Evaluate:

Solution:

(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 $\mathrm{i}$