The least positive integer n such that (2i)ⁿ/(1 - i)ⁿ⁻²,i = √(- 1) is a positive integer, is ___________.

Solution

Maths sir · 5 steps

Borrow (1-i)² to make both powers n, then the whole thing collapses to -2i(-1+i)ⁿ.

1. Match the powers. Write 1/(1-i)ⁿ⁻² = (1-i)²/(1-i)ⁿ. Then the expression becomes (1-i)²(2i/(1-i))ⁿ, with (1-i)² = 1 - 2i + i² = -2i.
2. Simplify the ratio. 2i/(1-i) = 2i(1+i)/(1-i)(1+i) = 2i(1+i)/2 = i + i² = -1 + i.

   (2i)ⁿ/(1-i)ⁿ⁻² = -2i (-1+i)ⁿ

3. Argument condition. -2i has argument -π/2 and -1+i has argument 3π/4. For a positive real number the total argument 3nπ/4 - π/2 must be a multiple of 2π, so 3n - 2 = 8k. Trying n = 1, 2, 3, …, the first n that works is n = 6.
4. Check n = 6. (-1+i)² = -2i, so (-1+i)⁶ = (-2i)³ = -8i³ = 8i. Then -2i × 8i = -16i² = 16, a positive integer.
5. So. The least positive integer is n = 6.

The common slip: 8. Dropping the extra factor (1-i)² = -2i and asking only for (-1+i)ⁿ to be positive gives 8. That extra factor shifts the argument by -π/2.

The idea to keep. When a power of one complex number is divided by a power of another, make the exponents match first. Then one base raised to n is left, and its argument decides when the result is real and positive.

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