Laurent expansion - Faster technique

Question

I'm currently preparing for an exam in complex analysis. There is a type of exercise, where I need to compute Laurent expansions about different places. However, my current technique does consume quite a lot of my available time, so my question is if anyone knows a faster technique.

I will show my technique on an example. Here's the task:

Compute all isolated singularities of the following function and specify their type. Also give the two terms with the smallest degree of the Laurent expansion about those singularities and their residuum. $$f(z) = \frac{z}{(z - 1)^2(z - 2)}$$

And now to my technique. The function has singularities where $(z - 1)^2(z - 2) = 0$, this is at $z_1 = 1$ and $z_2 = 2$. Thus, the singularities are also isolated. $$\lim_{z \to 1} |f(z)| = \frac{1}{\lim_{z \to 1} |(z - 1)^2(z - 2)|} = \infty\\ \lim_{z \to 2} |f(z)| = \frac{2}{\lim_{z \to 2} |(z - 1)^2(z - 2)|} = \infty$$ So, both singularities are poles.

For a Laurent expansion we would like the function to only consist of addends, so we do a partial fraction decomposition. $$\frac{z}{(z - 1)^2(z - 2)} = \frac{A}{z - 1} + \frac{B}{(z - 1)^2} + \frac{C}{z - 2} \Leftrightarrow\\ z = A(z - 1)(z - 2) + B(z - 2) + C(z - 1)^2 \Leftrightarrow\\ z = Az^2 - 3Az + 2A + Bz - 2B + Cz^2 - 2Cz + C \Leftrightarrow\\ z = z^2(A + C) + z(-3A + B - 2C) + 1(2A - 2B + C)$$ Okay, now we solve the equation system. $$\begin{pmatrix} 1 & 0 & 1 & 0\\ -3 & 1 &-2 & 1\\ 2 &-2 & 1 & 0\\ \end{pmatrix} \Leftrightarrow \begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 1 & 1\\ 0 &-2 & 1 & 0\\ \end{pmatrix} \Leftrightarrow \begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 2\\ \end{pmatrix} \Leftrightarrow\\ \begin{pmatrix} 1 & 0 & 0 & -2\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 2\\ \end{pmatrix} \Leftrightarrow \begin{pmatrix} 1 & 0 & 0 & -2\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2\\ \end{pmatrix}$$ With that we follow: $$f(z) = -\frac{2}{z - 1} - \frac{1}{(z - 1)^2} + \frac{2}{z - 2}$$

Let's first compute the Laurent expansion about $z_1 = 1$, we substitute with $w = z - 1 \Leftrightarrow z = w + 1$. $$f(z) = -w^{-2} - 2w^{-1} + \frac{2}{z - 2}$$

Now we need to adjust the last term in a form like $a_j w^j$. We do so by using the geometric series. $$\frac{2}{z - 2} = \frac{2}{w + 1 - 2} = 2 \cdot \frac{1}{w - 1} = -2 \cdot \frac{1}{1 - w}$$ As we expand about $z_1 = 1$, thus $|w| = |z - 1| < |1|$, the standard geometric series can be used. $$-2 \cdot \frac{1}{1 - w} = -2 \sum\limits_{n = 0}^\infty w^n$$

Back to the function, here's the Laurent expansion about $z_1 = 1$. $$f(z) = -w^{-2} - 2w^{-1} + -2 - 2w - 2w^2 - \ldots$$

The first two terms are $-w^{-2}$ and $-2w^{-1}$. The residue is the factor of $w^{-1}$, which is $-2$.

Last we would need to expand around $z_2 = 2$ but this works analogously, so I will not write it down here. There I brought $\frac{-2}{z - 1}$ and $\frac{-1}{(z - 1)^2}$ into a form with $w$ and then used the geometric series again. However, the later term is difficult because it's to the power of $2$. But it can be seen as the derivative of a similar geometric series. As we have compact convergence for the geometric series in this case, we can extract the derivative and so on.

Okay, so that was my technique. Very time consuming is the step where I need to compute the partial fraction decomposition. And after that, where I need to compute the exact series representation of nearly every addend of $f$. And this process repeats for every singularity. I wonder if there are faster ways as I only need the two terms with the smallest degree.

Thanks for your help :)

Answer 1

Here are some hints for faster calculation.

Task: Compute all isolated singularities of the following function and specify their type.

\begin{align*} f(z)=\frac{z}{(z-1)^2(z-2)}\tag{1} \end{align*}

Answer: From the representation (1) of $f(z)$ we see it has exactly two isolated singularities. One of them is a pole of second order at $z=1$ and the the other is a simple pole at $z=2$.

Note: Since you can deduce the isolated singularities directly from $f$ there is effectively nothing to calculate.

Task: Also give the two terms with the smallest degree of the Laurent expansion about those singularities and their residuum.

Answer: We start with the partial fraction decomposition \begin{align*} f(z)=\frac{z}{(z-1)^2(z-2)}=\frac{A}{z-1}+\frac{B}{(z-1)^2}+\frac{C}{z-2}\tag{2} \end{align*}

We can calculate the coefficients of terms with simple poles easily.

Answer: To obtain $C$ we multiply (2) with $z-2$ and put $z=2$ in the remaining expression.

\begin{align*} C=\left.f(z)(z-2)\right|_{z=2}=\left.\frac{z}{(z-1)^2}\right|_{z=2}=2 \end{align*}

Similarly we obtain $B$ by multplication with $(z-1)^2$ and putting $z=1$ in the remaining expression. \begin{align*} B=\left.f(z)(z-1)^2\right|_{z=1}=\left.\frac{z}{z-2}\right|_{z=1}=-1 \end{align*}

Putting $B$ and $C$ in (2) and isolating $A$ we obtain \begin{align*} A&=[z^0]\left(\frac{z}{(z-1)^2(z-2)}-\frac{1}{(z-1)^2}-\frac{2}{z-2}\right)\tag{3}\\ &=[z^0]\left(-\frac{1}{(z-1)^2}-\frac{1}{1-\frac{z}{2}}\right)\\ &=-1-1\\ &=-2 \end{align*}

Note: In order to calculate (3) we observe that the LHS is the constant $A$, so the RHS has also to be a constant. Let's consider the series expansion of the RHS around $z=0$ and let's check for the constant terms of the series.

The first term $\frac{z}{(z-1)^2(z-2)}$ is $z$ multiplied by the product of two binomial series, so there is no contribution to the constant term of the RHS, since the resulting series starts with $z$ having smallest power $1$.

The second term $[z^0]\frac{1}{(z-1)^2}=1$ since the binomial series starts with $1$ and also the the third term has $[z^0]\frac{1}{1-\frac{z}{2}}=1$ since the binomial series start with $1$.

We conclude the contributions to the constant term is \begin{align*} A=0-1-1=-2 \end{align*} as shown in (3).

Answer: We observe the partial fraction decomposition is \begin{align*} f(z)=\frac{z}{(z-1)^2(z-2)}=-\frac{2}{z-1}-\frac{1}{(z-1)^2}+\frac{2}{z-2}\tag{4} \end{align*}

and we are nearly finished. Note that the task is not to specifiy the Laurent series \begin{align*} \sum_{n=-\infty}^{\infty}a_n z^n \end{align*}

We only need to find the two terms with smallest degree of the Laurent series. But, if you look at the presentation (4) the first two terms are already the Laurent expansion of $f$ around $z=1$ while the third part is a Taylor series around $z=1$ and does nothing contribute to $a_{-2}$ or $a_{-1}$.

On the other hand, the third part is already the Laurent expansion around $z=2$ while the first two parts are just a Taylor series around $z=2$ and contribute nothing to the principal part. Here are the coefficients of the two terms with smallest degree $a_{-1}=2$ and $a_{0}$.

Therefore we can answer:

Answer: From the partial fraction decomposition (4) we see the two terms with the smallest degree around the poles are \begin{align*} &z=1\qquad\rightarrow\qquad a_{-2}=-1,& & a_{-1}=-2\\ &z=2\qquad\rightarrow\qquad a_{-1}=2,& & a_{0}=\ldots \end{align*}

It remains the calculation of $a_0$ when expanding $f$ around $z=2$.