Q90 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
The value of $(-\sqrt{-1})^{4 n-3}$, where $n \in N$, set of natural numbers is
- A1
- B-1
- Correct answerC-i
- Di
Solution
Given the expression $(-\sqrt{-1})^{4n-3}$, where $n \in N$.
Since $\sqrt{-1} = i$, the expression becomes $(-i)^{4n-3}$.
This can be written as $(-1)^{4n-3} \times i^{4n-3}$.
Since $4n-3$ is an odd number for any $n \in N$, $(-1)^{4n-3} = -1$.
Now, $i^{4n-3} = i^{4n} \times i^{-3} = (i^4)^n \times \dfrac{1}{i^3}$.
Since $i^4 = 1$ and $i^3 = -i$, we have $i^{4n-3} = 1^n \times \dfrac{1}{-i} = \dfrac{1}{-i} = \dfrac{i}{-i^2} = \dfrac{i}{1} = i$.
Therefore, $(-i)^{4n-3} = -1 \times i = -i$.
Answer: -i
Since $\sqrt{-1} = i$, the expression becomes $(-i)^{4n-3}$.
This can be written as $(-1)^{4n-3} \times i^{4n-3}$.
Since $4n-3$ is an odd number for any $n \in N$, $(-1)^{4n-3} = -1$.
Now, $i^{4n-3} = i^{4n} \times i^{-3} = (i^4)^n \times \dfrac{1}{i^3}$.
Since $i^4 = 1$ and $i^3 = -i$, we have $i^{4n-3} = 1^n \times \dfrac{1}{-i} = \dfrac{1}{-i} = \dfrac{i}{-i^2} = \dfrac{i}{1} = i$.
Therefore, $(-i)^{4n-3} = -1 \times i = -i$.
Answer: -i
Hence, the correct answer is option: 'C'.