# Multifunctions

• May 26th 2007, 06:39 PM
ThePerfectHacker
Multifunctions
I read about a nice way to think about that troubling complex logarithm function. Maybe some of you know this but I think it is helpful.

Given $a>0$ we know that $\ln a = b$ if and only if $e^b = a$. Now given a complex number $z\not =0$ we want to think about $\ln z=b$. Meaning $e^b = z$. But note that $e^{b+2\pi n i} = e^b \cdot e^{2\pi n i}=e^b$. So $b+2\pi n i$ also work. Which leads to a problem. It is not a function!

So one way to fix this is to restrict ourselves (as with inverse sine and cosine functions). First $\mbox{arg}(z)$ is the "argument" of $z$ it is defined to be the angle $z$ creates in the complex plane with $0\leq \theta <2\pi$. (Some people use the interval $-\pi < \theta\leq \pi$). To avoid confusion I will use $\ln$ as the standard function defined for real variables and $\log$ as its complex analogue. We given a complex number $z\not = 0$ we define,
$\log z = \ln |z| + i\cdot \mbox{arg}(z)$. Note that is $z$ is real and positive it produces the same thing as $\ln$.

The terrible thing is that the nice properties about $\ln$ do not generalize to $\log$. For example, $\log (z_1z_2) \not = \log (z_1) + \log (z_2)$.

This is were the "multifunction" comes in. We define a multifunction as $\mathbb{C}\mapsto \mathcal{P}(\mathbb{C})$. That is, a function which sents a complex number into a subset of the complex numbers. Hence it is a function, not a complex function, but it is a function. So we can define $\mbox{Log}(z) = \{ \log z + 2\pi n i\}$ (note the captial L). Meaning all the solutions to the equation $e^x = z$.
The good think about this is that the logarithm property is preserved. That is $\mbox{Log}(z_1z_2) = \mbox{Log}(z_1)+\mbox{Log}(z_2)$. (Addition here means adding terms in the set together).