Calculate Laurent Series for $\frac{\ln z}{(z-1)^3}$ about $z=1$

Question

Calculate the Laurent series of the function $g(z)= \frac{\ln z}{(z-1)^3}$ about the point $z=1$.

Well since the singularity and the centre of the circle we are expanding about collide, I can just bring about the $\frac{1}{(z-1)^3}$. But what do I do with the $\ln z$? Any help is appreciated.

Answer 1

Let $u = z - 1$. Then $$ g(z) = \frac{\ln(u + 1)}{u^3} $$ The series expansion of $$ \ln(u + 1) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}u^n $$ Now divide the expanded terms by $u^3$ and make the substitution $u = z - 1$.

Answer 2

You need the Laurent series for $\ln z = (z-1) - \frac{1}{2} (z-1)^2 + \frac{1}{3} (z-1)^3 -\cdots$. The rest should be easy.