Complex Analysis - Laurent Series & Singularities?

Question

(a) Locate and classify the singularities (giving the order of any poles) of the function:

f(z)= z/(1-e^z)

(b)Let f(z) = zsinh(1/(z+1))
(i) Find the Laurent series of f about -1 giving the general term of the series for odd and even powers of (z+1)
(ii) Write down a punctured open disk D, containing a circle C={z:|z+1|=1}, on which f is represented by this series
(iii) State the nature of th singularity of f at -1
(iv) Evaluate
integral zsinh (1/(z+1))dz where C = {z:|z+1|=1}
(c) Find the laurent series about 0 for the function

f(z)= 7z/((2z+1)(z-3)) on the set {z:|z|>3}, giving the general term of the series

Answer 1

I know that to find the Larent series you first have to know how to find the taylor series. Also you have to look at where you are centering the series, this means that you will need to rewrite your function before finding the series.

http://math.fullerton.edu/mathews/c2003/LaurentSeriesMod.html

Answer 2

An isolated singularity z0 implies that there exists some neighborhood of z0 that contains no other singularities of f. If a_k(z - z0)^k is the kth term of the Laurent series where k is some integer, then if z0 is a singularity of f, it is removable if every coefficient of negative k is 0. (If any coefficient of negative k is non-zero, we get an exploding term (z - z0)^k as z approaches z0 for that k, which corresponds to a pole.) Similarly, a pole of order n corresponds to a nonzero coefficient of negative k (n = -k) where all terms with exponents less than k have coefficient 0. Multiplying f by (z - z0)^n will thereby eliminate all negative exponent terms in the Laurent expansion giving a holomorphic function with at most a removable singularity at z0. It is then obvious that if f has infinitely many nonzero negative exponent terms, there is no n such that f*(z - z0)^n is holomorphic at z0. The isolated part is defined as in the first paragraph.