# Principal Value as an initial answer

Expression in questions involving complex analysis can often have multiple values. In such questions, should the answer provide the principal value as an answer first and then clarify that the expression has multiple values, or just give a formula from where multiple values of the expression can be extracted?

Take my answer to a particular question as an example, I give the principal value of $(-1)^\pi$ as an intial answer ($(-1)^\pi=\ldots$), but then clarify that this one of the multiple values of $(-1)^\pi$, and then give a formula for it.

Even though it is wrong to write "$(-1)^\pi=\ldots$" since it has infinitely many values, I feel that giving the principal value as an answer intially would make it easy for someone to understand the starting concept of the answer, before it goes to the idea of multiple values.

Thoughts?

• (Principal, rather than "principle".) – Andrés E. Caicedo Mar 17 '14 at 17:16

I think it's very important for students of complex analysis to understand early on that some concepts that they're familiar with on the real numbers extend in an quite counter-intuitive way to the complex numbers.

Taking roots is one of them. Over the real numbers, defining the $n$-th root of $y$ as the non-negative $x$ such that $x^n = y$ works pretty well. Over the complex numbers, not so much. First, unlike the reals where there only are 1 or 2 possible solutions of $x^n = y$, there now are $n$ possible solutions - which is particularly fun if you write something like $x^y$ where $y$ isn't yet known. And second, there's no natural order on $\mathbb{C}$ that would help you pick one principal value unambigiously.

The same goes for exponentation with non-integral exponents, since such an operation involves taking some root.

I therefore feel that you're doing a disservice to students by first computing a principal value - it creates the impression that these problems are a mere technicality, when they really are a fundamental property of complex numbers. I personally tend not to use the notation $x^y$ at all when I'm dealing with complex numbers, unless $y$ is know to be an integer.

One thing I stress in my computations that use complex methods is that one needs to understand when one has chosen a branch in which to work. Serious mistakes are made when the notion of branch is not deeply understood. My favorite example of this is in evaluating integrals using the residue theorem via a keyhole contour. We use such a contour to exploit a branch cut in, say, a logarithm in order to define a contour integral that will eventually provide the real integral we seek. For example, it is known that for sufficiently well-behaved functions $f$, we have

$$\int_0^{\infty} dx \, f(x) = -\sum_k \operatorname*{Res}_{z=z_k} f(z) \log{z}$$

This of course requires us to compute the log of some complex numbers. The log of course is multivalued, so for instance $-i$ can be $e^{-i \pi/2}$ or $e^{i 3 \pi/2}$, or something else., and it all shouldn't matter. But it does! The wrong choice gives us the wrong answer. And so what is the right choice? The answer is...you chose it already, when you specified the keyhole contour and the above equation. The arguments of your complex numbers are now forced to be between $0$ and $2 \pi$. Period. So the only right answer is $-i=e^{i 3 \pi/2}$.

In practical calculations using multivalued functions in complex analysis, you make the choice of branch in adopting some reference line dictated by the symmetry of the problem. Once that is done, you are now stuck with your "principal" value - any other is meaningless.

• Agreeing with what you wrote, I think "principal" (as in the OP) refers not to a particular branch chosen for the task, but specifically to the branch of $z^p$ and $\log z$ in $\mathbb C\setminus (-\infty,0]$ normalized by $1^p=1$ or $\log 1=0$. – user127096 Mar 17 '14 at 21:17