Evaluate:

Question:

Evaluate:

(i) $(\sqrt{-1})^{192}$

(ii) $(\sqrt{-1})^{93}$

(iii) $(\sqrt{-1})^{30}$.

 

Solution:

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 $-1$

 

Leave a comment