If α and β are the roots of the equation 2 z²-3 z-2 i=0, where i=√-1, then 16 · Re((α¹⁹+β¹⁹+α¹¹+β¹¹)/(α¹⁵+β¹⁵)) · lm((α¹⁹+β¹⁹+α¹¹+β¹¹)/(α¹⁵+β¹⁵)) is equal to

Solution

Maths sir · 4 steps

Since (αβ)⁴ = 1, the huge powers collapse and the whole fraction is just α⁴ + β⁴.

1. Sum and product. From 2z² - 3z - 2i = 0, we get α + β = 3/2 and αβ = -2i/2 = -i. So (αβ)⁴ = (-i)⁴ = 1, which means α⁻⁴ = β⁴ and β⁻⁴ = α⁴.
2. Factor the numerator. Write α¹⁹ + α¹¹ = α¹⁵(α⁴ + α⁻⁴) = α¹⁵(α⁴ + β⁴). In the same way, β¹⁹ + β¹¹ = β¹⁵(β⁴ + α⁴). Adding these, the numerator is (α⁴ + β⁴)(α¹⁵ + β¹⁵), and the fraction reduces to α⁴ + β⁴.

   (α¹⁹+β¹⁹+α¹¹+β¹¹)/(α¹⁵+β¹⁵) = α⁴ + β⁴

3. Build up the powers. α² + β² = (α+β)² - 2αβ = 9/4 + 2i. Then α⁴ + β⁴ = (9/4 + 2i)² - 2(αβ)² = 81/16 - 4 + 9i + 2 = 49/16 + 9i.
4. So. The real part is 49/16 and the imaginary part is 9. So 16 × 49/16 × 9 = 441, which is option (a).

The idea to keep. When large powers of the roots appear, first check whether some power of αβ equals 1. It lets you swap negative powers of one root for positive powers of the other, and the expression collapses.

Answer from the NTA final answer key. Solution by PrepWiser.