Mathematics of Physics and Engineering

(^224) Singularities of Complex Functions
singularity. Computation of the residue at a removable singularity or a pole
does not require the explicit knowledge of the Laurent series expansion.
If ZQ is a removable singularity, then, according to Exercise 4.4.2, ci =
Oand Res/(z) = 0.
2=20
If ZQ is a simple pole of /, then
Res/(z) = lim ((* - z 0 )f(z)), (4.4.14)
because the Laurent series around a simple pole is
oo
C-l
f(z) = ^^Z — Zn + J2cn(z-z 0 )k.
U k=0
An immediate consequence of (4.4.14) is the following result.
EXERCISE 4.4.8F Assume that f(z) = h(z)/g(z), where h,g are analytic at
ZQ, h(zo) ^ 0, g(z 0 ) = 0, g'(zo) / 0. Show that
Res/(z) = ^. (4-4.15)
Hint: use (4-4-14) and note that g'(zo) = limz»z 0 (<?(z) — g(zo))/(z — zo) =
\imz^zo g(z)/(z - z 0 ).
Usually, formula (4.4.15) is easier to use than (4.4.14). FOR EXAMPLE,
if f(z) = (z + 5)/(z^3 -z), then
2S
e

s

1 ^=^iL=

6

/

2

3

If z 0 is a pole of order N > 1, then the function (z — z 0 )N f(z) has a

removable singularity at ZQ and

(z - z 0 )Nf(z) = J2 ck(z - z 0 )k+N = C-N + C-N+1(z - zo)

k=-N

+ ... + c_ 2 (z - z 0 )N-^2 + d(z - zo)"-^1 + -..

We differentiate this equality N - 1 times with respect to z, so that all the
terms on the right that are before c_i(z — ZQ)N~X disappear, while the term
c-i(z — ZQ)N~X becomes (N — l)!c_i. We then set z = ZQ so that all the