Question

The value of ${\left(-\sqrt{-1}\right)}^{4n-3},$ where n ∈ N, is ____________.

Answer 1

$$\begin{gathered} {\left(-\sqrt{-1}\right)}^{4n-3} \\ ={\left(-i\right)}^{4n-3} \\ ={\left(-i\right)}^{4n}{\left(-i\right)}^{-3} \\ =1{\left(-i\right)}^{-3} \\ ={\left(\frac{1}{-i}\right)}^3 \\ =-\frac{1}{i^3} \\ =\frac{-1}{i^2.i} \\ =\frac{-1}{-1\left(i\right)} \\ =+\frac{1}{i}\times \frac{i}{i} \\ =\frac{i}{i^2} \\ \mathrm{Hence},{\left(-\sqrt{-1}\right)}^{4n-3}=-i \end{gathered}$$