# Evaluate:

Question:

Evaluate:

(i) $i^{19}$

(ii) $i^{62}$

(ii) $\mathrm{i}^{373} .$

Solution:

We all know that $i=\sqrt{(}-1)$.

And ${ }^{i^{4 n}}=1$

$\mathrm{i}^{4 \mathrm{n}+1}=\mathrm{i}$ (where $\mathrm{n}$ is any positive integer)

$\mathrm{i}^{4 \mathrm{n}+2}=-1$

$\mathrm{i}^{4 \mathrm{n}+3}=-1$

So,

(i) L.H.S $=\mathrm{i}^{19}$

$=i^{4 \times 4+3}$

$=i^{4 n+3}$

Since it is of the form $\mathrm{i}^{4 \mathrm{n}+3}$ so the solution would be simply $-\mathrm{i}$

Hence the value of $\mathrm{i}^{19}$ is $-\mathrm{i}$.

(ii) $\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{S}=\mathrm{i}^{62}$

$\Rightarrow \mathrm{i}^{4 \times 15+2}$

$\Rightarrow \mathrm{i}^{4 \mathrm{n}+2} \Rightarrow \mathrm{i}^{2}=-1$

so it is of the form $i^{4 n+2}$

so its solution would be $-1$

(iii) L.H.S. $=\mathrm{i}^{373}$

$\Rightarrow \mathrm{i}^{4 \times 93+1}$

$\Rightarrow \mathrm{i}^{4 \mathrm{n}+1}$

$\Rightarrow \mathrm{i}$

So, it is of the form of $\mathrm{i}^{4 \mathrm{n}+1}$ so the solution would be $\mathrm{i}$.