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 $z^i$?

  • $\begingroup$ The «ambiguity» you refer to is simply the fact that $(-1)^2=1$. What exactly does not referring to that even mean? $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '13 at 7:21
  • $\begingroup$ The ambiguity is in defining $i$ as "the" solution to an equation that has two solutions. I'm not satisfied with the argument here: math.stackexchange.com/questions/9770/…, which is based on the development here: feynmanlectures.caltech.edu/I_22.html#Ch22-S5, and both use the ambiguity to argue that $z^{-i} = (z^i)^*$. $\endgroup$ – Aaron Golden Oct 28 '13 at 7:29
  • $\begingroup$ 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. $\endgroup$ – Did Oct 28 '13 at 7:33
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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '13 at 7:37
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    $\begingroup$ The problem with that is more in assuming that the symmetry between $i$ and $-i$ will extend to every context. $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '13 at 7:53

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