Laurent Series Expansion about $z=1$

Question

If $p(z)=(z-1)^3(z-2)^4$ then how would I find the Laurent Expansion about $z=1$ for $1/p(z)$? I am having trouble computing expansions when it's not about $z=0$ so could someone walk me through the steps for this kind of question? I am trying to find the first three coefficients of the expansion.

Answer 1

$$\frac1{(z-1)^3(z-2)^4}=\frac1{(z-1)^3}\cdot\frac1{(1-(z-1))^4}=$$

$$=\frac1{(z-1)^3}\left(1+(z-1)+(z-1)^2+\ldots\right)^4=\frac1{(z-1)^3}\left(1+2(z-1)+3(z-1)^2+\ldots\right)^2=$$

$$=\frac1{(z-1)^3}\left(1+4(z-1)+10(z-10)^2+\ldots\right)=\frac1{(z-1)^3}+\frac4{(z-1)^2}+\frac{10}{z-1}+\ldots$$

If you need more elements then you develop more the temr to the fourth power. Of course, the above is true for $\;|z-1|<1\;$, which seems to be a safe assumption as you're interested in th Laurent series around $\;z=1\;$

Answer 2

here is way to do this with a change of variable $z = 1 + u.$ we have $$\begin{align} \frac 1{(z-1)^3(z-2)^4} = \frac 1{u^3(-1+u)^4} &= \frac 1 {u^3}(-1+u)^{-4}\\&=\frac 1{u^3}\left(1 - 4u + \frac{4 \cdot 5}{1 \cdot 2}u^2 - \frac{ 4 \cdot 5 \cdot 6}{1 \cdot 2 \cdot 3 }u^3 + \cdots\right)\end{align} $$