Laurent series expansion of Log(1+1/(z-1))

[Boorglar]

Homework Statement

Find the Laurent series expansion of f(z) = \log\left(1+\frac{1}{z-1}\right) in powers of \left(z-1\right).

Homework Equations

The function has a singularity at z = 1, and the nearest other singularity is at z = 0 (where the Log function diverges). So in theory there should be a Laurent series on the punctured disk centered at z = 1 with radius 1.

The Attempt at a Solution

I can't seem to rewrite the expression in a form which can be used to find the Laurent series for 0 &lt; |z - 1| &lt; 1. I can find the Laurent series for |z - 1| &gt; 1 by rewriting f(z) = -\log\left(\frac{1}{1+\frac{1}{z-1}}\right) and then expanding using the series expansion of \log\left(\frac{1}{1-z&#039;}\right) and substituting z&#039; = \frac{-1}{z-1}. This does not work in the region 0 < |z-1| < 1, unfortunately.

Wolfram and the solutions manual did not give any Laurent series for this region either, which leads me to believe that such a series does not exist. But I don't see why, since the function appears to be analytic in the punctured disk, and Laurent's theorem should apply.

 

[Boorglar]

I just realized I made a rookie mistake. The function is not analytic anywhere on the real line segment between 0 and 1, which means there is no punctured disk throughout which the function is analytic. The only region where the Laurent series exists is for |z-1| > 1.
Sorry for this :P