What is <any number>^i?

Question

I think I understand what imaginary numbers are, that $i$ is basically the name we give to $\sqrt{-1}$. Does $n^i$ have any sort of meaning? Is it used for anything? You can't really multiply $n$ by itself $i$ times, can you?

Answer 1

Please note that: $$e^{ix}=\cos(x)+i\sin(x)$$

Therefore: $$(e^x)^i=\cos(x)+i\sin(x)$$

Therefore: $$n^i=(e^{\ln(n)})^i=\cos(\ln(n))+i\sin(\ln(n))$$

Answer 2

For real $n$ we can define

$$n^i \equiv e^{i\log n}$$

where the exponential is usually defined via

$$e^z \equiv \sum_{k=0}^\infty \frac{z^k}{k!}$$

This gives us $$n^i = \left(\sum_{k=0}^\infty \frac{(\log n)^{2k}(-1)^k}{(2k)!}\right) + i\left(\sum_{k=0}^\infty \frac{(\log n)^{2k+1}(-1)^k}{(2k+1)!}\right)$$

We can also use the relation $e^{it} = \cos(t) + i\sin(t)$ to get the equivalent answer

$$n^i = \cos(\log n) + i \sin(\log n)$$