Let z=a+i b, b ≠ 0 be complex numbers satisfying z²=z̄ · 2^(1-|z|). Then the least value of n ∈ N, such that zⁿ=(z+1)ⁿ, is equal to __________.

Solution

Maths sir · 4 steps

Take the modulus first to pin down |z| = 1, and then z turns out to be a cube root of unity.

1. Find the modulus. Take modulus on both sides: |z|² = |z| · 2^(1-|z|). Since b ≠ 0, z ≠ 0, so |z| = 2^(1-|z|). The left side increases with |z| and the right side decreases, so there is only one solution, and |z| = 1 works.
2. Identify z. With |z| = 1 the equation becomes z² = z̄. Multiply by z: z³ = zz̄ = |z|² = 1. As b ≠ 0, z is not real, so z = ω or z = ω².
3. Reduce the condition. Take z = ω. Then ((z+1)/z)ⁿ = 1, and (z+1)/z = 1 + ω² = -ω. The case z = ω² gives -ω² and the same order.

   (-ω)ⁿ = (-1)ⁿ ωⁿ = 1

4. So. You need n even, so that (-1)ⁿ = 1, and a multiple of 3, so that ωⁿ = 1. The least such n is 6.

The common slip: 3. Using only ω³ = 1 and dropping the minus sign in 1 + ω² = -ω gives 3. At n = 3 you get (-ω)³ = -1, not 1.

The idea to keep. When |z| sits inside an exponent, take the modulus of both sides first. To solve zⁿ = wⁿ, look at (w/z)ⁿ = 1 and find the order of w/z.

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