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:
Show Hint
- \(\frac{3}{4}\)
- \(\frac{1}{2}\)
- \(\frac{2}{3}\)
- \(\frac{1}{4} \)
The Correct Option is A
Approach Solution - 1
To solve this problem, we want to find the probability that \( i^{k_1} + i^{k_2} \neq 0 \) for natural numbers \( k_1 \) and \( k_2 \). This is equivalent to finding the probability that \( i^{k_1} \neq -i^{k_2} \).
1. Determine the possible values of \( i^k \):
The possible values of \( i^k \) for any natural number \( k \) are \( \{i, -1, -i, 1\} \). Specifically, \( i^k \) depends on \( k \pmod{4} \):
\( i^k = \begin{cases} 1 & \text{if } k \equiv 0 \pmod{4} \\ i & \text{if } k \equiv 1 \pmod{4} \\ -1 & \text{if } k \equiv 2 \pmod{4} \\ -i & \text{if } k \equiv 3 \pmod{4} \end{cases} \)
2. Find when \( i^{k_1} + i^{k_2} = 0 \):
We have \( i^{k_1} + i^{k_2} = 0 \) if and only if \( i^{k_1} = -i^{k_2} \). This means that \( i^{k_1} \) and \( i^{k_2} \) must be opposite values. The possible pairs of opposite values are \( (i, -i) \) and \( (1, -1) \).
3. Calculate the probability of \( i^{k_1} = -i^{k_2} \):
We want to find the probability that \( i^{k_1} \) and \( i^{k_2} \) are not opposites. Consider the possible values of \( k_1 \pmod{4} \). For each value of \( k_1 \pmod{4} \), we determine the value of \( k_2 \pmod{4} \) that makes \( i^{k_1} = -i^{k_2} \).
- If \( k_1 \equiv 0 \pmod{4} \), then \( i^{k_1} = 1 \). We need \( i^{k_2} = -1 \), which means \( k_2 \equiv 2 \pmod{4} \).
- If \( k_1 \equiv 1 \pmod{4} \), then \( i^{k_1} = i \). We need \( i^{k_2} = -i \), which means \( k_2 \equiv 3 \pmod{4} \).
- If \( k_1 \equiv 2 \pmod{4} \), then \( i^{k_1} = -1 \). We need \( i^{k_2} = 1 \), which means \( k_2 \equiv 0 \pmod{4} \).
- If \( k_1 \equiv 3 \pmod{4} \), then \( i^{k_1} = -i \). We need \( i^{k_2} = i \), which means \( k_2 \equiv 1 \pmod{4} \).
In each case, there is exactly one value of \( k_2 \pmod{4} \) that makes \( i^{k_1} = -i^{k_2} \). Since there are 4 possible values for \( k_2 \pmod{4} \), the probability that \( i^{k_1} = -i^{k_2} \) is \( \frac{1}{4} \).
4. Calculate the probability of \( i^{k_1} \neq -i^{k_2} \):
The probability that \( i^{k_1} \neq -i^{k_2} \) is \( 1 - \frac{1}{4} = \frac{3}{4} \).
Final Answer:
The probability that \( i^{k_1} + i^{k_2} \neq 0 \) is \( {\frac{3}{4}} \).
Approach Solution -2
We are given \( i^{k_1} + i^{k_2} \), where \( i = \sqrt{-1} \).
The powers of \( i \) repeat in a cycle of 4 as follows:
\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and then it repeats.} \]
Step 2: Identify when the sum is zero.
We must find when \( i^{k_1} + i^{k_2} = 0 \).
That will happen when \( i^{k_1} = -i^{k_2} \).
From the cyclic pattern of powers of \( i \), we see that:
\[ i^{k_1} = -i^{k_2} \quad \text{if } k_1 \text{ and } k_2 \text{ differ by 2 (mod 4)}. \] That means:
\[ k_1 - k_2 \equiv 2 \pmod{4}. \]
Step 3: Find the total possible combinations.
Each \( k \) can take any of the 4 possible values mod 4:
\[ i, -1, -i, 1. \] Hence, there are 4 × 4 = 16 possible combinations for \((i^{k_1}, i^{k_2})\).
Step 4: Count the combinations where the sum is zero.
We list the cases where the sum is zero:
- \( i + (-i) = 0 \)
- \( -i + i = 0 \)
- \( 1 + (-1) = 0 \)
- \( -1 + 1 = 0 \)
So, there are 4 such combinations out of 16.
Step 5: Probability of non-zero sum.
The probability that the sum is zero is \( \frac{4}{16} = \frac{1}{4} \).
Therefore, the probability that the sum is non-zero is:
\[ 1 - \frac{1}{4} = \frac{3}{4}. \]
Final Answer:
\[ \boxed{\frac{3}{4}} \]
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