Question

Find all the values of:

a. (-i)^i

b. (-1)^i

please help we with this complex variable problem.

Answer 1

Answer:

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

Step-by-step 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) $(-i)^{i}=e^{i \log (-i)}=e^{i (\log \lvert -i \rvert +i(\arg (-i)+2\pi n))}=e^{i(0+i(-\frac{\pi }{2}+2\pi n))}=e^{-(-\frac{\pi}{2}+2\pi n)}=e^{\frac{\pi}{2}-2\pi n$, 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.