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708 CHAPTER 20 Complex Integration

For n = 0, this is Cauchy’s integral formula with the convention that f (0) (z) = f (z).Inthe
equation, n! (read n factorial) is the product of the integers 1 through n inclusive, and f (n) is
the nth derivative of f . This integral formula for the nth derivative of f (z) at z 0 is exactly what
we would get if we differentiated n times with respect to z 0 under the integral sign in Cauchy’s
integral formula.

EXAMPLE 20.14
We will evaluate
e
z 3
dz
γ (z − i) 3
with γ any closed path not passing through i.If γ does not enclose i, this integral is zero by
Cauchy’s theorem. Suppose that γ does enclose i. Because z − i occurs to the third power in
the denominator of the integral, let n = 2 in Cauchy’s formula for higher derivatives (20.3) with
3
z
f (z) = e . Compute
3
3
2 z
4
z

f (z) = 3z e and f (z) = (6z + 9z )e .
Then
e 2πi −i
z 3

dz = f (i) = (−6 + 9i)πe .
γ (z − i) 3 2!
Proof We will outline a proof of Cauchy’s integral formula. First, use the deformation theorem
to replace γ with a circle C of radius r about z 0 , as in Figure 20.11. Then
f (z) f (z) f (z) − f (z 0 ) + f (z 0 )

dz = dz = dz
γ z − z 0 C z − z 0 C z − z 0
1 f (z) − f (z 0 )

= f (z 0 ) dz + dz
C z − z 0 C z − z 0
f (z) − f (z 0 )

= 2πif (z 0 ) + dz
C z − z 0
in which we used the result of Example 20.11. We will have proved the Cauchy integral repre-
it
sentation if we can show that the last integral is zero. On C, write C(t)= z 0 +re for 0≤t ≤2π.
Then

y
C

z 0

x

FIGURE 20.11 A proof of the Cauchy inte-
gral formula.

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October 14, 2010 15:32 THM/NEIL Page-708 27410_20_ch20_p695-714