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A series with negative powers, valid on an annulus rather than a disc, and unique once the annulus is chosen — so one function has as many expansions about a point as there are rings between its singularities, and the coefficient on the power minus one is the only one an integral ever sees.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state Laurent's theorem, find the annuli a function has expansions on, expand a quotient as a geometric series the right way round for a given annulus, obtain a series with infinitely many negative powers by substitution, and read the residue off the coefficient of the power minus one.
Taylor series, with the radius of convergence read off the distance to the nearest singularity. This lesson expands about a point where the function itself misbehaves, so no disc will do and the region becomes a ring.
An annulus about $z_0$ is a set $\rho_1 < |z - z_0| < \rho_2$, where $\rho_1$ may be $0$ and $\rho_2$ may be unbounded. A Laurent series about $z_0$ is $\sum_{n=-\infty}^{\infty} a_n(z - z_0)^{n}$, with negative powers allowed; its principal part is the terms of negative index. The residue is the coefficient $a_{-1}$. A singularity is isolated when the function is analytic on some punctured disc about it.
Laurent's theorem. If $f$ is analytic on the annulus $\rho_1 < |z - z_0| < \rho_2$, then on that annulus
$$f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^{n}, \qquad a_n = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{n+1}}\,dz,$$
with $C$ any circle in the annulus, and the expansion is unique.
The coefficient formula is the one from Taylor's theorem, unchanged — but now $n$ is allowed to be negative, and the integral still makes sense because $f$ need not be analytic inside $C$, only on the annulus containing it.
The region is part of the answer. A function has one Taylor series about a point and as many Laurent series as there are annuli. $\dfrac{1}{(z - 1)(z - 2)}$ about $0$ has three: on $|z| < 1$, on $1 < |z| < 2$, and on $|z| > 2$. They are genuinely different series, all correct, and asking for the Laurent series without naming a region is asking an incomplete question.
How they are found. Almost never from the integral. The method is to write each factor as a geometric series in whichever ratio is small on the chosen annulus:
Multiplying a known series by a power of $z - z_0$ shifts every exponent, and substituting $1/z$ into a series valid everywhere (the exponential, sine, cosine) reflects them.
Why anyone cares. Of the whole doubly infinite list of coefficients, one matters for integration: $a_{-1}$. Integrating the series term by term round a circle kills every term except that one, because $\oint (z - z_0)^{n}\,dz$ is zero for every $n$ but $-1$. So
$$\oint_C f(z)\,dz = 2\pi i\,a_{-1}.$$
That is the residue theorem in embryo, and it is why the next three lessons exist.
Another way: picture
A Taylor series is a disc of validity with the centre included. A Laurent series is a ring, and the hole in the middle is exactly where the function misbehaves. Both boundaries of the ring are circles that run through a singularity — the series stops converging when it reaches one, in either direction, and a different ring on the other side of it carries a different series for the same function.
Another way: steps
Work $\dfrac{1}{(z - 1)(z - 2)}$ about $0$ all three ways, because the comparison is the lesson.
Partial fractions first: $\dfrac{1}{(z - 1)(z - 2)} = \dfrac{1}{z - 2} - \dfrac{1}{z - 1}$.
On $|z| < 1$. Both singularities are outside, so both factors are expanded with the point on top: powers of $z/1$ and $z/2$. Every exponent is non-negative — it is the Taylor series, and the function is analytic on that disc, so it had to be.
On $1 < |z| < 2$. The singularity at $1$ is now inside the ring, so that factor is expanded the other way, in powers of $1/z$; the one at $2$ is still outside, so it keeps powers of $z/2$. The result has infinitely many negative powers and infinitely many positive ones.
On $|z| > 2$. Both singularities are inside, both factors are expanded in powers of $1/z$, and every exponent is negative.
Three series, one function, and the residue read off each is different — which is the point. A residue is attached to a singularity, and reading it off the wrong annulus's series answers a different question. The annulus that matters for a residue at $z_0$ is always the innermost one, $0 < |z - z_0| < \rho$, the punctured disc reaching out to the next singularity.
The commonest error is to speak of the Laurent series of a function about a point. There is one for each annulus, and they are not small variations on one another: one may have no negative powers at all and another nothing but negative powers.
This matters because the residue is read off a series, and the series has to be the one on the punctured disc immediately around the singularity in question. Reading a residue at $z_0 = 0$ off the expansion valid for $|z| > 2$ gives a number, and it is not the residue.
The second error is arithmetic and just as common: expanding $\dfrac{1}{z - a}$ in powers of $z/a$ when the region is $|z| > a$. The series produced diverges everywhere in the region asked about. The check that catches it is to ask which of the two quantities is the small one, every time, before writing anything.
A third: the order of a pole is read from the most negative power present, and the residue from the power $-1$ specifically. Those are different terms and they coincide only for a simple pole, which is why the example above has order $2$ and residue $1$, and the faded one has order $2$ and residue $0$.
$\dfrac{e^{z}}{z^{2}}$ about $0$: expand the numerator, $e^{z} = 1 + z + \dfrac{z^{2}}{2} + \cdots$.
Expand the analytic factor.
Divide through by $z^{2}$: $\dfrac{1}{z^{2}} + \dfrac{1}{z} + \dfrac{1}{2} + \cdots$.
Dividing shifts every exponent down by two.
Two negative powers, so a pole of order $2$; the residue is the coefficient of $1/z$, which is $1$.
Order from the lowest power, residue from the power minus one.
$\dfrac{1}{z - 3}$ about $0$ on $|z| < 3$: $-\dfrac{1}{3}\left(1 + \dfrac{z}{3} + \dfrac{z^{2}}{9} + \cdots\right)$.
The point is the small one inside.
On $|z| > 3$: $\dfrac{1}{z}\left(1 + \dfrac{3}{z} + \dfrac{9}{z^{2}} + \cdots\right)$.
The singularity is the small one outside.
Both are correct; neither converges where the other does. The region was part of the question.
$\sin z = z - \dfrac{z^{3}}{6} + \dfrac{z^{5}}{120} - \cdots$.
Expand the analytic factor.
Dividing by $z^{3}$: $\dfrac{1}{z^{2}} - \dfrac{1}{6} + \dfrac{z^{2}}{120} - \cdots$. A pole of order $2$, with residue $0$.
Expand $f(z) = \dfrac{1}{z\left(1 - 5z\right)}$ about $0$, on the annulus where the geometric series converges, and fill in the table.
| Value | |
|---|---|
| The coefficient of the power minus one | |
| The coefficient of the constant term | |
| The coefficient of the first power | |
| The coefficient of the second power | |
| The residue at the origin |
Expand $\dfrac{1}{z\left(1 - 4z\right)}$ about $z = 0$. What is the coefficient of $z^{1}$?
Answer:
$f(z) = \dfrac{1}{(z - 1)(z - 6)}$ is expanded about $0$ as a Laurent series with both positive and negative powers. Give the set of values of $s = |z|$ on which that expansion converges.
This task has no paper form; do it on a device.
Put the steps of finding a Laurent expansion of a quotient in order.
Number the steps in order (write the number in the box):
The function is $\dfrac{1}{z - 4}$, expanded about $0$. Match each region to the expansion that is valid there.
| Factor out the singularity and expand in powers of the point over it | Factor out the point and expand in powers of the singularity over it | Nothing to expand: the function is already a single negative power | No expansion exists, because a series cannot cross the singularity | |
|---|---|---|---|---|
| $|z| < 4$ | ||||
| $|z| > 4$ | ||||
| A small disc about $z = 4$ itself | ||||
| The whole plane at once |
Expand $e^{1/z}$ about $0$. What is the coefficient of $z^{-3}$? Give a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Expand $f(z) = \dfrac{1}{z\left(1 - 4z\right)}$ about $0$, on the annulus where the geometric series converges, and fill in the table.
| Value | |
|---|---|
| The coefficient of the power minus one | |
| The coefficient of the constant term | |
| The coefficient of the first power | |
| The coefficient of the second power | |
| The residue at the origin |
You can produce a Laurent expansion on a named annulus and read a residue off it. Say in your own words why one function has more than one Laurent series about the same point. Next: what the negative powers tell you about the singularity they surround.
9. Your turn: the Laurent expansion of $\dfrac{\sin z}{z^{3}}$ about $0$, step 2