Complex Analysis - Maths KU

276 Chapter 5 Integration in the Complex Plane
In addition, let L be the length of C and let δ denote the shortest distance
between points on C and the point z0. Thus for all points z on C we have
|z − z0| ≥δ or
1 1
2 ≤ .
|z − z0| δ2 Furthermore, if we choose |∆z| ≤ 1
2δ, then by (10) of Section 1.2,
and so,
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Now,
C
|z − z0 − ∆z| ≥||z − z0|−|∆z||≥δ −|∆z| ≥ 1
2 δ
1 2
≤
|z − z0 − ∆z| δ .
�
f(z)
f(z)
dz −
(z − z0) 2
C (z − z0 − ∆z)(z − z0) dz
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= �
−∆zf(z)
�
dz
(z − z0 − ∆z)(z − z0) 2
C
�
�
�
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≤ 2ML|∆z|
δ 3
.
Because the last expression approaches zero as ∆z → 0, we have shown that
f ′ f(z0 +∆z) − f(z0)
(z0) = lim
=
∆z→0 ∆z
1
�
f(z)
dz,
2πi C (z − z0) 2
which is (6) for n =1. ✎
Like (1), formula (6) can be used to evaluate integrals.
EXAMPLE 3 Using Cauchy’s Integral Formula for Derivatives
�
Evaluate
C
z +1
z4 dz, where C is the circle |z| =1.
+2iz3 Solution Inspection of the integrand shows that it is not analytic at z =0
and z = −2i, but only z = 0 lies within the closed contour. By writing the
integrand as
z +1
z4 =
+2iz3 z +1
z +2i
z 3
we can identify, z0 =0,n = 2, and f(z) =(z +1)/ (z +2i). The quotient
rule gives f ′′ (z) =(2− 4i)/(z +2i) 3 and so f ′′ (0) = (2i − 1)/4i. Hence from
(6) we find
�
C
z +1
z4 2πi
dz =
+4z3 2! f ′′ (0) = − π π
+
4 2 i.