Let S={z ∈ ℂ-{i, 2 i}: (z²+8 i z-15)/(z²-3 i z-2) ∈ ℝ}. If α-13/11 i ∈ S, α ∈ ℝ-{0}, then 242 α² is equal to _________.

Solution

Maths sir · 5 steps

Split off a 1 from the fraction, and the given imaginary part makes the leftover numerator purely imaginary.

1. Split the fraction. The two quadratics differ by 11iz - 13, so (z²+8iz-15)/(z²-3iz-2) = 1 + (11iz-13)/(z-i)(z-2i). The whole expression is real exactly when this last fraction is real.
2. Use the given z. Put z = α - 13/11i. Then 11iz = 11iα + 13, so the numerator is 11iz - 13 = 11iα, which is purely imaginary because α ≠ 0.
3. Condition on denominator. An imaginary number divided by D is real only when D is also purely imaginary. So the real part of (z-i)(z-2i) must be zero.

   (z-i)(z-2i) = (α - 24/11i)(α - 35/11i), real part = α² - (24 × 35)/121

4. Solve for alpha. Setting the real part to zero gives α² = 840/121. The denominator is not zero, because its imaginary part -59/11α is non-zero.
5. So. 242α² = 2 × 121 × 840/121 = 2 × 840 = 1680.

The idea to keep. When a ratio of two similar quadratics must be real, subtract them first. The leftover numerator is usually simple, and the condition reduces to one real or imaginary part being zero.

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