Question

Find all the values of:

a. (-i)^i

b. (-1)^i

please help we with this complex variable problem.

Answer 1

(a)
`(-i)^(i)=e^(i \log (-i))=e^(i (\log \lvert -i \rvert +i(\arg (-i)+2\pi n)))=e^{i(0+i(-(\pi )/(2)+2\pi n))}=e^{-(-(\pi)/(2)+2\pi n)}=e^{(\pi)/(2)-2\pi n`, where
`n\in \mathbb{Z}`

(b)
`(-1)^(i)=e^(i \log (-1))=e^(i (\log 1 +i(\pi+2\pi n)))=e^(-(\pi+2\pi n))`, where
`n\in \mathbb{Z}`.

Explanation:

Let's remember the definition of complex exponents. If
`z` is a nonzero complex number and
`c` is a complex number, we define
`z^(c)` by

`z^(c)=e^(c \log z)`

Where the
`\log` function is the complex logarithm function. That is to say,
`\log z=\log \lvert z\rvert + i(\arg z+2\pi n)`, where
`n\in \mathbb{Z}`. With this in mind we can calculate the given powers as follows:

(a)
`(-i)^(i)=e^(i \log (-i))=e^(i (\log \lvert -i \rvert +i(\arg (-i)+2\pi n)))=e^{i(0+i(-(\pi )/(2)+2\pi n))}=e^{-(-(\pi)/(2)+2\pi n)}=e^{(\pi)/(2)-2\pi n`, where
`n\in \mathbb{Z}`

(b)
`(-1)^(i)=e^(i \log (-1))=e^(i (\log 1 +i(\pi+2\pi n)))=e^(i(0+i(\pi +2\pi n)))=e^(-(\pi+2\pi n))`, where
`n\in \mathbb{Z}`.

This are all the values of the given powers.