Laurent Series 221EXERCISE 4.4.3.c Find the Laurent series for the function
(2z + 5\
f(z) =JK cos
' \z + 2 J
around the point ZQ = —2. What is the type of the singularity of f at ZQ?
When writing the Laurent series expansion, one should pay attention to
the domain in which this expansion should hold. Recall that the expansion
is written in an annulus {z : Ri < \z — ZQ\ < R}, where ZQ is a fixed complex
number, and none of the singular points of / should be inside this annulus.
The expansion in such an annulus has the form Y^kL-oo ck(z ~ zo)k- The
point ZQ does not have to be a singular point of f, but at least one singular
point of f must be in the set {z : \z — ZQ\ = Ri}, and, unless R is infinite, at
least one singular point of f must be in the set {z : \z — ZQ\ = R} (otherwise,
we are able to expand the annulus). The Laurent series that converges in
the disk {z : \z — ZQ\ < R} is the same as the Taylor series at ZQ. As a result,
the same function can have different expansions in different domains, even
when the point ZQ is the same: recall that (1 - z)~l = Ylk>o zk ^or \z\ < 1
and (1 - z)-^1 = Efc^o--1 for \z\ > 1.
As A DIFFERENT EXAMPLE, consider the function
The Laurent series of this function at ZQ = 0 is f(z) = (2z)~^1 +
Ylk>o 2~k~^2 zk (check it), and ZQ is a simple pole (a pole of order one).
The expansion is true for 0 < \z\ < 2. Similarly, the Laurent series in the
domain {z : 2 < \z — 2|} is computed as follows: 1/z = (z — 2 + 2)""^1 =
(z - 2)-x(l + 2/(z - 2))-^1 , and so
!{Z) = _ 2(1^2) +^
0 (z- 2)fc+! = W=Y) + g (z - 2)*+i • (4A9)
EXERCISE 4.4.4.C Find the Laurent series of the function f from (4-4-8) in
the domain {z : 0 < \z — 2| < 2}.
It follows from (4.4.8) that ZQ = 2 is a pole of order one; this is also
what you should conclude from the previous exercise. On the other hand,
the expansion in (4.4.9) contains infinitely many negative powers of (z — 2).
Should we conclude from (4.4.9) that ZQ = 2 is an essential singularity of /,
and how do we reconcile these seemingly contradictory conclusions? After
looking more closely at (4.4.8), we realize that (4.4.8) is not a Laurent series