Could anyone show me how to find the Laurent series of
$$ f(z) = \frac{\operatorname{Log}(z)}{(z-1)(z-5)} $$
centered at $5$? I know how to find it for $\operatorname{Log}(z)$, using the geometric series and then integrating. I'm not sure how to deal with the denominator though.
It is easier to me to find the Laurent series of $f(z+5)$ in $z=0$ then translate it back. We have: $$ g(z)=f(z+5)=\frac{1}{z}\cdot\frac{\log(z+5)}{z+4}=\frac{h(z)}{z},$$ where $h(z)$ is a holomorphic function in a neighbourhood of zero: $$ h(z)=(\log 5 + \log(1+z/5))\cdot\frac{1}{4}\cdot\frac{1}{1+\frac{z}{4}}$$ and: $$\frac{1}{1+\frac{z}{4}}=\sum_{j=0}^{+\infty}\frac{(-1)^j}{4^j}z^j,$$ $$\log\left(1+\frac{z}{5}\right)=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k\cdot5^k}z^k,$$ so: $$ h(z)=\sum_{j=0}^{+\infty}\frac{(-1)^j\log 5}{4^{j+1}}z^j+\sum_{n=1}^{+\infty}(-1)^{n+1}\left(\sum_{m=1}^{n}\frac{1}{m\cdot 5^m 4^{1+n-m}}\right)z^n$$ and: $$ g(z)=\frac{\log 5}{4z}+\left(\frac{1}{20}-\frac{\log 5}{16}\right)+\sum_{n=1}^{+\infty}(-1)^{n+1}K_n z^n,$$ where: $$ K_n = \sum_{m=1}^{n}\frac{1}{m\cdot 5^{m}4^{1+n-m}},$$ giving:
$$ f(z)=\frac{\log 5}{4(z-5)}+\left(\frac{1}{20}-\frac{\log 5}{16}\right)+\sum_{n=1}^{+\infty}(-1)^{n+1}K_n (z-5)^n.$$
The coefficients are given by contour integrals, which are easily evaluated by the Cauchy integral formula. Our function can be written $h(z)/(z-5)$. Then $$a_n=\int_\gamma (z-5)^{-n} h(z)\, dz=\int_\gamma \frac {f(z)}{(z-5)^{n+1}}\, dz. $$ In particular $a_n=0$ for $n\leq -2$, $a_{-1}=h(5) = {{\ln(5)}\over 4}$, $a_0=h'(5)$, and generally $a_n=h^{(n+1)}(5)$ for $n\geq- 1$. All of which you can verify agrees with Jack's values.
I find this usually much simpler than trying to expand everything into series and multiplying out, though in some cases the derivatives are just ugly enough that either approach seems reasonable.
The following is a way of getting the first few terms easily. The general term requires us to find a generating function, and is arguably no easier than Jack's approach, but nevertheless workable and arguably no more difficult.
Let $$f(z)=\frac {\log(z)}{(z-1)(z-5)}.$$ The Laurent series about 5 can be written $$f(z)=\sum a_n (z-5)^{n}.$$ Clearing denominators and writing $z-1=4+(z-5)$ we have $$\log(z)=\sum (4a_n+a_{n-1})(z-5)^{n+1}.$$ On the other hand we easily find that $$\log(z)= \ln(5)+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k 5^k} (z-5)^k.$$ The pole of $f$ is simple, so $a_n=0$ for $n<-1$. Comparing coefficients gives the recurrence relation $$a_{-1}=\ln(5)/4$$ $$4a_{n-1}+a_{n-2}= \frac{(-1)^{n+1}}{n 5^n}, \ n \geq 1.$$ If all you want is the first few terms, then you can solve by hand. For lots of terms, or all the terms, we have to solve the relation. In this case we use a generating function.
Define the generating function $A(x)=\sum_{n=-1}^\infty a_n x^{n+1}.$ The recurrence relation gives $$(x+4)A(x)=\ln(5)+\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n 5^n}x^n.$$ Converting $1/(x+4)$ into its series about 0 we conclude $$A(x)= \frac{\ln 5}{4}+\left({1\over{20}}-{\ln5 \over 16}\right)x+\sum_{m+n=k}^\infty \frac{(-1)^{k+1}}{n 5^n 4^{m+1}} x^{k},$$ which gives us our coefficients. Indeed it's pretty much exactly the Laurent series, since the series is $A(z-5)/(z-5)$, and with a slight adjustment to the definition we would get it precisely.
