262 CHAPTER 11. FUNCTIONS OF ONE COMPLEX VARIABLE

•a1

γ1

•a2

γ2

γ0

Assume f ′ (z) exists for all z in some open set which includes all the contours and theirinsides other than at the two points shown, a1 and a2. Then applying the above to thesecond piecewise smooth closed curve, we find that on adding the contour integrals overthe top and the bottom, yields that the contour integrals over the straight lines cancel andso what finally results is∫

γ0

f (z)dz+∫−γ1

f (z)dz+∫−γ2

f (z)dz = 0

In other words, ∫γ0

f (z)dz =∫

γ1

f (z)dz+∫

γ2

f (z)dz

where γ i for i = 1,2 are oriented counter clockwise.Obviously you could have many more exceptional points ai and small circular contours

surrounding these, and the same argument would work. This example and its obviousgeneralizations is a paradigm for the procedure of evaluating contour integrals with themethod of residues. First is a definition of what is meant by a pole and a residue.

Definition 11.5.1 A function f has a pole at a if

f (z) = g(z)+n

∑k=1

bk

(z−a)k

where bn ̸= 0 and g is analytic near a. The order of the pole is n. Denote by S (a,z) the termdescribed by the sum. It is called the singular part of f at a. The residue at a is defined tobe b1 and is denoted as res( f ,a). See Theorem 11.3.2 to see how this takes place.

How can we find it? Multiply by (z−a)n. This gives

f (z)(z−a)n = g(z)(z−a)n +b1 (z−a)n−1 +n

∑k=2

bk (z−a)n−k

Now take the derivative n−1 times and then take a limit as z→ a. The differentiation willzero out all the terms in the sum on the right. Then the limit will zero out all other termsexcept the b1 term which will be (n−1)!b1. This justifies the following procedure.

Procedure 11.5.2 When f has a pole at a of order n, to find the residue, multiplyby (z−a)n , take n−1 derivatives, and finally take the limit as z→ a. The residue will bethis number b divided by (n−1)!. Thus

res( f ,a) =1

(n−1)!limz→a

dn−1

dzn−1 ( f (z)(z−a)n)