Let z and ω be two complex numbers such that ω = zz̅ - 2z + 2,| (z + i)/(z - 3i) | = 1 and Re(ω) has minimum value. Then, the minimum value of n ∈ N for which ωⁿ is real, is equal to ______________.

Solution

Maths sir · 5 steps

Read the modulus condition as a locus first. It pins Im(z), and then Re(ω) becomes a simple quadratic.

1. Find the locus. |(z+i)/(z-3i)| = 1 means |z+i| = |z-3i|. So z is equally far from -i and 3i, and it lies on their perpendicular bisector, the line y = 1. Write z = x + i.
2. Build ω. zz̅ = |z|² = x² + 1 and 2z = 2x + 2i. So ω = (x² + 1) - (2x + 2i) + 2 = (x² - 2x + 3) - 2i.

   Re(ω) = x² - 2x + 3 = (x-1)² + 2

3. Minimise Re(ω). The minimum value is 2, at x = 1. Then ω = 2 - 2i = 2√2(cos π/4 - i sin π/4), so arg ω = -π/4.
4. Make ω^n real. arg(ωⁿ) = -nπ/4, and ωⁿ is real when this is a multiple of π. That needs n to be a multiple of 4. As a check, ω² = -8i is not real, but ω⁴ = (-8i)² = -64 is real.
5. So. The smallest natural number is n = 4.

The common slip: 8. Demanding that ωⁿ be a positive real number gives 8. But ω⁴ = -64 is already real, and a negative number counts.

The idea to keep. |z - a| = |z - b| is the perpendicular bisector of a and b. For powers, use the argument: ωⁿ is real exactly when n arg ω is a multiple of π.

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