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