# Laurent series for $\frac{1}{z(z+3)(z-1)^2}$

Find the Laurent Series for

$$\frac{1}{z(z+3)(z-1)^2}$$ in $1 < |z-1| < 4$

So I did the partial fraction decomposition which yields:

$$\frac{1}{4(-1+z)^2} - \frac{5}{16(-1+z)} + \frac{1}{3z} - \frac{1}{48(3+z)}$$

Can anyone help me finish this problem?

You are on the right track. Partial fraction expansion reveals

$$\frac{1}{z(z+3)(z-1)^2}=\frac{1/3}{z}-\frac{1/48}{z+3}-\frac{5/16}{z-1}+\frac{1/4}{(z-1)^2}\tag1$$

The last two terms on the right-hand side of $(1)$ constitute part of the Laurent expansion in the annulus $1<|z-1|<4$.

We now expand the first term, $\frac{1/3}{z}$, for $|z-1|>1$ as

\begin{align} \frac{1/3}{z}&=\frac{1/3}{z-1}\left(\frac{1}{1+\frac{1}{z-1}}\right)\\\\ &=\frac13\sum_{n=0}^\infty (-1)^n (z-1)^{-(n+1)}\tag2 \end{align}

Similarly, we expand the term $\frac{1/48}{z+3}$ for $|z-1|<4$ as

\begin{align} \frac{1/48}{z+3}&=\frac{1/48}{4+(z-1)}\\\\ &=\frac1{192}\sum_{n=0}^\infty (-1)^n\left(\frac{z-1}{4}\right)^n\tag3 \end{align}

Now, assemble the expansion using $(2)$ and $(3)$ in $(1)$.

• I can do themn all except the $\frac{1/4}{(z-1)^2}$ what would be the geometric form of that? Commented May 4, 2018 at 14:32
• That term is already part of the Laurent expansion in powers of $\frac1{z-1}$. Commented May 4, 2018 at 14:33
• I got $-1/4 \sum_{i=0}^{\infty} (-1)^n (4/(z-1))^{-(n+1)}$ Commented May 4, 2018 at 14:37
• Thank you for the help! Commented May 4, 2018 at 19:56
• You're welcome. My pleasure. Commented May 4, 2018 at 22:48