4. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f (t) of a real variable t: f (t)= u(t)+ iv (t), which is assumed to be a piecewise continuous function defined in the closed interval a ≤ t ≤ b. The integral of f (t) from t = a to t = b, is defined as b a f (t) dt = b a u(t) dt + i b a v (t) dt. 1
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4. Complex integration: Cauchy integral theorem and Cauchy
integral formulas
Definite integral of a complex-valued function of a real variable
Consider a complex valued function f(t) of a real variable t:
f(t) = u(t) + iv(t),
which is assumed to be a piecewise continuous function defined in
the closed interval a ≤ t ≤ b. The integral of f(t) from t = a to
t = b, is defined as
∫ b
af(t) dt =
∫ b
au(t) dt + i
∫ b
av(t) dt.
1
Properties of a complex integral with real variable of integration
1.
Re
∫ b
af(t) dt =
∫ b
aRe f(t) dt =
∫ b
au(t) dt.
2.
Im
∫ b
af(t) dt =
∫ b
aIm f(t) dt =
∫ b
av(t) dt.
3.∫ b
a[γ1f1(t) + γ2f2(t)] dt = γ1
∫ b
af1(t) dt + γ2
∫ b
af2(t) dt,
where γ1 and γ2 are any complex constants.
4.∣∣∣∣∣∣
∫ b
af(t) dt
∣∣∣∣∣∣
≤∫ b
a|f(t)| dt.
2
To prove (4), we consider∣∣∣∣∣∣
∫ b
af(t) dt
∣∣∣∣∣∣
= e−iφ∫ b
af(t) dt =
∫ b
ae−iφf(t) dt,
where φ = Arg
(∫ b
af(t) dt
)
. Since
∣∣∣∣∣∣
∫ b
af(t) dt
∣∣∣∣∣∣
is real, we deduce that
∣∣∣∣∣∣
∫ b
af(t) dt
∣∣∣∣∣∣
= Re
∫ b
ae−iφf(t) dt =
∫ b
aRe [e−iφf(t)] dt
≤∫ b
a|e−iφf(t)| dt =
∫ b
a|f(t)| dt.
3
Example
Suppose α is real, show that
|e2απi − 1| ≤ 2π|α|.
Solution
Let f(t) = eiαt, α and t are real. We obtain∣∣∣∣∣∣
∫ 2π
0eiαt dt
∣∣∣∣∣∣
≤∫ 2π
0|eiαt| dt = 2π.
The left-hand side of the above inequality is equal to∣∣∣∣∣∣
∫ 2π
0eiαt dt
∣∣∣∣∣∣
=
∣∣∣∣∣∣
eiαt
iα
∣∣∣∣∣∣
2π
0
∣∣∣∣∣∣
=|e2απi − 1|
|α|.
Combining the results, we obtain
|e2απi − 1| ≤ 2π|α|, α is real.
4
Definition of a contour integral
Consider a curve C which is a set of points z = (x, y) in the complex
plane defined by
x = x(t), y = y(t), a ≤ t ≤ b,
where x(t) and y(t) are continuous functions of the real parameter
t. One may write
z(t) = x(t) + iy(t), a ≤ t ≤ b.
• The curve is said to be smooth if z(t) has continuous derivative
z′(t) 6= 0 for all points along the curve.
• A contour is defined as a curve consisting of a finite number
of smooth curves joined end to end. A contour is said to be a
simple closed contour if the initial and final values of z(t) are
the same and the contour does not cross itself.
5
• Let f(z) be any complex function defined in a domain D in the
complex plane and let C be any contour contained in D with
initial point z0 and terminal point z.
• We divide the contour C into n subarcs by discrete points z0, z1, z2,
. . ., zn−1, zn = z arranged consecutively along the direction of in-
creasing t.
• Let ζk be an arbitrary point in the subarc zkzk+1 and form the
sum
n−1∑
k=0
f(ζk)(zk+1 − zk).
6
Subdivision of the contour into n subarcs by discrete points z0, z1, · · · ,zn−1, zn = z.
7
We write △zk = zk+1 − zk. Let λ = maxk
|△zk| and take the limit
limλ→0n→∞
n−1∑
k=0
f(ζk) △zk.
The above limit is defined to be the contour integral of f(z) along
the contour C.
If the above limit exists, then the function f(z) is said to be inte-
grable along the contour C.
If we write
dz(t)
dt=
dx(t)
dt+ i
dy(t)
dt, a ≤ t ≤ b,
then∫
Cf(z) dz =
∫ b
af(z(t))
dz(t)
dtdt.
8
Writing f(z) = u(x, y) + iv(x, y) and dz = dx + idy, we have∫
Cf(z) dz =
∫
Cu dx − v dy + i
∫
Cu dy + v dx
=
∫ b
a
[
u(x(t), y(t))dx(t)
dt− v(x(t), y(t))
dy(t)
dt
]
dt
+ i∫ b
a
[
u(x(t), y(t))dy(t)
dt+ v(x(t), y(t))
dx(t)
dt
]
dt.
The usual properties of real line integrals are carried over to their
complex counterparts. Some of these properties are:
(i)
∫
Cf(z) dz is independent of the parameterization of C;
(ii)∫
−Cf(z) dz = −
∫
Cf(z) dz, where −C is the opposite curve of C;
(iii) The integrals of f(z) along a string of contours is equal to the
sum of the integrals of f(z) along each of these contours.
9
Example
Evaluate the integral∮
C
1
z − z0dz,
where C is a circle centered at z0 and of any radius. The path is
traced out once in the anticlockwise direction.
Solution
The circle can be parameterized by
z(t) = z0 + reit, 0 ≤ t ≤ 2π,
where r is any positive real number. The contour integral becomes
∮
C
1
z − z0dz =
∫ 2π
0
1
z(t) − z0
dz(t)
dtdt =
∫ 2π
0
ireit
reitdt = 2πi.
The value of the integral is independent of the radius r.
10
Example
Evaluate the integral
(i)
∫
C|z|2 dz and (ii)
∫
C
1
z2dz,
where the contour C is
(a) the line segment with initial point −1 and final point i;
(b) the arc of the unit circle Im z ≥ 0 with initial point −1 and final
point i.
Do the two results agree?
11
Solution
(i) Consider
∫
C|z|2 dz,
(a) Parameterize the line segment by
z = −1 + (1 + i)t, 0 ≤ t ≤ 1,
so that
|z|2 = (−1 + t)2 + t2 and dz = (1 + i) dt.
The value of the integral becomes
∫
C|z|2 dz =
∫ 1
0(2t2 − 2t + 1)(1 + i) dt =
2
3(1 + i).
12
(b) Along the unit circle, |z| = 1 and z = eiθ, dz = ieiθdθ. The initial
point and the final point of the path correspond to θ = π and
θ = π2, respectively. The contour integral can be evaluated as
∫
C|z|2 dz =
∫ π2
πieiθ dθ = eiθ
∣∣∣∣∣∣
π2
π
= 1 + i.
The results in (a) and (b) do not agree. Hence, the value of this
contour integral does depend on the path of integration.
13
(ii) Consider
∫
C
1
z2dz.
(a) line segment from −1 to i
∫
C
1
z2dz =
∫ 1
0
1 + i
[−1 + (1 + i)t]2dt = − 1
−1 + (1 + i)t
∣∣∣∣∣
1
0
= −1−1
i= −1+i.
(b) subarc from −1 to i
∫
C
1
z2dz =
∫ π2
π
1
e2iθieiθ dθ = −e−iθ
]π2
π= −1 + i.
14
Estimation of the absolute value of a complex integral
The upper bound for the absolute value of a complex integral can
be related to the length of the contour C and the absolute value of
f(z) along C. In fact,∣∣∣∣∣∣
∫
Cf(z) dz
∣∣∣∣∣∣
≤ ML,
where M is the upper bound of |f(z)| along C and L is the arc length
of the contour C.
15
We consider∣∣∣∣∣∣
∫
Cf(z) dz
∣∣∣∣∣∣
=
∣∣∣∣∣∣
∫ b
af(z(t))
dz(t)
dtdt
∣∣∣∣∣∣
≤∫ b
a|f(z(t))|
∣∣∣∣∣∣
dz(t)
dt
∣∣∣∣∣∣
dt
≤∫ b
aM
∣∣∣∣∣∣
dz(t)
dt
∣∣∣∣∣∣
dt
= M∫ b
a
√√√√
(
dx(t)
dt
)2
+
(
dy(t)
dt
)2
dt = ML.
16
Example
Show that
∣∣∣∣∣∣
∫
C
1
z2dz
∣∣∣∣∣∣
≤ 2, where C is the line segment joining −1 + i and 1 + i.
Solution
Along the contour C, we have z = x + i, −1 ≤ x ≤ 1, so that 1 ≤|z| ≤
√2. Correspondingly,
1
2≤ 1
|z|2≤ 1. Here, M = max
z∈C
1
|z|2= 1
and the arc length L = 2. We have∣∣∣∣∣∣
∫
C
1
z2dz
∣∣∣∣∣∣
≤ ML = 2.
17
Example
Estimate an upper bound of the modulus of the integral
I =
∫
C
Log z
z − 4idz
where C is the circle |z| = 3.
Now,
∣∣∣∣∣∣
Log z
z − 4i
∣∣∣∣∣∣
≤
∣∣∣∣∣ln |z|
∣∣∣∣∣+ |Arg z|
||z| − |4i||so that
maxz∈C
∣∣∣∣∣∣
Log z
z − 4i
∣∣∣∣∣∣
≤ ln 3 + π
|3 − 4|= ln3 + π; L = (2π)(3) = 6π.
Hence,
∣∣∣∣∣∣
∫
C
Log z
z − 4idz
∣∣∣∣∣∣
≤ 6π(π + ln3).
18
Example
Find an upper bound for
∣∣∣∣∣
∫
Γez/(z2 + 1) dz
∣∣∣∣∣, where Γ is the circle
|z| = 2 traversed once in the counterclockwise direction.
Solution
The path of integration has length L = 4π. Next we seek an upper
bound M for the function ez/(z2 + 1) when |z| = 2. Writing z =
x + iy, we have
|ez| = |ex+iy| = ex ≤ e2, for |z| =√
x2 + y2 = 2,
and by the triangle inequality
|z2 + 1| ≥ |z|2 − 1 = 4 − 1 = 3 for |z| = 2.
Hence, |ez/(z2 + 1)| ≤ e2/3 for |z| = 2, and so∣∣∣∣∣
∫
Γ
ez
z2 + 1dz
∣∣∣∣∣≤ e2
3· 4π.
19
Path independence
Under what conditions that∫
C1
f(z) dz =∫
C2
f(z) dz,
where C1 and C2 are two contours in a domain D with the same
initial and final points and f(z) is piecewise continuous inside D.
The property of path independence is valid for f(z) = 1z2 but it fails
when f(z) = |z|2. The above query is equivalent to the question:
When does∮
Cf(z) dz = 0
hold, where C is any closed contour lying completely inside D? The
equivalence is revealed if we treat C as C1 ∪ −C2.
We observe that f(z) = 1z2 is analytic everywhere except at z = 0
but f(z) = |z|2 is nowhere analytic.
20
Cauchy integral theorem
Let f(z) = u(x, y)+iv(x, y) be analytic on and inside a simple closed
contour C and let f ′(z) be also continuous on and inside C, then∮
Cf(z) dz = 0.
Proof
The proof of the Cauchy integral theorem requires the Green theo-
rem for a positively oriented closed contour C: If the two real func-
tions P(x, y) and Q(x, y) have continuous first order partial deriva-
tives on and inside C, then∮
CP dx + Q dy =
∫∫
D(Qx − Py) dxdy,
where D is the simply connected domain bounded by C.
21
Suppose we write f(z) = u(x, y) + iv(x, y), z = x + iy; we have∮
Cf(z) dz =
∮
Cu dx − v dy + i
∮
Cv dx + u dy.
One can infer from the continuity of f ′(z) that u(x, y) and v(x, y)
have continuous derivatives on and inside C. Using the Green the-
orem, the two real line integrals can be transformed into double
integrals.∮
Cf(z) dz =
∫∫
D(−vx − uy) dxdy + i
∫∫
D(ux − vy) dxdy.
Both integrands in the double integrals are equal to zero due to the
Cauchy-Riemann relations, hence the theorem.
In 1903, Goursat was able to obtain the same result without assum-
ing the continuity of f ′(z).
22
Goursat Theorem
If a function f(z) is analytic throughout a simply connected domain
D, then for any simple closed contour C lying completely inside D,
we have∮
Cf(z) dz = 0.
Corollary 1
The integral of a function f(z) which is analytic throughout a simply
connected domain D depends on the end points and not on the
particular contour taken. Suppose α and β are inside D, C1 and C2
are any contours inside D joining α to β, then∫
C1
f(z) dz =
∫
C2
f(z) dz.
23
Example
If C is the curve y = x3−3x2+4x−1 joining points (1,1) and (2,3),
find the value of∫
C(12z2 − 4iz) dz.
Method 1. The integral is independent of the path joining (1,1)
and (2,3). Hence any path can be chosen. In particular, let us
choose the straight line paths from (1,1) to (2,1) and then from
(2,1) to (2,3).
Case 1 Along the path from (1,1) to (2,1), y = 1, dy = 0 so that
z = x + iy = x + i, dz = dx. Then the integral equals