Complex Exponents

Solution 1:

Dear Nate, first, why is $e$ the preferred base for exponentials? Imagine that you have 1 dollar and your bank gives you 100% interest rate. After 1 year, you will have 2 dollars.

Now it offers you to add interests 100 times in a year but the interest is 1% at each moment. How much will you get? You will get $$ (1+0.01)^{100} \approx 2.704 $$ What if they add you $1/N \times $ 100% at $N$ moments of the year and you send $N$ to infinity? Well, you will have $e\approx 2.71828$ dollars after one year.

In fact, the general exponential - power with the base of $e$ - may be defined by this "repeated small interest" formula as $$ e^X = \exp(X) = \lim_{N\to \infty} \left(1+\frac {X}{N}\right)^N $$ It only has this simple form if the base is $e$. A more general power may be defined as $$ Y^X = \exp(X \ln Y) .$$ Here, $\ln$ is the natural logarithm so that $\exp\ln X = X$. If I replaced the base $e$ by another base such as $2$ or $10$, the "repeated small interest" formula above would have to contain $\ln 2$ or $\ln 10$ or other awkward factors at various places. It wouldn't be natural.

So instead of powers $Y^X$ and logarithms with general bases, you should think that in mathematics, only $\exp(X)$ and $\ln(X)$ are really needed, and all the other powers and logarithms may be expressed as composite functions. Also, $\exp(X)$ has the advantage that its derivative is exactly equal to the very same function $\exp(X)$. In particular, the derivative evaluated at $X=0$ is equal to one, very nice and simple. It would be $\ln(Y)$ if you used a different base $Y$ instead of $e$.

Now, what is the exponential of an imaginary exponent? Again, you may write $$\exp(iX) = \lim_{N\to\infty} \left(1+\frac {iX}{N}\right)^N $$ You multiply $N$ copies of a number that is very close to one. What do you get?

Well, the multiplication by a complex number has the effect of magnifying (or reducing) the plane, and rotating it. In particular, the absolute value of the number $(1+iX/N)$ is essentially one, up to second-order corrections that disappear in the $N\to \infty$ limit. So in the limit, $(1+iX/N)$ is effectively a number whose absolute value equals one.

Multiplying by complex numbers whose absolute value is equal to one looks like a rotation of the complex plane. The angles are preserved - those are some things one should know about the complex numbers. Moreover, it's clear that multiplying by $(1+iX/N)$ is equivalent to the rotation by $X/N$ in radians. If you multiply the same factor $N$ times, you simply get a rotation by $X/N$ in radians.

So the $N$th power of $1+iX/N$, in the limit $N\to\infty$, is the number that you get by rotating $1$ in the counter-clockwise direction by the angle $X$ in radians. Clearly, the answer is $$ \exp(iX) = \cos (X) + i \sin(X) $$ where the trigonometric functions have arguments in radians, of course. Once again, the mathematically natural unit of an angle is in radians for very similar reasons why the natural base of the powers or exponentials is $e$. Only in radians, it's true that the derivative of $\sin X$ equals $\cos X$ and many other things.

In fact, the previous formula makes it natural to say that $\cos X$ and $\sin X$ are not "independent" functions, either. They may be defined as $$\cos (X) = \frac 12 ( e^{iX} + e^{-iX} ) $$ $$\sin(X) = \frac{1}{2i} (e^{iX} - e^{-iX} ) $$ You may substitute the last two equations into the previous one or vice versa to check that everything is consistent.

Just to be sure, general complex numbers may also be exponentiated via $\exp(A+iB) = \exp(A)\exp(iB)$ where both factors are known.

Solution 2:

There are various possible approaches, but I prefer to start with the exponential function $e^z$ as solution to the differential equation $f'(z) = f(z)$ with $f(0)=1$. The other exponentials are derived from this: $a^z = e^{z \log(a)}$ is a solution to $f'(z) = \log(a) f(z)$ with $f(0)=1$.