When I raise some integer to some imaginary power I obtain a complex number. E.g., $$2^i = 0.7692 + 0.6390i.$$
Why does this happen? What does it mean to raise an integer to an imaginary power?
Asked
Active
Viewed 169 times
0
-
1Note $$2^i=e^{i\ln(2)}=\cos(\ln(2))+i\sin(\ln(2))$$ – Matthew H. Apr 06 '21 at 15:21
-
Please don't use pictures. – Dietrich Burde Apr 06 '21 at 15:21
-
I guess I need an intuitive answer – Ulziibayar Apr 06 '21 at 15:44
1 Answers
0
Let $n\in\mathbb Z\setminus\{0\}\,$ and $\,p\in\mathbb R.$
By definition, $$n^{p\,i} = \exp\left[p\,i\,\log\left(n\right)\right] \\ = \begin{cases}\exp\left[p\,i\,\left(\ln(n)+i\,2k\pi\right)\right] &\text{if }n>0; \\ \exp\left[p\,i\,\left(\ln(-n)+i(2k+1)\pi\right)\right] &\text{if }n<0 \end{cases} \\ = \begin{cases}\exp\left[-2pk\pi\right] \,\exp\left[i\,p\ln(n)\right] &\text{if }n>0; \\ \exp\left[-p(2k+1)\pi\right] \,\exp\left[i\,p\ln(-n)\right] &\text{if }n<0 \end{cases},$$ where $k\in\mathbb Z.$
ryang
- 38,879
- 14
- 81
- 179
-
@Ulziibayar On a related note: What happens when an integer is raised to an irrational power? – ryang Apr 07 '21 at 14:02
-
P.S. For readability, $\exp()$ denotes $e^{()},$ and is not necessarily single-valued. – ryang May 09 '21 at 07:33