Composition of meromorphic functions

[fortissimo]

Homework Statement

If f is meromorphic at a in ℂ^ (extended complex plane), and g is meromorphic at b = f(a) in ℂ^, show that h := g ° f is meromophic at a, with an exception. If f takes on the value b in ℂ^ with multiplicity m at a and g assumes the value c = g(b) in ℂ^ with multiplicity n at b, what is the multiplicity of c at a?

Homework Equations

The Attempt at a Solution

For the first part I don't know exactly what to do. Could you use the fact that f and g can be expanded into Laurent series?
For the second part, at least if a, b, c ≠ ∞, we have that g(f(z)) - c = g1(f(z))*(f(z) - b)^n where g1(b) ≠ 0, and f(z) - b = f1(z)*(z-a)^m (where f1(a) ≠ 0), so the multiplicity is nm. It seems tedious if you have to make different cases for a, b, c = ∞. Or do you need to do this?

 

[mighty2000]

Thank you for your question. In order to show that h := g ° f is meromorphic at a, we can use the fact that both f and g are meromorphic at a and b respectively. This means that both functions have poles at a and b, but they are removable. Therefore, the composition of these two functions will also have a removable pole at a, making it meromorphic at a.

For the second part of the question, we can use the fact that the multiplicity of a root in a function's Laurent series is equal to the order of the root. Therefore, if f takes on the value b with multiplicity m at a, and g assumes the value c with multiplicity n at b, the multiplicity of c at a will be equal to the product of m and n, as you correctly stated.

I hope this helps. Let me know if you have any further questions.