Answer with Step-by-step explanation:
Part 1)
we know that
[tex]e^{i\theta }=cos(\theta )+isin(\theta )[/tex]
thus [tex]-i=e^{\frac{-i\times (4n-1)\pi }{2}}[/tex]
thus [tex](-i)^i=(e^{\frac{-i\times (4n-1)\pi }{2}})^i\\\\(-i)^i=e^{\frac{-i^2\times (4n-1)\pi }{2}}=e^{\frac{(4n-1)\pi }{2}}\\\\\therefore (-i)^i=e^{\frac{(4n-1)\pi }{2}}[/tex] where 'n' is any integer
Part 2)
We have [tex]-1=e^{(2n+1)\pi }\\\\\therefore (-1)^{i}=(e^{i(2n+1)\pi })^{i}\\\\(-1)^i=(e^{i^2(2n+1)\pi })\\\\(-1)^i=e^{-(2n+1)\pi }[/tex] where 'n' is any integer
