Lecture 11: Laurent Series for Complex Functions

Laurent series help in analyzing the nature of singularities of a function. By examining the Laurent series expansion, you can determine whether a singularity is removable, a pole, or an essential singularity. In Taylor series, we write an analytic function inside a disk as a series $f(z) =\sum_{j=0}^{j=+\infty} c_j (z-z_0)^j.$ However, in Laurent series, we write an analytic function in an annulus as a series $f(z) =\sum_{j=-\infty}^{j=+\infty} C_j (z-z_0)^j.$

Laurent series

Annulus centered at z0 between radii R1 and R2, with point z and a radial crosscut. — Contour deformation in the proof of Theorem Theorem 1.

Theorem 1 (Laurent Series). Let $R_2>R_1>0$ be given and $f(z)$ be analytic in the annulus $\{z\in\mathbb C\mid R_1\le |z-z_0|\le R_2\}$ . Then we can write $f(z)$ into the series $\label{EQ:Laurent} f(z) =\sum_{j=-\infty}^{j=+\infty} C_j (z-z_0)^j$ in the region $\{z\in\mathbb C\mid R_1<R_a\le |z-z_0|\le R_b< R_2\}$ with $\label{EQ:Laurent Coeff} C_j=\dfrac{1}{2\pi i} \int_{C} \dfrac{f(z)}{(z-z_0)^{j+1}}dz,$ $C$ being any simple closed contour in the region of analyticity that encloses the inner circle $|z-z_0|=R_1$ . Moreover, the series converges uniformly in $\{z\in\mathbb C\mid R_1<R_a\le |z-z_0|\le R_b< R_2\}$ .

Proof. Let $\mathcal B_j=\{z\in\mathbb C\mid |z-z_0|=R_j\}$ , $j=1, 2$ . By the Method of Contour Deformation (See Figure 1) and Cauchy’s Integral Formula, we have that $\label{EQ:Laurent Proof1} f(z)=\dfrac{1}{2\pi i}\int_{\mathcal B_2} \dfrac{f(\xi)}{\xi-z} d\xi-\dfrac{1}{2\pi i}\int_{\mathcal B_1} \dfrac{f(\xi)}{\xi-z} d\xi \,.$ When $\xi\in \mathcal B_1$ , we have $|z-z_0|>|\xi-z_0|$ , and therefore $\dfrac{1}{\xi-z}=\dfrac{1}{\xi-z_0-(z-z_0)}=\dfrac{1}{(\xi-z_0)(1-\frac{z-z_0}{\xi-z_0})}=\dfrac{1}{(\xi-z_0)}\sum_{k=0}^\infty (\frac{z-z_0}{\xi-z_0})^k\,.$ When $\xi\in \mathcal B_2$ , we have $|z-z_0|<|\xi-z_0|$ , and therefore $\dfrac{1}{\xi-z}=\dfrac{1}{\xi-z_0-(z-z_0)}=\dfrac{1}{(z-z_0)(1-\frac{\xi-z_0}{z-z_0})}=\dfrac{1}{(z-z_0)}\sum_{k=0}^\infty (\frac{\xi-z_0}{z-z_0})^k\,.$ Plugging the above into [EQ:Laurent Proof1] then gives $f(z) = \sum_{k=0}^\infty A_k(z-z_0)^k+\sum_{k=0}^\infty B_k(z-z_0)^{-k-1}$ with $A_k=\dfrac{1}{2\pi i}\int_{\mathcal B_2} \dfrac{f(\xi)}{(\xi-z_0)^{k+1}} d\xi,\ \ \ B_k=\dfrac{1}{2\pi i}\int_{\mathcal B_1} f(\xi)(\xi-z_0)^{k} d\xi\,.$ The next step is to make a change of index $-(k+1)\to k$ in the second summation to have $f(z) = \sum_{k=0}^\infty A_k(z-z_0)^k+\sum_{k=-\infty}^{k=-1} B_{-k-1}(z-z_0)^k$ with $A_k=\dfrac{1}{2\pi i}\int_{\mathcal B_2} \dfrac{f(\xi)}{(\xi-z_0)^{k+1}} d\xi,\ \ \ B_{-k-1}=\dfrac{1}{2\pi i}\int_{\mathcal B_1} \dfrac{f(\xi)}{(\xi-z_0)^{k+1}} d\xi\,.$ The last step is to realize that due to analyticity of $f(z)$ in the annulus, the integrals over $\mathcal B_1$ and $\mathcal B_2$ can be replaced with the same integrals over the contour $C$ in the theorem. That is, if we define $C_k=\dfrac{1}{2\pi i}\int_{C} \dfrac{f(\xi)}{(\xi-z_0)^{k+1}} d\xi\,,$ then $C_k=A_k$ when $k\ge 0$ and $C_k=B_{-k-1}$ when $k\le -1$ . The proof is complete. ◻

Definition 2. The series [EQ:Laurent] we defined in Theorem Theorem 1 is called the Laurent series of $f(z)$ in $\{z\in\mathbb C\mid R_1\le |z-z_0|\le R_2\}$ . The coefficient $C_{-1}$ is called the residue of the function $f(z)$ . The sum of the negative powers in the Laurent series is called the principal part of the function $f(z)$ .

Remark 3. Due to the uniqueness of a Taylor series within its radius of convergence, we would expect that if the function $f(z)$ is analytic inside the big disk in Theorem Theorem 1, $\{z\in\mathbb C\mid |z-z_0|\le R_2\}$ , the coefficients in the Laurent series should be $C_j=0$ if $j\le -1$ and $C_j=\dfrac{f^{(j)}(z_0)}{j!}$ . Indeed, this is true. In other words, the Laurent series degenerates to the Taylor series of $f$ in this case. To see this, we observe that when $j\le -1$ , $\frac{f(z)}{(z-z_0)^{j+1}}$ is analytic (since $f(z)$ and $\frac{1}{(z-z_0)^{j+1}}$ are both analytic). Therefore, the contour integral vanishes. When $j\ge 0$ , by Cauchy’s Derivative Formula, the integral gives $C_j=\dfrac{f^{(j)}(z_0)}{j!}$ .

Example 4. Let $f(z)=\dfrac{1}{1-z}$ . Find the Laurent series of $f(z)$ for $|z|>1$ .

Let $C$ be simple closed contour enclosing the circle $|z|=1$ . We compute the coefficient of the Laurent series as follows. (i) When $j= 0$ , we have $C_0 = \dfrac{1}{2\pi i} \int_{C} \dfrac{f(z)}{z}dz = \dfrac{1}{2\pi i} \int_{C} \dfrac{1}{(1-z)z}dz =0.$ (ii) When $j\ge 1$ , we have $C_j = \dfrac{1}{2\pi i} \int_{C} \dfrac{f(z)}{z^{j+1}}dz = \dfrac{1}{2\pi i } \int_{C} \dfrac{1}{(1-z)z^{j+1}}dz = \dfrac{1}{2\pi i} \int_{C} \dfrac{1}{z^j} \left(\dfrac{1}{(1-z)}+\dfrac{1}{z}\right) dz = - C_{j-1},$ which gives us $C_j=0$ .

(iii) When $j\le -1$ , we have $C_j = \dfrac{1}{2\pi i} \int_{C} \dfrac{f(z)}{z^{j+1}}dz = -\dfrac{1}{2\pi i} \int_{C} \dfrac{z^{-(j+1)}}{(z-1)}dz =-1.$

Therefore, the Laurent series is $\dfrac{1}{1-z} =-\sum_{j=-\infty}^{-1} z^j=\sum_{j=1}^\infty (\dfrac{1}{z})^j =-\dfrac{1}{z}-\dfrac{1}{z^2}-\cdots .$

Laurent series via Taylor series

To find the Laurent series of a given function in a given annulus, we can start from the definition and perform necessary computation to find the coefficients. It is very often more convenient to find Laurent series directly from known Taylor series, using the $z\to 1/z$ transform. Here are a few examples.

Example 5. Let $f(z)=\dfrac{1}{1-z}$ . Find the Laurent series of $f(z)$ for $|z|>1$ .

We know that we can expand $f(z)$ in Taylor series when $|z|<1$ . To find the series for $|z|>1$ , we notice that $|z|>1$ is equivalent to $1/|z|<1$ . Therefore, we can find the Laurent series for $|z|>1$ by the following calculation: $f(z)=\dfrac{1}{1-z}=\dfrac{1}{z(1/z-1)}=-\dfrac{1}{z}\dfrac{1}{1-\frac{1}{z}} = -\dfrac{1}{z}\sum_{k=0}^\infty (\dfrac{1}{z})^k = -\dfrac{1}{z}-\dfrac{1}{z^2}-\cdots .$ This is exactly what we obtained in the previous example.

Example 6. Let $f(z)=\dfrac{1}{(z-3)(z-6)}$ . Find the Laurent series of $f(z)$ for $3< |z|<6$ .

We observe that $f(z)=\dfrac{1}{(z-3)(z-6)} = \dfrac{1}{3}[\dfrac{1}{(z-6)}-\dfrac{1}{(z-3)}]=\dfrac{1}{3}[-\dfrac{1}{6(1-z/6)}-\dfrac{1}{z(1-3/z)}]$ since $3<|z|<6$ , we have that $3/|z|<1$ and $|z|/6<1$ . Therefore $f(z)=\dfrac{1}{3}[-\dfrac{1}{6(1-z/6)}-\dfrac{1}{z(1-3/z)}] =\dfrac{1}{3}[-\dfrac{1}{6}\sum_{k=0}^\infty (z/6)^k-\dfrac{1}{z}\sum_{k=0}^\infty (3/z)^k]$ We can then rearrange a little bit to have $f(z)=\sum_{k=0}^\infty (-2) 6^{-(k+2)} z^k+\sum_{k=-\infty}^{-1} (-1) 3^{-k-1} z^k .$

Guided review and additional examples

Learning goals

Separate a Laurent series into analytic and principal parts.
Choose an expansion that matches the specified annulus.
Read the residue from the coefficient of $(z-z_0)^{-1}$ .
Use Laurent series to classify isolated singularities.

Concept connection: the region selects the series

Unlike a Taylor series, a Laurent expansion depends on an annulus. The same function can have different valid expansions about the same center because a geometric-series factor must be expanded using either a small quantity such as $z$ or a small quantity such as $1/z$ .

Worked example 1: two annuli for one function

Consider

$f(z)=\frac1{z(z-1)}.$

For $0<|z|<1$ ,

$f(z)=-\frac1{z(1-z)} =-\sum_{n=0}^{\infty}z^{n-1} =-\frac1z-1-z-z^2-\cdots.$

For $|z|>1$ , expand in powers of $1/z$ instead:

$f(z)=\frac1{z^2(1-1/z)} =\sum_{n=0}^{\infty}z^{-n-2} =\frac1{z^2}+\frac1{z^3}+\cdots.$

The first annulus has residue $-1$ at zero; the outer expansion has no $z^{-1}$ term because it describes behavior outside both singularities.

Worked example 2: read a residue from a product

Expand $e^z/z^2$ about zero:

$\frac{e^z}{z^2}=\frac1{z^2}+\frac1z+\frac1{2!}+\frac{z}{3!}+\cdots.$

The principal part is $z^{-2}+z^{-1}$ , so zero is a pole of order $2$ . The coefficient of $z^{-1}$ is $1$ , which is the residue at zero.

Check your understanding

Why is it not enough to find a formal series without stating its annulus of convergence?

Show the answer

Different geometric-series manipulations yield different coefficients and are valid in different regions. The annulus tells where the algebraic expansion actually converges and which singularities lie on either side of the region.