Laurent Series of $\frac{1}{(z^2+1)(z-1)^2}$

Question

Find the Laurent series of $\dfrac{1}{(z^2+1)(z-1)^2}$ and use this to identify singularities and their residues.

$$\dfrac{1}{(z^2+1)(z-1)^2}=\dfrac{i}{2}\dfrac{1}{z^2+1}+\dfrac{i-1}{2}\dfrac{1}{z-1}+\dfrac{1}{2}\dfrac{1}{(z-1)^2}$$ by partial fractions. Trying to get a Laurent series by seeing if a Taylor series (via geometric series) can be used doesn't appear to have gotten much result for me.

The issue for me comes with that there doesn't immediately appear to be an annulus for this fraction that makes sense, as $|i| = 1$, so limiting it to a circle centred at $0$ such that one is outside and one is inside doesn't appear possible.

Once I get the Laurent series, I can go after defining the singularities at $z=i$ [EDIT: AND $z=-i$, wow how did I miss that?] and $z=1$ and their residues, so I just need help with that first hurdle.

Answer 1

Edited to clarify motivation - see GEdgar's comment below.

I think the issue here is that the question is not worded clearly.

Your function has three singularities: at the points $z = i$, $z = -i$ and $z = 1$. Your objective is to characterise each of these singularities, and to determine the residues at each of these singularities.

There are many different annuli inside $\mathbb C \setminus \{ i, -i, 1 \}$. For each choice of annulus, there will be a different Laurent expansion for your function that is valid on that annulus.

However, your task is to characterise the singularities at $z = i$, $z = -i$ and $z = 1$. The three Laurent series that are most directly relevant to your task are the following:

• The Laurent expansion in powers of $(z-1)$, valid on the punctured disk $\{ z \in \mathbb C : 0 < |z - 1 | < \sqrt{2} \}$. This is directly relevant for characterising the singularity at $z = 1$.
• The Laurent expansion in powers of $(z-i)$, valid on the punctured disk $\{ z \in \mathbb C : 0 < |z - i | < \sqrt{2} \}$. This is directly relevant for characterising the singularity at $z = i$.
• The Laurent expansion in powers of $(z+i)$, valid on the punctured disk $\{ z \in \mathbb C : 0 < |z + i | < \sqrt{2} \}$. This is directly relevant for characterising the singularity at $z = -i$.

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Why don't we work through one example: the Laurent expansion in powers of $(z-1)$ on the punctured disk $\{ z \in \mathbb C : 0 < |z - 1 | < \sqrt{2} \}$.

Notice that $1 / (z^2 + 1)$ is holomorphic on this punctured disk. (The singularities of $1 / (z^2 + 1)$ are situated at $z = i$ and $z = -i$, which lie outside of this punctured disk.)

So on this punctured disk, we can write $1 / (z^2 + 1)$ as a Taylor series in $(z - 1)$. \begin{align}\frac{1}{z^2 + 1} & = \frac 1 2 - \frac 1 2 (z - 1) + \frac 1 4 (z - 1)^2 + \dots \end{align} I calculated this using Wolfram Alpha, but you can compute this by hand by. For example, you can evaluate the all the $n$th order derivatives of $\frac{1}{z^2 + 1}$ at $z = 1$. Alternatively, you can write $\frac{1}{z^2 + 1}$ as $\frac{1}{(z - 1) + (1 - i)} \times \frac{1}{(z - 1) + (1 + i)}$, then expand both of the factors in this expression as geometric series.

Therefore, \begin{align}\frac{1}{(z^2 + 1)(z-1)^2} & = \frac 1 2 (z-1)^{-2}- \frac 1 2 (z - 1)^{-1} + \frac 1 4 + \dots \end{align} is the Laurent series expansion for $\frac{1}{(z^2 + 1)(z-1)^2}$ on the punctured disk around $z = 1$.

Thus the point $z = 1$ is a double pole, and the residue at $z = 1$ is $ - \frac 1 2$.

You can characterise the other two singularities in a similar way.