Two numbers $\mathbf{k}_1$ and $\mathbf{k}_2$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_1} + i^{k_2}$, $(i = \sqrt{-1})$ is non-zero, equals

Step 1: Understand powers of i

The powers of $i$ follow a cycle of four values: $i$, $-1$, $-i$, and $1$. This cycle repeats for higher powers. We can determine the value of $i^k$ by finding the remainder when $k$ is divided by 4.

Step 2: Analyze the sum $i^{k_1} + i^{k_2}$

The sum $i^{k_1} + i^{k_2}$ will be zero if and only if $i^{k_1}$ and $i^{k_2}$ are additive inverses of each other. This means they must be equal in magnitude but opposite in sign.

Step 3: Identify conditions for the sum to be zero

For the sum to be zero, the values of $i^{k_1}$ and $i^{k_2}$ must be opposite. There are four such pairs: $(i, -i)$, $(-i, i)$, $(1, -1)$, and $(-1, 1)$. Each pair corresponds to specific remainders when $k_1$ and $k_2$ are divided by 4.

Step 4: Calculate the probability of the sum being zero

Since $k_1$ and $k_2$ are randomly chosen natural numbers, the probability that $k$ leaves any specific remainder when divided by 4 (0, 1, 2, or 3) is $1/4$. We sum the probabilities of the four cases where the sum is zero.

Step 5: Calculate the probability of the sum being non-zero

The probability that the sum is non-zero is the complement of the probability that the sum is zero. We subtract the probability of the sum being zero from 1.

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