1
CAUCHY INTEGRAL FORMULA AND
RELATED THEOREMS
Theorem 1.1 If f (z) is analytic inside and on the boundary of the curve C of a
simply connected region R, then
I
1
f (z)
f (a) =
dz
2πi C z − a
that is,
I
C
f (z)
dz = 2πif (a).
z−a
This is known as Cauchy integral formula and if z = a is a pole of order n + 1 and
the nth derivative of f (z) exists then
I
n!
f (z)
(n)
f (a) =
dz
2πi C (z − a)n+1
that is,
2πif (n) (a)
=
n!
I
C
f (z)
dz.
(z − a)n+1
Proof We provide the proof here for a simple pole within a region which is simply
connected. The generalised proof follows very easily likewise that of multiply con(z)
nected regions. The integrand function fz−a
is analytic inside and on C except at
z = a. Thus we can construct a small region around a bounded by the curve Γ of
radius centred at a. It therefore follows that
I
I
f (z)
f (z)
dz =
dz.
z
−
a
z
−a
C
Γ
Here Γ is the circle of | z − a |= which implies that z − a = eiθ , 0 ≤ θ ≤ 2π. So if
we let z = a + eiθ it implies that dz = ieiθ dθ and hence
I
∫ 2π
f (z)
f (a + eiθ ) iθ
dz =
ie dθ
eiθ
Γ z−a
0
∫ 2π
= i
f (a + eiθ )dθ.
0
This implies that
I
C
f (z)
dz = i
z−a
∫
2π
f (a + eiθ )dθ
0
and taking limit as → 0 we obtain
∫ 2π
∫
lim i
f (a + eiθ )dθ = i
→0
0
2π
f (a)dθ = 2πif (a).
0
2
Example 1.2 Evaluate
I
C
sin πz 2 + cos πz 2
dz, C :| z |= 3
(z − 1)(z − 2)
1
Solution
1
Since (z−1)(z−2)
=
I
C
1
z−2
−
1
z−1
sin πz 2 + cos πz 2
dz =
(z − 1)(z − 2)
we have
I
C
sin πz 2 + cos πz 2
dz −
(z − 2)
I
C
sin πz 2 + cos πz 2
dz.
(z − 1)
The first with a = 2 integral gives
2πif (a) = 2πi(sin π4 + cos π4) = 2πi.
The second integral with a = 1 gives
2πif (a) = (sin π + cos π) = −2πi.
Hence the integral is 2πi − −2πi = 4πi.
Example 1.3 Evaluate
I
C
e2z
dz.
(z + 1)4
Here C :| z |= 3.
Solution
Here f (a) = e2z while a = −1 and n = 3. Thus
I
e2z
f 000 (z)
dz =
· 2πi.
4
3!
C (z + 1)
Thus
f 0 (z) = 2e2z , f 00 = 4e2z , f 000 (z) = 8e2z .
Therefore the integral is 83 πie−2 .
Exercise 1: Evaluate
I
C
z
dz, C :| z − 2 |≤ 2.
(z − 2)2 (z + 4)
Exercise 2: Using Cauchy integral formula for a simple pole, prove the Gauss mean
value theorem
∫ 2π
1
f (a) =
f (a + reiθ )dθ
2π 0
if f (z) is analytic inside and on a circle C with centre at a and radius r.
2