Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do we know how many branches the inverse function of an elementary function has ? For instance Lambert W function. How do we know how many branches it has at e.g. $z=-0.5$ , $z=0$ , $z=0.5$ or $z=2i$ ?

share|cite|improve this question
up vote 3 down vote accepted

Suppose your elementary function $f$ is entire and has an essential singularity at $\infty$ (as in the case you mention, with $f(z) = z e^z$). Then Picard's "great" theorem says that $f(z)$ takes on every complex value infinitely often, with at most one exception. Thus for every $w$ with at most one exception, the inverse function will have infinitely many branches at $w$. Determining whether that exception exists (and what it is) may require some work. In this case it is easy: $f(z) = 0$ only for $z=0$ because the exponential function is never $0$, so the exception is $w=0$.

share|cite|improve this answer
How do we go up the branch ? What is the invariant of $W(z)$ ? $W(z) = W(f(z))$ ? I seem to have trouble ' visualizing '. @Robert Israel: Thanks for the answer. Could you help me with the questions above ? I seem to have read a bad/confusing paper apparantly. – mick Sep 19 '12 at 16:51
$W$ is supposed to be the (multivalued) inverse of $f$, so $f(W(z)) = z$. I'm not sure what you mean by "go up the branch", but I suppose you may be asking how to connect the values at one point with those at another. In brief, the answer is: analytic continuation. – Robert Israel Sep 19 '12 at 18:13
Im aware of analytic continuation. I see now I asked my question incorrect. I did not type what i meant. Sorry. Before i rephrase my question i will consider it again first and maybe i can find the answer myself. – mick Sep 20 '12 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.