Find all values of given degrees

Question

I have $1^i$. To find all values, I do $$1^i=e^{\operatorname{Ln}1^i}=e^{i\operatorname{Ln}1}$$ Since $\operatorname{Ln}Z=\ln|Z| + i(\operatorname{arg}Z + 2\pi k)$, therefore $\operatorname{Ln}1=2\pi ki$.

Then the above is equal to $$e^{i\times2\pi ki} = e^{i^22\pi k} = e^{-2\pi k}$$

But the answer on the book is $e^{2\pi k}$

What am I doing wrong?

Answer 1

You are right, and so is your book. It turns out that$$\{e^{2\pi k}\,|\,k\in\mathbb{Z}\}=\{e^{-2\pi k}\,|\,k\in\mathbb{Z}\}.$$

Answer 2

We have that

$$1=e^{2\pi hi} \implies 1^i=e^{-2h\pi}=e^{2(-h)\pi}=e^{2k\pi} \quad k\in \mathbb{Z}$$