# What is the value of (-sqrt(-1))^(4n+3)?

Oct 22, 2016

${\left(- \sqrt{- 1}\right)}^{4 n + 3} = i$ or $\sqrt{- 1}$

#### Explanation:

${\left(- \sqrt{- 1}\right)}^{4 n + 3}$

= ${\left(- i\right)}^{4 n + 3}$, where ${i}^{2} = - 1$

= ${\left(- 1 \times i\right)}^{4 n + 3}$

= ${\left(- 1\right)}^{4 n + 3} \times {i}^{4 n + 3}$

= $- 1 \times {i}^{4 n} \times {i}^{3}$, as odd power of $- 1$ is $- 1$

= $- 1 \times 1 \times {i}^{2} \times i$, as ${i}^{2} = - 1$ ${i}^{4} = {\left({i}^{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1$

= $- 1 \times 1 \times \left(- 1\right) \times i$

= $i$ or $\sqrt{- 1}$