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The coefficient on the power minus one is the only part of a Laurent series a contour integral can see, and it can be found without the series: cancel and substitute at a simple pole, multiply and differentiate at a higher one, and expand when the singularity is essential.
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 say why the coefficient on the power minus one is the one that matters, compute a residue at a simple pole by cancelling and substituting, use the quotient rule version when the denominator does not factorise, compute a residue at a pole of higher order with the multiply-differentiate-substitute formula, and choose between a formula and the series.
Laurent series and the classification of isolated singularities. The residue is one coefficient of a series you already know how to write; this lesson is about getting it without writing the series, which is what makes it a usable tool.
The residue of $f$ at an isolated singularity $z_0$, written $\operatorname{Res}_{z_0} f$, is the coefficient $a_{-1}$ of its Laurent series on a punctured disc about $z_0$. A pole is simple when its order is $1$. A rational function is a quotient of polynomials, and all of its singularities are poles.
Why this coefficient. Integrating a Laurent series term by term round a small circle about $z_0$ kills every term except one, because $\oint (z - z_0)^{n}\,dz$ is zero for every integer $n$ but $-1$, where it is $2\pi i$. So
$$\oint_C f(z)\,dz = 2\pi i\,a_{-1} = 2\pi i \operatorname{Res}_{z_0} f.$$
Out of a doubly infinite list of coefficients, exactly one survives contact with an integral. That is what makes it worth a name.
Getting it without the series.
Simple pole. $\operatorname{Res}_{z_0} f = \lim_{z \to z_0} (z - z_0)f(z)$: cancel the offending factor and substitute.
Simple pole of a quotient. If $f = g/h$ with $g(z_0) \ne 0$ and $h$ having a simple zero there, then $\operatorname{Res}_{z_0} f = g(z_0)/h'(z_0)$ — which avoids factorising $h$ at all, and is the formula to reach for when the denominator is something like $\sin z$.
Pole of order $m$.
$$\operatorname{Res}_{z_0} f = \frac{1}{(m-1)!}\lim_{z \to z_0}\frac{d^{m-1}}{dz^{m-1}}\Bigl[(z - z_0)^{m}f(z)\Bigr].$$
Multiply first, differentiate second, substitute third, divide by the factorial last. For $m = 1$ it collapses to the simple-pole rule.
Essential singularity. No formula exists. Expand the Laurent series and read the coefficient off, as $e^{1/z}$ forced in the last unit.
A warning about the order. The formula for order $m$ differentiates $m - 1$ times. Over-estimating the order — reading the exponent off the denominator without checking what the numerator cancels — differentiates too many times and returns a wrong number with no sign that anything went wrong. Finding the order is part of the calculation, not a preliminary to it.
Another way: picture
Think of the Laurent series as an infinite ledger of coefficients, running from very negative powers up through zero into the positive ones. Wrap a contour round the singularity and every entry is cancelled except one line, the power $-1$. A residue is what a contour integral can see, and everything else about the function is invisible to it.
Another way: steps
Three routes exist and the formulas are not always the fastest.
Read it off. If the function is already written as a sum of powers of $z - z_0$ — $\dfrac{3}{z^{2}} + \dfrac{5}{z} + 7$, say — it is its own Laurent series and the residue is visible: $5$. No formula needed, and applying one anyway wastes a differentiation.
Cancel and substitute. For a simple pole of a quotient of polynomials this is unbeatable, and the $g(z_0)/h'(z_0)$ version is better still when the denominator does not factorise: the residue of $\dfrac{e^{z}}{\sin z}$ at $0$ is $\dfrac{e^{0}}{\cos 0} = 1$, in one line.
Expand. For a pole of high order the differentiation gets unpleasant fast, and multiplying two known series is often easier. To find the residue of $\dfrac{e^{z}}{z^{4}}$ at $0$, the formula asks for three derivatives of $e^{z}$; the series asks for the coefficient of $z^{3}$ in $e^{z}$, which is $\tfrac16$, and that is the answer.
The rule of thumb: formulas for order one and two, series for order three and above, and always look first to see whether the answer is already written down.
Three things sit at the same point and get confused with one another: the order of the pole, the residue, and the value of whatever is left after cancelling. They are different numbers and only one of them is what an integral wants.
$\dfrac{5}{z^{2}}$ has order $2$ and residue $0$. $\dfrac{5}{z}$ has order $1$ and residue $5$. A function can have a pole of very high order and residue zero, and any contour integral round it will return zero regardless of how badly it behaves.
The second error is the sign. Cancelling the factor $z$ from $\dfrac{1}{z(z - a)}$ leaves $\dfrac{1}{z - a}$, and substituting $z = 0$ gives $-\dfrac{1}{a}$, not $\dfrac{1}{a}$. Every residue at a pole to the left of another one carries this sign, and it is lost more often than any other step.
The third is the order, again. The formula differentiates $m - 1$ times, so believing a pole is of order $3$ when it is of order $1$ differentiates twice too many and returns a number that is usually zero — plausible, wrong, and silent. Expand the numerator before deciding.
$\operatorname{Res}_{z = 2} \dfrac{z + 1}{(z - 2)(z + 3)}$: the pole at $2$ is simple.
Check the order first.
Multiply by $z - 2$ and substitute: $\dfrac{2 + 1}{2 + 3} = \dfrac{3}{5}$.
Cancel, then substitute.
$\operatorname{Res}_{z = 0} \dfrac{e^{z}}{z^{3}}$. The formula would ask for two derivatives of $e^{z}$ and a division by $2!$.
The formula is available.
The series is quicker: $\dfrac{e^{z}}{z^{3}} = \dfrac{1}{z^{3}} + \dfrac{1}{z^{2}} + \dfrac{1}{2z} + \cdots$.
Expand and divide.
The coefficient on the power $-1$ is $\tfrac12$, which is the residue. Both routes agree, and one of them took a line.
$\cos z = 1 - \dfrac{z^{2}}{2} + \cdots$, so the quotient is $\dfrac{1}{z^{2}} - \dfrac{1}{2} + \cdots$.
Expand the numerator and divide.
There is no term in $1/z$ at all, so the residue is $0$ — a pole of order two with residue zero.
Fill in the table. Every entry concerns the point $z = 2$.
| Value | |
|---|---|
| The order of the pole of a over the linear factor | |
| The residue of a over the linear factor | |
| The residue of a over the squared linear factor | |
| The residue of a times z over the linear factor |
What is the residue of $\dfrac{1}{z\left(z - 4\right)}$ at $z = 0$? Give a fraction.
Answer:
Match each function to its residue at $z = 0$.
| $7$ | $0$ | $-7$ | $1$ | |
|---|---|---|---|---|
| $\dfrac{7}{z}$ | ||||
| $\dfrac{7}{z^{2}}$ | ||||
| $\dfrac{7}{z(z - 1)}$ | ||||
| $\dfrac{z + 7}{z^{2}}$ |
Put the steps of computing a residue at a pole of order $3$ in order.
Number the steps in order (write the number in the box):
What is the residue of $\dfrac{8z}{\left(z - 3\right)^{2}}$ at $z = 3$?
Answer:
$f$ is a rational function whose denominator has degree at least $2$ more than its numerator. Add together the residues at all of its poles. What is the total?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Fill in the table. Every entry concerns the point $z = 6$.
| Value | |
|---|---|
| The order of the pole of a over the linear factor | |
| The residue of a over the linear factor | |
| The residue of a over the squared linear factor | |
| The residue of a times z over the linear factor |
You can compute a residue at a pole of any order and can say why the order has to be found before a formula is chosen. Say in your own words why only one coefficient of the series survives an integral. Next: adding up the residues inside a contour, which is the theorem this was all for.
9. Your turn: $\operatorname{Res}_{z = 0} \dfrac{\cos z}{z^{2}}$, step 2