Question

The principal argument of $i^{-1097}$ is ______

Answer 1

Let $z=i^{-1097}$

$\Rightarrow z=\frac{1}{i^{1097}}$

$\Rightarrow z=\frac{1}{i^{(4\times 274+1)}}$

$\Rightarrow z=\frac{1}{i^1}$

$\because i^{(4\times n+k)}=i^k$

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

$\Rightarrow z=\frac{i}{-1}$

$\Rightarrow z=-i$

Clearly, $z$ lies on the negative $y-$axis

$\therefore arg\left(z\right)=-\frac{\pi }{2}$

The principal argument of $i^{-1097}$ is$-\frac{\pi }{2}$