# What is i^97-i?

## (a) $- 2 i$ (b) $0$ (c) $1 - i$ (d) $- 1 - i$

Sep 11, 2016

${i}^{97} - i = 0$ that is answer is (b)

#### Explanation:

As $i = \sqrt{- 1}$, while ${i}^{1} = i$,

${i}^{2} = - 1$

and ${i}^{3} = {i}^{2} \times i = - 1 \times i = - i$

and ${i}^{4} = {\left({i}^{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1$

Similarly we can work out ${i}^{5} = i$, ${i}^{6} = - 1$, ${i}^{7} = - i$ and ${i}^{8} = 1$

This goes on in a cycle of $4$,

hence ${i}^{4 n + 1} = i$, ${i}^{4 n + 2} = - 1$, ${i}^{4 n + 3} = - i$ and ${i}^{4 n} = 1$

Hence as $97 = 4 \times 24 + 1$

${i}^{97} = {\left({i}^{4}\right)}^{24} \times i = {1}^{24} \times i = i$ and ${i}^{97} - i = i - i = 0$ that is answer is (b).