## Homework 9

So, does the Laurent series of an analytic function f allow convergence outside of the RoC for the normal power series of f?--Rgilhamw 19:50, 25 November 2009 (UTC)

I don't think so. It seems like the laurent series is just another power series representation of the function with another ROC. Like the example in the book 1/(1-z) can be represented by a power series with negative powers of z but with ROC abs(z)>1 instead of less than 1. The Laurent series seems like it is used to represent an analytic function in the annulus $ r<|z-c|<R $ where c is the center of the annulus.< --Adrian Delancy

A group of us got stuck on problem VII.18.3, as well as VIII.12.2.d. Does anyone have any tips for these? --Andy Bohn

For VIII.12.2 d, he showed in class that 1/(sin z) has poles of order one on it's singularity at 0 (Which can be done for all singularities which happen at 0+2PIn where n is an integer) by looking at it's power series expansion and factoring out a 1/z. Since the pole is of degree one, we can safely use the formula g(z)/h'(z), yielding 1/(cos z) for z = 0+2PIn, n an integer, which is simply 1/1 = 1. For VII.18.3, most of the problem was solved in lecture (after the question was posted) so refer to the December 2 lecture for clarification on the notation and how to solve it. --Whoskins

for VII.14.1, I have a solution, but I don't see how it relates to the identity theorem. I just took the zeroes of the function $ f=z^m g(z) $ where m is the multiplicity of the zero, and showed that the $ \lim_{n \to \infty}g(\frac{1}{n}) $ either goes to zero or does not converge depending on the value of m. Was there an easier way to do this?--Rgilhamw 22:02, 30 November 2009 (UTC)

I have an idea for VIII.12.2.d. Use the result of VIII.12.1., with g and h defined for all z in the disk of radius pi centered at one of the singularities. Then, the residue is 1/cos(z), where z is a singularity. I am starting to second-guess if this will work.--Phebda 23:22, 30 November 2009 (UTC)

VII.14.1: an easier solution can be done using the identity theorem. Consider the functions $ g\left(\frac{1}{n}\right) = \frac{1}{n^2} $ for even $ n $ and $ h\left(\frac{1}{n}\right) = -\frac{1}{n^2} $ for odd $ n $.

VII.18.3: my first thought was maximum modulus, but I haven't worked out all the details and it may not work. --Dan Stratman

VII.18.3: I'm confused as to what this problem is even asking. how can a function be unbounded when you input a single point? --Rgilhamw 20:19, 1 December 2009 (UTC)

There have been enough questions about HWK 9 that I will give everyone an extension until Friday, Dec. 4 to turn it in. --Steve Bell

VII.18.3: It's not the "function" that is unbounded here, it is the set u^-1(c), which is quite different. --Ysuo 23:42, 1 December 2009 (UTC)