Main Maths Test- 1 Free Online Test 2026

Given the equation α² + α + 1 = 0, we recognise that α is a non-real cube root of unity since α³ = 1. To find α³¹:

• Express α³ in terms of lower powers:
  • Multiply both sides by α: α³ + α² + α = 0
  • Substitute α² from the original equation: α³ - α - 1 + α = 0 ⇒ α³ = 1
• Simplify α³¹ using modular arithmetic:
  • Since α³ = 1, any exponent of α can be reduced modulo 3.
  • Compute 31 mod 3: 31 ÷ 3 = 10 remainder 1
  • Therefore, α³¹ = α³₁₀ + 1 = (α³)¹⁰ * α = 1¹⁰ * α = α

Thus, the answer is A. α

The polar form of a complex number is given by r(cos^θ + i sin^θ), where:

• r is the modulus
• θ is the argument

Among the options, only option C correctly represents this standard form.

Step 1: Write each complex number in polar form r[cosθ + i sinθ].

• For z₁ = −1 + i√3:
  r = √(1 + 3) = 2; angle θ = 120° (Quadrant II, tan60° = √3).
  ⇒ z₁ = 2[cos120° + i sin120°].

• For z₂ = −1 − i√3:
  r = 2; angle θ = 240° (Quadrant III) or −120°.
  ⇒ z₂ = 2[cos240° + i sin240°] = 2[cos(−120°) + i sin(−120°)].

• For 1 − i:
  r = √2; angle = −45°.
  ⇒ 1 − i = √2[cos(−45°) + i sin(−45°)].

• For 1 + i:
  r = √2; angle = 45°.
  ⇒ 1 + i = √2[cos45° + i sin45°].

Step 2: Raise to the required powers (De Moivre).

• (−1 ± i√3)¹⁵:
  = 2¹⁵ [cos(15·±120°) + i sin(15·±120°)]
  = 2¹⁵ [cos(±1800°) + i sin(±1800°)]
  Since 1800° is a multiple of 360°, cos = 1, sin = 0.
  ⇒ (−1 ± i√3)¹⁵ = 2¹⁵.

• (1 ∓ i)²⁰ (note the signs in denominators):
  (1 − i)²⁰ = (√2)²⁰ [cos(20·(−45°)) + i sin(20·(−45°))]
  = 2¹⁰ [cos(−900°) + i sin(−900°)]
  = 2¹⁰ [cos180° + i sin180°] (since −900° ≡ 180° mod 360)
  = 2¹⁰ (−1) = −2¹⁰.

  (1 + i)²⁰ = (√2)²⁰ [cos(20·45°) + i sin(20·45°)]
  = 2¹⁰ [cos900° + i sin900°]
  = 2¹⁰ [cos180° + i sin180°]
  = −2¹⁰.

Step 3: Evaluate each fraction and add.

• First fraction: 2¹⁵ / (−2¹⁰) = −2⁵ = −32.

• Second fraction: 2¹⁵ / (−2¹⁰) = −2⁵ = −32.

• Sum S = −32 + (−32) = −64.

The given expression z = a + b_i represents a complex number, where:

• a and b are real numbers
• i is the imaginary unit satisfying i² = -1

Complex numbers consist of:

• A real part (a)
• An imaginary part (b_i)

Therefore, the correct classification of z among the provided options is a complex number.

When you replace _i with -_i in the complex number z = a + b_i, you obtain a - b_i. This operation:

• Changes the sign of the imaginary part
• Keeps the real part unchanged

The resulting complex number is known as the complex conjugate of z, denoted as z*. The other options are incorrect because:

• A. The additive inverse would change both the real and imaginary parts to their negatives: -a - b_i.
• B. Prime factor is unrelated to complex numbers.
• D. The multiplicative inverse of z is 1/z, which involves both real and imaginary parts.

Number of ways of placed six X's in 8 squares

Here two squares of first and third row are empty

. ^.. Required arrangements = 28 - 2 = 26

¹⁴C₁ + ¹⁴C₃ + ¹⁴C₅ +.......¹⁴C₁₁
For odd numbers : 2^(n-1)
2¹³ -14