14—Complex Variables 4241 /z(z−1)has a zero in the denominator for bothz= 0andz= 1. What is the full behavior near these
two points?
1
z(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 −z
2
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 +···]
(5)
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 /z term 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 problem 17