Evaluate:
(i) $i^{-50}$
(ii) $\mathrm{i}^{-9}$
(ii) $\mathrm{i}^{-131}$.
(i) L.H.S. $=\mathrm{i}^{-50}$
$\Rightarrow \mathrm{i}^{-4 \times 13+2}$
$\Rightarrow \mathrm{i}^{4 \mathrm{n}+2}$
$\Rightarrow-1$
Since it is of the form ${ }^{i^{4 n+2}}$ so the solution would be $-1$
(ii) L. H.S. $=\mathrm{i}^{-9}$
$\Rightarrow \mathrm{i}^{-4 \times 3+3}$
$\Rightarrow \mathrm{i}^{4 \mathrm{n}+3}$
$\Rightarrow \mathrm{i}^{3}=-\mathrm{i}$
Since it is of the form of $\mathrm{i}^{4 \mathrm{n}+3}$ so the solution would be simply -i.
(iii) L.H.S. $=\mathrm{i}^{-131}$
$\Rightarrow \mathrm{i}^{-4 \times 33+1}$
$\Rightarrow \mathrm{i}^{4 \mathrm{n}+1}$
$\Rightarrow \mathrm{i}^{1}=\mathrm{i}$
Since it is of the form $\mathrm{i}^{4 \mathrm{n}+1}$. so the solution would be $\mathrm{i}$
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