Question

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

• A. \(\frac{3}{4}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{1}{4} \)

Hint:

Understanding the cyclic properties of complex numbers can significantly simplify probability calculations involving their powers.

Answer 1

The Correct Option is A

To address this problem, we aim to ascertain the probability that \( i^{k_1} + i^{k_2} eq 0 \) for natural numbers \( k_1 \) and \( k_2 \). This is equivalent to determining the probability that \( i^{k_1} eq -i^{k_2} \).

1. Identify the possible values of \( i^k \):
For any natural number \( k \), the set of possible values for \( i^k \) is \( \{i, -1, -i, 1\} \). The specific value of \( i^k \) is contingent upon \( k \pmod{4} \):
\( i^k = \begin{cases} 1 & \text{if } k \equiv 0 \pmod{4} \\ i & \text{if } k \equiv 1 \pmod{4} \\ -1 & \text{if } k \equiv 2 \pmod{4} \\ -i & \text{if } k \equiv 3 \pmod{4} \end{cases} \)

2. Determine when \( i^{k_1} + i^{k_2} = 0 \):
The condition \( i^{k_1} + i^{k_2} = 0 \) holds if and only if \( i^{k_1} = -i^{k_2} \). This necessitates that \( i^{k_1} \) and \( i^{k_2} \) are additive inverses. The pairs of additive inverses are \( (i, -i) \) and \( (1, -1) \).

3. Compute the probability of \( i^{k_1} = -i^{k_2} \):
We need to calculate the probability that \( i^{k_1} \) and \( i^{k_2} \) are not additive inverses. Considering the possible values of \( k_1 \pmod{4} \), we can identify, for each value, the required value of \( k_2 \pmod{4} \) that satisfies \( i^{k_1} = -i^{k_2} \).

• If \( k_1 \equiv 0 \pmod{4} \), then \( i^{k_1} = 1 \). The requirement is \( i^{k_2} = -1 \), which corresponds to \( k_2 \equiv 2 \pmod{4} \).
• If \( k_1 \equiv 1 \pmod{4} \), then \( i^{k_1} = i \). The requirement is \( i^{k_2} = -i \), which corresponds to \( k_2 \equiv 3 \pmod{4} \).
• If \( k_1 \equiv 2 \pmod{4} \), then \( i^{k_1} = -1 \). The requirement is \( i^{k_2} = 1 \), which corresponds to \( k_2 \equiv 0 \pmod{4} \).
• If \( k_1 \equiv 3 \pmod{4} \), then \( i^{k_1} = -i \). The requirement is \( i^{k_2} = i \), which corresponds to \( k_2 \equiv 1 \pmod{4} \).

For each case of \( k_1 \pmod{4} \), there is exactly one value of \( k_2 \pmod{4} \) that fulfills \( i^{k_1} = -i^{k_2} \). Given that there are 4 possible values for \( k_2 \pmod{4} \), the probability of \( i^{k_1} = -i^{k_2} \) is \( \frac{1}{4} \).

4. Compute the probability of \( i^{k_1} eq -i^{k_2} \):
The probability that \( i^{k_1} eq -i^{k_2} \) is calculated as \( 1 - \frac{1}{4} = \frac{3}{4} \).

Final Answer:
The probability that \( i^{k_1} + i^{k_2} eq 0 \) is \( {\frac{3}{4}} \).