TITLE: How can I intuitively understand complex exponents?
QUESTION [60 upvotes]: Could you help me fill in a gap in my understanding of maths?
As long as the exponent is rational, I can decompose it into something that always works with primitives I know.
Say, I see a power: $x^{-{a\over b}}$ for natural a,b. 
Using the basic rules: 
$$\begin {align}
x^{-a} =& {1 \over x^a} \\
x^{1\over b} =& \sqrt[b]{x} \\
({x^a})^b =& {x^{a b}}
\end{align}$$
I can always decompose rational exponents into $ x^{{1\over b} \cdot -a} = \sqrt[b]{x} ^ {-a} = {1 \over \sqrt[b]{x}^a }$- and integer degree roots are something well within my grasp. Integer root is always just simply a number of multiplications, $\sqrt[a]{x}$ is just some $y \cdot y \cdot y \cdot ... \cdot y$, as many $y$'s as $a$ says. 
If the exponent is irrational, it's trickier, but I can always take $e^{\pi} \approx e^{314159265358979323 \over 100000000000000000}$ and take more $\pi$ digits if needed, and even though the number of underlying multiplications becomes ridiculous, that's still something I can understand.
But I'm completely at loss how to understand stuff like $x^{i\pi \over 4}$. I just can't perform an imaginary number of multiplications. I know how to perform conversion of complex numbers between the exponential and $a+ib$ form, but I perform it like a mysterious voodoo magic recipe without ability to wrap my mind around how that kind of exponentiation is supposed to work. I can still apply the old conversion formulas, $x^a \cdot x^b = x^{a+b}$; $({x^a})^b = {x^{a b}};$ and obtain correct results but I just don't understand the underlying mechanism - as $x^n$ is just $n$ multiplications of $x$, how can understand how that works when $n$ is imaginary?

REPLY [4 votes]: Actually, I think your fundamental mistake is to think too fundamentally without thinking more fundamentally :P. The problem is deep rooted in how we define exponentiation. 
This used to confuse me for some time :) I hope this answer helps you and anybody else that might stumble upon my answer!
A Simpler Problem
You seem comfortable calculating the value of $x^{1 \over b}$, but given a paper, a pen and infinite time, how would you evaluate:
$$3^{0.3}$$
How do you produce the digits? Most calculators reduce this expression to:
$$ e ^ {\log 3^{0.3}} = e ^ {0.3 \log 3}$$
And then we can use lookup tables to calculate $log3$ and indeed the exponential of the resulting number. The lookup tables are defined from limits and calculus though, so :
$$ e^x = \lim_{n\rightarrow \infty} \left( 1 + {x\over n} \right)^n $$
and:
$$\log x = \int_1^x {1 \over x} dx $$
So we can go ahead and work out this integral directly using a lookup as I mentioned, or by using something like Simpsons rule. 
Notice that we've lost the original definition here, we are no longer defining:
$$ a^b = \underbrace{a * a * a * \ldots }_\text{n times} $$
but instead, we are now defining exponentiation by:
$$ a^b = e ^ {b \log a }$$
This represents a fundamental shift in how we should perceive exponentiation. Because the idea of multiplying something by itself 0.3 times is not well defined, and we shouldn't think of it in that way anymore :)
Extending to Complex Numbers
Now if we apply this definition to complex numbers, with the aide of Eulers amazing equation, that shows us multiplication of two complex numbers represents a multiplication of magnitudes and a rotation by their arguments, we can see that this equation is exactly what we would expect it to be:
$$ x ^ {i {\pi \over 4}} = e ^ {i \left( {\pi \over 4 }\log x \right)} = \cos\left( {\pi \over 4 }\log x \right) + i \sin \left( {\pi \over 4 }\log x \right)$$
This just uses our new definition of exponentiation, and it extends very naturally to complex numbers as you can hopefully see :)
Some Words
I think some of the others put this better than I could. As you learn more and more mathematics, you start to learn to explain things that you once could only explain with a certain pattern of thinking, in terms of new patterns that would have been absurd to show a new learner.
I indeed asked myself this same question some time ago, and after some careful thought, I ended up with this :)
Ps.
So I guess now you have to figure out how to calculate sin and cos ;) 
Have a good one!