I'm trying to gain some perspective on complex analysis, and there's this concept of multivalued function which is weird to me because I'm used to functions having only 1 value for an argument, though I think that a multivalued function is not technically a function. So the question is, as indicated in the title: Can a multivalued function have an infinite number of values for a given argument?
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$\begingroup$ You're indeed correct in your thinking that multi-valued functions aren't really functions in the strict sense. The term is a misnomer as mentioned in the Wikipedia page. $\endgroup$– learnerDec 6, 2015 at 21:22
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$\begingroup$ The simplest example of a multi-valued function (a relation, to be pedantic) would be $f\colon\Bbb R\mapsto\Bbb R$ defined by $f(x)=\textrm{Arctan}(x)$ which outputs all the values $y$ for which $\tan(y)=x$. This is multi-valued since $\tan(x)$ is a periodic function. $\endgroup$– learnerDec 6, 2015 at 21:27
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$\begingroup$ Yes. For instance, the logarithm is a multivalued function, $\log z = \log|z|+i \text{Arg}z + 2\pi n i$, for any $z\in\mathbb C$ and $n\in\mathbb Z$. This is simply due to the fact that it is the inverse of $e^z$, and its multivaluedness reflects the fact that $e^{z}=e^{z+2\pi ni}$. You can gain more perspective if you write $z=re^{i\theta}$. The exponential function is periodic, you can add any multiple of $2\pi$ to the phase and you end up with the same number. $\endgroup$– krvolokDec 6, 2015 at 21:28
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$\begingroup$ The 'point' of multivalued functions is this: as you trace a circle around the origin, starting from some point and ending on the same point, a multivalued function doesn't end up on the same point, but "climbs up" to a new branch, continuously. $\endgroup$– krvolokDec 6, 2015 at 21:28
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3 Answers
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In some sense yes. $\ln(1)=2\pi ik$ for every $k \in \mathbb{Z}$.
answered Dec 6, 2015 at 21:27
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To be specific, you're used to functions, which by definition have only one output value associated with each input.
For multi-valued functions, each input is not associated with a single output. For example, consider the complex logarithm, and note that we can represent any point on the complex plane as $z = re^{i\phi} = re^{i(\phi +2\pi k)}$ for any integer $k$.
$$\begin{align*} \log z &= \log (re^{i(\phi +2\pi k)}) \\ &= \log (r) + \log (e^{i(\phi +2\pi k)}) \\ &= \log (r) + i(\phi +2\pi k) \\ \end{align*}$$
Inputting any one point $z$ will output many different points due to $k$. This is not a function in the traditional sense. This is why we often restrict a multi-valued function to a particular branch. For example, define $\log_{0} z$ to the branch where $\theta\in [0,2\pi)$. Then, we have a function
$$\log_{0} z= \log (r) + i\theta\quad\,\quad \theta\in [0,2\pi)$$
In a sense, you've already dealt with multivalued functions before. Recall the inverse trig functions. Take $y=\sin^{-1} x$, which has a restricted range $y\in [-\pi/2,\pi/2]$ analogous to a particular branch. If this range weren't restricted, the expression $\sin^{-1}(0)$ would be multivalued (it would evaluate to $\pi k$ instead of $0$).
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It may help to think of a multivalued function a little differently. If $f:X\to Y$ is a multivalued function, there is a corresponding (actual) function $f':X\to\mathcal{P}(Y)$ defined by $f'(x)=\{y\in Y | y \text{ is in the image of }x\text{ under }f\}$. In fact, this is a bijection from the set of multivalued functions from $X$ to $Y$ and the set of functions from $X$ to $\mathcal{P}(Y)$ as long as you count regular functions as a special case of multivalued functions. Thinking about it this way, it isn't hard to see that an element can have an infinite set of values as long as the codomain is an infinite set.
