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# 8.7: Laurent Series

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##### Theorem $$\PageIndex{1}$$ Laurent series

Suppose that $$f(z)$$ is analytic on the annulus

$A: r_1 < |z - z_0| < r_2.$

Then $$f(z)$$ can be expressed as a series

$f(z) = \sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n} + \sum_{n = 0}^{\infty} a_n (z - z_0)^n.$

The coefficients have the formulus

$\begin{array} {l} {a_n = \dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(w)}{(w - z_0)^{n + 1}}\ dw} \\ {b_n = \dfrac{1}{2\pi i} \int_{\gamma} f(w) (w - z_0)^{n - 1}\ dw} \end{array}$

where $$\gamma$$ is any circle $$|w - z_0| = r$$ inside the annulus, i.e. $$r_1 < r < r_2$$.

Furthermore

• The series $$\sum_{n = 0}^{\infty} a_n (z - z_0)^n$$ converges to an analytic function for $$|z - z_0| < r_2$$.
• The series $$\sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n}$$ converges to an analytic function for $$|z - z_0| > r_1$$.
• Together, the series both converge on the annulus $$A$$ where $$f$$ is analytic.

The proof is given below. First we define a few terms.

##### Definition: Laurent Series

The entire series is called the Laurent series for $$f$$ around $$z_0$$. The series

$\sum_{n = 0}^{\infty} a_n (z - z_0)^n$

is called the analytic or regular part of the Laurent series. The series

$\sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n}$

is called the singular or principal part of the Laurent series.

##### Note

Since $$f(z)$$ may not be analytic (or even defined) at $$z_0$$ we don’t have any formulas for the coefficients using derivatives.

Proof

(Laurent series). Choose a point $$z$$ in $$A$$. Now set circles $$C_1$$ and $$C_3$$ close enough to the boundary that $$z$$ is inside $$C_1 + C_2 - C_3 - C_2$$ as shown. Since this curve and its interior are contained in $$A$$, Cauchy’s integral formula says

$f(z) = \dfrac{1}{2\pi i} \int_{C_1 + C_2 - C_3 - C_2} \dfrac{f(w)}{w - z}\ dw$

The integrals over $$C_2$$ cancel, so we have

$f(z) = \dfrac{1}{2\pi i} \int_{C_1 - C_3} \dfrac{f(w)}{w - z}\ dw.$

Next, we divide this into two pieces and use our trick of converting to a geometric series. The calculuations are just like the proof of Taylor’s theorem. On $$C_1$$ we have

$\dfrac{|z - z_0|}{|w - z_0|} < 1,$

so

$\begin{cases} {rcl} {\dfrac{1}{2\pi i} \int_{C_1} \dfrac{f(w)}{w - z}\ dw} & = & {\dfrac{1}{2\pi i} \int_{C_1} \dfrac{f(w)}{w - z_0} \cdot \dfrac{1}{(1 - \dfrac{z - z_0}{w - z_0})} \ dw} \\ {} & = & {\dfrac{1}{2\pi i} \int_{C_1} \sum_{n = 0}^{\infty} \dfrac{f(w)}{(w - z_0)^{n + 1}} (z - z_0)^n \ dw} \\ {} & = & {\sum_{n = 0}^{\infty} (\dfrac{1}{2\pi i} \int_{C_1} \dfrac{f(w)}{(w - z_0)^{n + 1}} \ dw) (z - z_0)^n} \\ {} & = & {\sum_{n = 0}^{\infty} a_n (z - z_0)^n.} \end{cases}$

Here $$a_n$$ is defined by the integral formula given in the statement of the theorem. Examining the above argument we see that the only requirement on $$z$$ is that $$|z - z_0| < r_2$$. So, this series converges for all such $$z$$.

Similarly on $$C_3$$ we have

$\dfrac{|w - z_0|}{|z - z_0|} = 1.$

so

$\begin{array} {rcl} {\dfrac{1}{2\pi i} \int_{C_3} \dfrac{f(w)}{w - z} dw} & = & {\dfrac{1}{2\pi i} \int_{C_3} -\dfrac{f(w)}{z - z_0} \cdot \dfrac{1}{(1 - \dfrac{w - z_0}{z - z_0})} \ dw} \\ {} & = & {-\dfrac{1}{2\pi i} \int_{C_3} \sum_{n = 0}^{\infty} f(w) \dfrac{(w - z_0)^n}{(z - z_0)^{n + 1}} \ dw} \\ {} & = & {-\dfrac{1}{2\pi i} \sum_{n = 0}^{\infty} (\int_{C_1} f(w) (w - z_0)^n \ dw) (z - z_0)^{-n - 1}} \\ {} & = & {-\sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n}.} \end{array}$

In the last equality we changed the indexing to match the indexing in the statement of the theorem. Here $$b_n$$ is defined by the integral formula given in the statement of the theorem. Examining the above argument we see that the only requirement on $$z$$ is that $$|z - z_0| > r_1$$. So, this series converges for all such $$z$$.

Combining these two formulas we have

$f(z) = \dfrac{1}{2\pi i} \int_{C_1 - C_3} \dfrac{f(w)}{w - z}\ dw = \sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n} + \sum_{n = 0}^{\infty} a_n (z - z_0)^n$

The last thing to note is that the integrals defining $$a_n$$ and $$b_n$$ do not depend on the exact radius of the circle of integration. Any circle inside $$A$$ will produce the same values. We have proved all the statements in the theorem on Laurent series. QED

## Examples of Laurent Series

In general, the integral formulas are not a practical way of computing the Laurent coefficients. Instead we use various algebraic tricks. Even better, as we shall see, is the fact that often we don’t really need all the coefficients and we will develop more techniques to compute those that we do need.

##### Example $$\PageIndex{1}$$

Find the Laurent series for

$f(z) = \dfrac{z + 1}{z} \nonumber$

around $$z_0 = 0$$. Give the region where it is valid.

Solution

$f(z) = 1 + \dfrac{1}{z}. \nonumber$

This is a Laurent series, valid on the infinite region $$0 < |z| < \infty$$.

##### Example $$\PageIndex{2}$$

Find the Laurent series for

$f(z) = \dfrac{z}{z^2 + 1} \nonumber$

around $$z_0 = i$$. Give the region where your answer is valid. Identify the singular (principal) part.

Solution

Using partial fractions we have

$f(z) = \dfrac{1}{2} \cdot \dfrac{1}{z - i} + \dfrac{1}{2} \cdot \dfrac{1}{z + i}. \nonumber$

Since $$\dfrac{1}{z + i}$$ is analytic at $$z = i$$ it has a Taylor series expansion. We find it using geometric series.

$\dfrac{1}{z + i} = \dfrac{1}{2i} \cdot \dfrac{1}{1 + (z - i)/(2i)} = \dfrac{1}{2i} \sum_{n = 0}^{\infty} (-\dfrac{z - i}{2i})^n \nonumber$

So the Laurent series is

$f(z) = \dfrac{1}{2} \cdot \dfrac{1}{z - i} + \dfrac{1}{4i} \sum_{n = 0}^{\infty} (-\dfrac{z -i}{2i})^n \nonumber$

The singular (principal) part is given by the first term. The region of convergence is $$0 < |z - i| < 2$$.

##### Note

We could have looked at $$f(z)$$ on the region $$2< |z - i|< \infty$$. This would have produced a different Laurent series. We discuss this further in an upcoming example.

##### Example $$\PageIndex{3}$$

Compute the Laurent series for

$f(z) = \dfrac{z + 1}{z^3 (z^2 + 1)} \nonumber$

on the region $$A$$: $$0 < |z| < 1$$ centered at $$z = 0$$.

Solution

This function has isolated singularities at $$z = 0, \pm i$$. Therefore it is analytic on the region $$A$$.

At $$z = 0$$ we have

$f(z) = \dfrac{1}{z^3} (1 + z)(1 - z^2 + z^4 - z^6 +\ ...). \nonumber$

Multiplying this out we get

$f(z) = \dfrac{1}{z^3} + \dfrac{1}{z^2} - \dfrac{1}{z} - 1 + z + z^2 - z^3 -\ ... \nonumber$

The following example shows that the Laurent series depends on the region under consideration.

##### Example $$\PageIndex{4}$$

Find the Laurent series around $$z = 0$$ for $$f(z) = \dfrac{1}{z(z - 1)}$$ in each of the following regions:

$\begin{array} {rl} {\text{(i)}} & {\text{the region } A_1: 0 < |z| < 1} \\ {\text{(ii)}} & {\text{the region } A_2: 1 < |z| < \infty.} \end{array} \nonumber$

Solution

For $$\text{(i)}$$

$f(z) = -\dfrac{1}{z} \cdot \dfrac{1}{1 - z} = -\dfrac{1}{z} (1 + z + z^2 +\ ...) = -\dfrac{1}{z} - 1 - z - z^2 - \ ... \nonumber$

For $$\text{(ii)}$$: Since the usual geometric series for $$1/(1 - z)$$ does not converge on $$A_2$$ we need a different form,

$f(z) = \dfrac{1}{z} \cdot \dfrac{1}{z (1 - 1/z)} = \dfrac{1}{z^2} (1 + \dfrac{1}{z} + \dfrac{1}{z^2} +\ ...) \nonumber$

Since $$|1/z| < 1$$ on $$A_2$$ our use of the geometric series is justified.

One lesson from this example is that the Laurent series depends on the region as well as the formula for the function.

8.7: Laurent Series is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.