The value of

Question:

The value of $(-\sqrt{-1})^{4 n-3}$, where $n \in N$, is

Solution:

$(-\sqrt{-1})^{4 n-3}$

$=(-i)^{4 n-3}$

$=(-i)^{4 n}(-i)^{-3}$

$=1(-i)^{-3}$

$=\left(\frac{1}{-i}\right)^{3}$

$=-\frac{1}{i^{3}}$

$=\frac{-1}{i^{2} \cdot i}$

$=\frac{-1}{-1(i)}$

$=+\frac{1}{i} \times \frac{i}{i}$

$=\frac{i}{i^{2}}$

Hence, $(-\sqrt{-1})^{4 n-3}=-i$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now