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734 CHAPTER 22 Singularities and the Residue Theorem

This means that

f (z)dz = 2πic −1 .
γ
We will know the integral if we just know c −1 ! This one term of the Laurent expansion wins the
game!

The coefficient of 1/(z − z 0 ) in the Laurent expansion of f about z 0 is called the residue
of f at z 0 and is denoted Res( f, z 0 ).

What we have so far is that f (z)dz = 2πiRes( f, z 0 ) if z 0 is the only singularity of f
γ
enclosed by γ . This is of limited value. The power of the residue theorem is that it allows for any
finite number of singularities of f to be enclosed by γ .

THEOREM 22.4 The Residue Theorem
Let γ be a closed path, and suppose f is differentiable on γ and all points enclosed by γ , except
for z 1 ,··· , z n , which are all of the isolated singularities of f enclosed by γ . Then
n

f (z)dz = 2πi Res( f, z j ).
γ
j=1
The proof is immediate. Enclose each z j by a small circle C j that does not intersect any of

the other circles or γ . Because C j encloses just one singularity, z j , f (z)dz = 2πiRes( f, z j ).
C j
By the extended deformation theorem,
n
n

f (z)dz = f (z)dz = 2πi Res( f, z j ).
γ j=1 C j j=1
The residue theorem is as effective as our ability to evaluate residues. We do not want to have
to write a Laurent series about each singularity to pick off the coefficient of each 1/(z − z j ).We
will now develop efficient ways to compute residues at poles.

THEOREM 22.5 Residue at a Simple Pole

If f has a simple pole at z 0 , then
Res( f, z 0 ) = lim(z − z 0 ) f (z).
z→z 0
To see why this is true, the Laurent expansion of f about a simple pole z 0 has the form
∞
c −1 n
f (z) = + c n (z − z 0 )
z − z 0
n=0
in some annulus about z 0 . Then
∞
n+1
(z − z 0 ) f (z) = c −1 + c n (z − z 0 ) ,
n=0
(z − z 0 ) f (z).
so c −1 = lim z→z 0

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October 14, 2010 15:37 THM/NEIL Page-734 27410_22_ch22_p729-750