NDA Mathematics Question (2025) — Solution

Step-by-step solution

First, we simplify the expressions inside the parentheses by rationalizing the denominators: 1-i 1+i = (1-i)(1-i) (1+i)(1-i) = 1 - 2i + i^2 1 - i^2 = 1 - 2i - 1 1 - (-1) = -2i 2 = -i 1+i 1-i = (1+i)(1+i) (1-i)(1+i) = 1 + 2i + i^2 1 - i^2 = 1 + 2i - 1 1 - (-1) = 2i 2 = i Substitute these simplified forms back into the given equation: (-i)^ 2m (i)^ 2n = 1 Using the property of exponents, we can rewrite this as: ((-i)^2)^m (i^2)^n = 1 Since i^2 = -1 and (-i)^2 = i^2 = -1, we get: (-1)^m (-1)^n = 1 (-1)^ m+n = 1 For (-1)^ m+n to be equal to 1, the exponent (m+n) must be an even integer. We know that for any two integers m and n, their sum (m+n) and their difference (m-n) always share the same parity (both are even or both are odd). Since m+n is even, m-n must also be an even integer. The question asks for the smallest positive value of (m-n). The smallest positive even integer is 2. Answer: 2