1

Is there a way to develop the definition of the imaginary exponent, $z^i$, for complex $z$, that does not appeal to the notion that $i$ and $-i$ are "qualitatively indistinct" and that does not rely on defining $a^b = e^{b\log{a}}$ using a pre-defined exponentiation function?

My motivation is that I would like to be able to develop complex exponentiation in a "natural" way, in a sequence like:

  1. $z^n = z * z \cdots * z$, $n$ times for $n\in\mathbb{N}$
  2. $\sqrt[m]{z}$ is the non-negative real $y$ such that $y^m = z$.
  3. For rational $q = n/m$, $z^{n/m} = \sqrt[m]{z^n}$.
  4. For $r\in\mathbb{R}$, $z^r = \lim_{q->r}{z^q}$, where $q$ is rational.
  5. $z^i$ is "something."

Finally, the usual rules for dealing with products and sums in the exponent allow us to evaluate $z^w$ for complex $z$ and $w$.

The payoff for this development is that $e$ could be "discovered" as the answer to the question: For which positive real number, $a$, does $f_a(\theta) = a^{i \theta}$ traverse the unit circle at just the right speed so that the period of $f_a$ is the circumference of the circle? And the connection between $e^{i \theta}$ and the trigonometric functions follows intuitively from there.

The way I've seen complex exponentiation developed before fall into two categories:

  1. The exponential function is simply defined as $\exp{z} = \sum{ z^k/k!}$, and the fact that this function happens to behave on suitable inputs just like taking powers of $e$ might be interpreted as a convenient miracle.

  2. The fact that $i$ and $-i$ are equally valid solutions to $x^2 = -1$ is invoked to "prove" that $z^i$ lies on the unit circle, some hand waving about approximating $z^{s i}$ for small $s$ occurs, and that approximation is used to build up a full definition. The problem is that the whole argument hinges on the ambiguity between $i$ and $-i$ when $i$ is defined as "the" solution to $x^2 = -1$. If the complex numbers are developed more carefully, so $i$ is merely the (unambiguous) pair $(0, 1)$, then the argument can't be developed. Besides, $z^i$ doesn't lie on the unit circle for all $z\in\mathbb{C}$.

So is there a more natural (yet rigorous) way to develop $z^i$?

Aaron Golden
  • 1,031
  • 6
  • 12
  • The «ambiguity» you refer to is simply the fact that $(-1)^2=1$. What exactly does not referring to that even mean? – Mariano Suárez-Álvarez Oct 28 '13 at 07:21
  • The ambiguity is in defining $i$ as "the" solution to an equation that has two solutions. I'm not satisfied with the argument here: http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents, which is based on the development here: http://www.feynmanlectures.caltech.edu/I_22.html#Ch22-S5, and both use the ambiguity to argue that $z^{-i} = (z^i)^*$. – Aaron Golden Oct 28 '13 at 07:29
  • The point 2. requires $z$ to be REAL, no? More generally, you might want to make more precise where in your post you assume that $z$ is real (positive) and where you do not. – Did Oct 28 '13 at 07:33
  • 1
    No, one does not define $i$ to be «the» solution to the equation but «a» solution. The first would indeed be ambiguous, and that is precisely why we don't do it. – Mariano Suárez-Álvarez Oct 28 '13 at 07:37
  • I don't see how you plan to continue after defining $z^i$: how will you define $z^{ir}$ for real $r$? – Martin Argerami Oct 28 '13 at 07:38
  • @MarianoSuárez-Alvarez If you read the linked answer and document you'll see what I'm getting at. The argument goes: $i$ and $-i$ are equally valid imaginary units, so any valid expression involving $i$ is equally valid if we replace all the $i$'s with $-i$. So if $z^i = x + i y$ then $z^{-i} = x - i y$ and $1 = z^i z^{-i} = (x + i y)(x - i y) = x^2 + y^2$ so $z^i$ is on the unit circle. And it's just not true. Take $z = -1$ for example. – Aaron Golden Oct 28 '13 at 07:47
  • 1
    The problem with that is more in assuming that the symmetry between $i$ and $-i$ will extend to every context. – Mariano Suárez-Álvarez Oct 28 '13 at 07:53
  • A much simpler instance of the problem mentioned in my last comment is that from $\Re(i)>0$ you cannot conclude that $\Re(-i)>0$. – Mariano Suárez-Álvarez Oct 28 '13 at 18:06
  • @MarianoSuárez-Alvarez I think I'm still missing something. What's the context that justifies the $z^{-i} = x - i y$ step? I can see that the formula is "true" when $z$ is a positive real number, but then I'm relying on a pre-existing definition for complex exponentiation. The conditions under which it's valid to substitute $-i$ for every instance of $i$ in an expression need to be developed carefully, right? I was hoping complex exponentiation could be developed without relying on that substitution. – Aaron Golden Oct 28 '13 at 19:12
  • The difficulty is that there are many extensions of the $\mathbb R\to\mathbb R$ function $x\mapsto a^x$ to a $\mathbb C\to\mathbb C$ function, even if we require the extension to satisfy $a^{w+z}=a^wa^z$. The assumption that $a^{\overline z}=\overline{a^z}$ picks out a particularly nice family of extensions. That is, this condition is not provable, but is imposed. My answer to a similar question has more details. –  Jun 15 '14 at 17:38
  • @StevenTaschuk Thanks for the comment, Steven. I'd love to read the similar question and answer, but the link you posted goes to this same question. – Aaron Golden Jun 15 '14 at 17:45
  • Sorry, I'm klutzing my links today. It was supposed to be: my answer to a similar question. –  Jun 15 '14 at 17:46

0 Answers0