Understanding imaginary exponents

Consider a real number $A$, and take it to the power $i$. If our system of complex numbers is to be consistent, then $A^i$ must be a complex number; in other words, there must be two real numbers $x$ and $y$, which depend on $A$, such that:

$A^i=x+iy$

Furthermore, we can write $A^{-i}=x-iy$ for the same $x$ and $y$. Hence:

$x^2+y^2=(x+iy)(x-iy)=A^iA^{-i}=A^{i-i}=A^0=1$

We have shown that for any real number $A$, $|A^i|=1$, and therefore $A^i$ corresponds to a complex number which lies some angle $\theta$ along the unit circle.

Now consider the sine and cosine functions for extremely small angles $\epsilon$. A tiny angle $\epsilon$ cuts out a slice of the unit circle, and the curvature of the circumference over this small angle is negligible. We can therefore think of this slice as a right triangle with angle $\epsilon$, and the hypotenuse and adjacent sides are both length one since they correspond to the radius of the unit circle.

Using the formula for the arc length of a circle, it's easy to determine that in the right triangle formed by the small angle approximation, the length of the side opposite to the angle $\epsilon$ is equal to $\epsilon$. We can read off the $(x,y)$ coordinates from this diagram (which are $(cos(\epsilon),sin(\epsilon))$), and therefore we conclude that for very small angles $\epsilon$:

$sin(\epsilon) \approx \epsilon \hspace{10mm} cos(\epsilon) \approx 1$

therefore $cos(\epsilon) + isin(\epsilon) \approx 1+i\epsilon$, and hence for real numbers $A$ which are extremely close to one (so that $lnA$ is small), the complex number $A^i$ lies approximately at an angle $lnA$ along the unit circle, since $A^i=e^{ilnA}\approx 1+i(lnA)$.

Lets say you want to figure out what $x^{a + ib}$ is, then like you mentioned, you start by writing

$x^{a + ib} = e^{a \ln(x) + i b \ln(x)}$

This can be split up though as

$ e^{a \ln(x) + i b \ln(x)} = e^{a \ln(x)} e^{i b \ln(x)} = x^a e^{i b \ln(x)} $

and then you can use euler's formula to take care of the imaginary exponent, so that

$x^{a + ib} = x^a \cos(b \ln(x)) + i x^a \sin(b \ln(x)) $

This formula does not provide much intuition as to what is really happening though. You mentioned that you can understand integer exponentiation as simple repeated multiplication, but I don't think that that is the right way to look at it in complex analysis. I think it is much better to look at exponentiation geometrically.

When exponentiating a real number by a complex value, we found that

$x^{a + ib} = x^a e^{i b \ln(x)} $

so exponentiating $x$ by $a + ib$ gives you the point in the plane with magnitude $x^a$ and angle $b \ln(x)$.

The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$.

By definition

$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+i\sin y)$$

which agrees with the real exponential function when $y=0$.

The principal logarithm of $z=x+iy$ is the complex number

$$w=\text{Log }z=\log |z|+i\arg z$$

so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$.

The complex power is

$$z^w=e^{w\text{ Log} z}.$$

In your example $z=x,w=i$ is therefore $x^i=e^{i \text{ Log}x}$.

If $x>0$, $\text{Log }x=\log x$. If $x<0$, $\text{Log }x=\log |x|+i\pi$.

Examples:

$(-1)^i=e^{i\text{Log }(-1)}=e^{i(i\pi)}=e^{-\pi}$.

$2^i=e^{i\text{Log }(2)}=e^{i\log 2}=\cos (\log 2)+i\sin (\log 2)$.

$(-2)^i=e^{i\text{Log }(-2)}=e^{i(\log 2+i\pi)}=e^{i\log 2}e^{-\pi}=(\cos (\log 2)+i\sin (\log 2))e^{-\pi}.$