5:07 PM

A comment on my post about complex numbers has made me want to go back and discuss exponentials in more detail even though it feels off-topic to this series as a whole. There are two problems this discussion will run into. The first is that I've been trying to build up the math I've been using systematically. Even though I leave out a lot of the details, I've avoided statements along the lines of "this is true, but I can't show that right now. You'll just have to trust me." I think I will get to all of the math involved in this subject eventually, but it may not be until June or so. Even then, there's one step I'm not sure I will be able to fully justify.

The second issue is more practical. I have been writing all of my math in plain HTML. I think this has been basically working in previous posts, although I haven't always been able to write everything as clearly as I would have liked. In this post, there will be some expressions where I can either write them in a way that's comprehensible or a way that's legible, but I can't do both. I'm going to aim for legible and hope that anyone who is following along will be able to decode the meaning.

Going back to raising negative numbers to rational exponents, there's a problem I glossed over. The equation (−1)^1/2=x is equivalent to −1=x², and there are two (complex) solutions to that equation. In general, for any real number a (other than 0) and any natural number n, there are n complex solutions to the equation a^1/n=x. In some cases there is an obvious preferred solution. If a is a positive real number, then there is always a positive real number x which satisfies the equation. If a is a negative real number and n=2, the two solutions are of the form bi and −bi, where b is a positive real number, so it makes sense to prefer the positive value.

However, the four solutions to (−1)^1/4 do not have an obvious preferred solution. The solutions are 2^−1/2+i2^−1/2, −2^−1/2+i2^−1/2, 2^−1/2−i2^−1/2, and −2^−1/2−i2^−1/2. You could choose the solution with positive components, but that poses a problem, because (−1)^3/4 also has four solutions, and if we also choose the solution to that with positive components, then [(−1)^1/4]³≠(−1)^3/4.

One resolution to this is to write the complex numbers in polar form. Given a complex number a+bi, the magnitude (or absolute value) r of the number is √a²+b². If we think of a+bi as a right triangle with legs a and b and hypotenuse r, then there is an angle θ such that cos θ=a/r and sin θ=b/r. Then a+bi=r(a/r)+ir(b/r)=r cos θ + ir sin θ=r(cos θ + i sin θ).

There is not a unique value of θ in this representation. If a+bi=r(cos θ + i sin θ), if θ is a solution then any value of the form θ+2πk is a solution for any integer k. We can define the preferred value of θ such that 0≤θ<2π. (If it's not obvious, we are measuring the angle θ in radians. We could measure it in degrees, but things would get really messy soon.)

Here's the step I can't fully justify. I am going to assert that cos θ + i sin θ = e^iθ. I can offer a partial justification for this, but this justification has a hole, at least in the scope of my math experience. The values for e^x, cos x, and sin x can be written as Taylor series, or infinite sums of polynomial terms. I expect to cover Taylor series in depth eventually, but for now I'm just going to assert that this works. However, in my experience, Taylor series are only defined for real values of x.

If we write e^iθ as a Taylor series, I'm not convinced that's valid since iθ is not a real number. Regardless, the series has a bunch of real terms and a bunch of imaginary terms. We can separate the series into two series, one consisting of real terms and one consisting of imaginary terms. The real series is the Taylor series of cos θ. The imaginary series is i times the Taylor series of sin θ. So e^iθ=cos θ + i sin θ, as long as we accept the Taylor series expansion of e^iθ as valid.

Then we can write any complex number a+bi in the form re^iθ, where r is a non-negative real number and θ is between 0 and 2π. θ can equal 0, but cannot equal 2π. As stated earlier, re^iθ=re^i(θ+2πk) for any integer k. With the laws of exponents for real numbers and the identity a=e^{ln a}, we can now find a unique value for any complex number raised to any complex power, except for 0⁰.

Looking at complex numbers raised to rational powers, this gives us a preferred unique solution for (a+bi)^1/n, but it also allows us to find the other solutions. (a+bi)^1/n=(re^iθ)^1/n=r^1/ne^iθ/n. However, re^iθ is equivalent to re^i(θ+2πk) for any integer k, and (re^i(θ+2πk))^1/n=r^1/ne^{i(θ/n+2πk/n)}. This has the same magnitude as the original solution, but only has the same angle if k is a multiple of n, so by taking all k from 0 to n-1, we can find all of the solutions.

In the general case of any real number x for (re^i(θ+2πk))^x, if x is irrational, then the set of angles produced by all integers k is infinite. However, the magnitude is always the same. We can still be assured of a unique value by requiring θ to be between 0 and 2π and k to be 0. Weirder, if the exponent is a complex number, then there is also an infinite set of magnitudes. I find this disconcerting, even though we can again specify a unique magnitude (and angle) by constraining the range of θ and requiring k to be 0.

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