Complex analysis - multivalued functions

  • #1
What are the implications for holomorphicity of a function being a multifunction.

take [itex]f(z)=\ln{z}=\ln{r}+i arg(z)[/itex],

here [itex]z=z_0+2k \pi[/itex] all correspond to the same value of z but give different values of f(z) i.e. its a multifunction.

how does this affect its holomorphicity?
as far as i can tell, it doesn't and this function is holomorphic everywhere in [itex]\mathbb{C}[/itex]. Is this true?
Or is it the case that it isn't holomorphic on the negative real axis since (just as for f(z)=z), if we make a branch cut and choose only to work with the principal argument then we have a discontinuity in the principal argument on the negative real axis. Why does this discontinuity affect holomorphicity though?

Are there any useful things to note about multifunctions? All my notes give is a definition. I was just wondering if there was any properties I should know about for my exam?

Answers and Replies

  • #2

The function [tex]1/z[/tex] does not have an analytic antiderivative on [tex]\mathbb{C}-\{0\}[/tex], but an analytic antiderivative can be defined, for example, on [tex]\mathbb{C}-R[/tex] where [tex]R[/tex] is any closed ray emanating from the origin. A suitably chosen branch of [tex]\ln z[/tex] works on [tex]\mathbb{C}-R[/tex].

The ray [tex]R=\{x:x\in\mathbb{R},x\le0\}[/tex] is a commonly chosen ray, but others work, as well as "non-rays."

You just can't have an analytic logarithm on all of [tex]\mathbb{C}-\{0\}[/tex] at once, so to speak.
  • #3

If you choose a particular branch cut to define a logarithm, f1(z), then you can analytically continue f1(z) across that branch cut to obtain another logarithm f2(z) . And then you can analytically continue f2(z) accross its branch cut to obtain a logarithm f3(z), etc. etc.

If you need to do a contour integration involving a contour that goes arounds the origin and would necessarily cross any branch cuts, then you can divide that contour into contours that don't cross branch cuts and each contour integral involves a different fi(z).

E.g. try to evaluate the integral of x^p/(1+x) from zero to infinity (for -1<p<0).

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