Evaluate:
(i) $(\sqrt{-1})^{192}$
(ii) $(\sqrt{-1})^{93}$
(iii) $(\sqrt{-1})^{30}$.
Since $\mathrm{i}=\sqrt{-1}$ so
(i) L.H.S. $=(\sqrt{-1})^{192}$
$\Rightarrow \mathrm{i}^{192}$
$\Rightarrow \mathrm{i}^{4 \times 48}=1$
Since it is of the form ${ }^{i^{4 n}}=1$ so the solution would be 1
(ii) L.H.S. $=(\sqrt{-1})^{93}$
$\Rightarrow \mathrm{i}^{4 \times 23+1}$
$\Rightarrow \mathrm{i}^{4 \mathrm{n}+1}$
$\Rightarrow \mathrm{i}^{1}=\mathrm{i}$
Since it is of the form of $\mathrm{i}^{4 \mathrm{n}+1}=\mathrm{i}$ so the solution would be simply $\mathrm{i}$.
(iii) L.H.S $=(\sqrt{-1})^{30}$
$\Rightarrow \mathrm{i}^{4 \times 7+2}$
$\Rightarrow \mathrm{i}^{4 \mathrm{n}+2}$
$\Rightarrow \mathrm{i}^{2}=-1$
Since it is of the form ${ }^{i^{4 n+2}}$ so the solution would be $-1$