Two number k₁ and k₂ are randomly chosen from the set of natural numbers. Then, the probability that the value of i^(k₁)+i^(k₂),(i=√-1) is non-zero, equals

Solution

Maths sir · 4 steps

Powers of i repeat every 4, so only the remainders of k₁ and k₂ on division by 4 matter.

1. Reduce to remainders. Since i⁴ = 1, the value of iᵏ depends only on the remainder of k when divided by 4. Each remainder 1, 2, 3, 0 is equally likely, so there are 4 × 4 = 16 equally likely pairs of remainders.
2. When the sum vanishes. i^(k₁) + i^(k₂) = 0 means i^(k₂) = -i^(k₁). Since -1 = i², this happens exactly when k₂ is 2 more than k₁ in the remainder cycle.

   k₁ mod 4	i^(k₁)	Needed i^(k₂)	k₂ mod 4 that gives 0
   1	i	-i	3
   2	-1	1	0
   3	-i	i	1
   0	1	-1	2

3. Count. For each of the 4 choices of k₁, exactly one of the 4 remainders of k₂ makes the sum zero. That gives 4 bad pairs out of 16, so P(sum = 0) = 4/16 = 1/4.
4. So. P(non-zero) = 1 - 1/4 = 3/4, which is option (a).

The idea to keep. Whenever powers of i appear with random exponents, work with remainders mod 4. Then count the bad case and subtract from 1.

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