14—Complex Variables 351Theorder of the pole is the size of the largest negative power. These have respectively first order and
second order poles.
If the function has an infinite number of negative powers, and the series converges all the way
down to (but of course not at) the singularity, it is said to have anessential singularity.
e^1 /z=
∑∞
01
k!zk
or sin[
t
(
z+
1
z
)]
=··· or1
1 −z
=
1
z
− 1
1 −^1 z=−
∑∞
1z−k
The first two have essential singularities; the third does not.
It’s worth examining some examples of these series and especially in seeing what kinds of
singularities they have. In analyzing these I’ll use the fact that the familiar power series derived for real
variables apply here too. The binomial series, the trigonometric functions, the exponential, many more.
1 /z(z−1)has a zero in the denominator for bothz= 0andz= 1. What is the full behavior
near these two points?
1z(z−1)
=
− 1
z(1−z)
=
− 1
z
(1−z)−^1 =
− 1
z
[
1 +z+z^2 +z^3 +···
]
=
− 1
z
− 1 −z−z^2 −···
1
z(z−1)
=
1
(z−1)(1 +z−1)
=
1
z− 1
[
1 + (z−1)
]− 1
=
1
z− 1
[
1 + (z−1) + (z−1)^2 + (z−1)^3 +···
]
=
1
z− 1
+ 1 + (z−1) +···
This shows the full Laurent series expansions near these points. Keep your eye on the coefficient
of the inverse first power. That term alone plays a crucial role in what will follow.
csc^3 znearz= 0:
1
sin^3 z
=
1
[
z−z
3
6 +z^5
120 −···] 3 =
1
z^3
[
1 −z2
6 +z^4
120 −···] 3
=
1
z^3
[
1 +x
]− 3
=
1
z^3
[
1 − 3 x+ 6x^2 − 10 x^3 +···
]
=
1
z^3
[
1 − 3
(
−
z^2
6
+
z^4
120
−···
)
+ 6
(
−
z^2
6
+
z^4
120
−···
) 2
−···
]
=
1
z^3
[
1 +
z^2
2
+z^4
(
1
6
−
3
120
)
+···
]
=
1
z^3
[
1 +
1
2
z^2 +
17
120
z^4 +···
]
(14.6)
This has a third order pole, and the coefficient of 1 /zis 1 / 2. Are there any other singularities for
this function? Yes, every place that the sine vanishes you have a pole: atnπ. (What is the order of
these other poles?) As I commented above, you’ll soon see that the coefficient of the 1 /zterm plays a
special role, and if that’s all that you’re looking for you don’t have to work this hard. Now that you’ve
seen what various terms do in this expansion, you can stop carrying along so many terms and still get
the 1 / 2 zterm. See problem14.17
The structure of a Laurent series is such that it will converge in an annulus. Examine the absolute
convergence of such a series.
∑∞−∞∣
∣akzk
∣
∣=
∑−^1
−∞∣
∣akzk
∣
∣+
∑∞
0