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SRIMAAN COACHING CENTRE
PGTRB-MATHEMATICS STUDY MATERIAL
COMPLEX INTEGRATION
In this section we shall study complex integrations of complex functions and
established fundamental theorem of calculus for line integral.
We show that an analytic function has a power series expansion as a Taylor’s
theorem. Form then we established Cauchy’s estimate to prove Cauchy’s theorem.
We begin with elementary definitions.
Definition: A path in region G ⊂ is a continuous function γ :[ a , b ] → for some
interval [a , b] in .
If γ '(t) exists for each t in [a , b] and γ ':[a , b] → is continuous then γ is called smooth
path. Also γ is called piecewise smooth if there is partition of [a , b], a = t 0 < t1 < ... < t n = b ,
such that γ is smooth on each subinterval [t j −1,t j ], 1≤ j ≤ n .
Definition : Let γ :[a , b ] → G be a path then trace of γ is { γ (t ):t ∈[a , b] } and it is
denoted by
γ
=
γ (t ):t ∈[a , b]
. Note that trace of a path is always compact.
i.e. {γ}
Definition :A function γ :[a , b] → , for [a , b] ⊂ , is of bounded variation if there is a
constant M > 0 such that for any partition P = {a = t 0 < t1 < ... < t m = b} of [a , b]
m
υ ( γ ; P ) = ∑ γ (t k ) − γ (t k −1) ≤ M . k =1
The total variation of γ , denoted by V (γ ) is defined as
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V (γ ) = sup{υ ( γ ; P ) : P a partition of [a,b] } .
Definition :A path γ :[a , b] → is rectifiable if γ is a function of bounded variation.
Theorem: Let γ :[a , b] → be of bounded variation. Then:
a) If P and Q are partitions of [a , b] and P ⊂ Q then υ ( γ ; P ) ≤ υ ( γ ;Q).
b) If σ :[a , b] → is also of bounded variation and α , β ∈ then αγ + βσ is of bounded
variation and V (αγ + βσ ) ≤
α
V (γ ) +
β
V (σ ).
Theorem : If γ :[a , b] → is piecewise smooth then γ is of bounded variation and
b
V ( γ ) = ∫
γ '(t )
dt .
a
Theorem :
Let f and g be continuous functions on [a , b] and let γ and σ be functions
of bounded variation on [a , b]. Then for any scalar α and β
b
a) ∫ ( α f + β g ) d γ = α
b
a
b
a
gdyfdy
a
b) b
a
fd ( αγ + βσ ) = α
b
a
b
a
gdyfdy .
a
Theorem :
If γ is piecewise smooth and f :[a , b] →is continuous then
b b
fdy = ∫ f (t ) γ '(t ) dt . a a
Definition: If γ :[a , b] → is a rectifiable path and f is a function defined and continuous
b
on the trace of γ then (line) integral of f along γ is ∫ f ( γ (t ) ) d γ ( t ). This line integral is a
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also denoted by ∫ f = ∫ f (z ) dz .
γ γ
The idea is to recognize all the paths having same trace as identical. The
above definition brings about a partition of the class of all paths.
A curve is an equivalence class of paths. The trace of a curve is the trace of
any one of its members.
A curve is smooth (piecewise smooth) if and only if one of its representative
is smooth (piecewise smooth). A curve C is called simple if it does not cross over itself.
That is γ ( t1 ) ≠ γ (t2 ) whenever t1 ≠ t2 . A curve C is called a simple closed
curve if i) γ ( a ) = γ (b)
(ii) γ ( t1 ) ≠ γ (t2 ) whenever t1 ≠ t2 , except when t1 = a and t 2 = b . It is
also called Jordan curve. Theorem :
Let γ be a rectifiable curve and suppose that f is a function continuous on {γ } .
Then:
a) ∫ f = − ∫ f ; γ −γ
b) ∫
f ≤ ∫
f dz
f (z )
≤ V (γ )sup : z ∈ γ ;
γ γ
c) If c∈ then ∫ f (z ) dz = ∫ f (z − c ) dz . γ γ +c
Theorem : Let G be an open set in and let γ be a rectifiable path in G with initial and
end points α and β respectively. If f :G → is a continuous function with primitive
F :G → , then ∫ f = F ( β ) − F(α ) . γ
Theorem Let ϕ :[a , b ] × [c , d] → be a continuous function. Define g :[c , d] → by
b
g (t ) = ∫ϕ(s , t ) ds , ∀ t ∈ [c , d] a
Then,
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i) g is continuous function and
ii)
∂ϕ
If
exists and continuous, then g is continuously differentiable. ∂t
b ∂
Moreover, g '(t ) = ∫ ∂t ϕ(s , t ) ds . a
Theorem Let f :G → be analytic and suppose B ( a , r ) ⊂ G ( r > 0) . If γ (t ) = a + reit ,
0 ≤ t ≤ 2π , then f (z ) = 1
∫ f ( w)
dw , for
z − a
< r .
2πi w − z
γ
Lemma Let γ be a rectifiable curve in and suppose that Fn and F are continuous
functions on{γ } . If F = u − lim Fn on {γ } then ∫ F = lim ∫ Fn . γ γ
∞
Theorem: Let f be analytic in B ( a , R) . Then f (z ) = ∑an (z − a)n for
z − a
< R n=0
where a n = 1
f ( n)
(a) and this series has radius of convergence ≥ R . n!
∞
Corollary: If f :G → is analytic and a ∈G then f (z ) = ∑an (z − a)n
for
z − a
< R n=0
where R = d ( a , ∂G).
Proof. Since R = d ( a , ∂G ) = inf {d ( a , ∂G ) : z ∈ ∂G } = inf {
z − a
: z ∈ ∂G} , B ( a , R ) ⊂ G
Since f is analytic in G , f is analytic on B ( a , R)
Hence by Taylor’s theorem,
∞
f (z ) = ∑an (z − a)n
for
z − a
< R
n=0
Corollary : If f :G → is analytic then f is infinitely differentiable.
Proof. Suppose f :G → is analytic, then for any a ∈G , f has Taylor’s series expansion
about a
∞
f (z ) = ∑an (z − a)n
for
z − a
< R , R = d ( a , ∂G)
n=0
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Then by theorem, f is infinitely differentiable.
Corollary: If f :G → is analytic and
f ( n)
(a ) = n ! ∫
f ( w)
dw B ( a , r ) ⊂ G then
2πi n+1
γ (w − a)
where γ (t ) = a + reit , 0 ≤ t ≤ 2π
Example : Evaluate the integral ∫ e iz
dz where γ (t ) = reit , 0 ≤ t ≤ 2π .
z 2 γ
Solution: Let f (z ) = e
iz , then f is analytic function
We have
f (1)
(0) = 1! ∫ f (z)
dz
2πi 1+1
γ (z − 0)
iz
ie i 0 =
1! ∫ e dz
2πi 2
γ z
∫ eiz
dz = −2π . z
2
γ
Example: Evaluate the integral ∫ sin z dz where γ (t ) = reit , 0 ≤ t ≤ 2π .
3
γ z
Solution: Let f (z ) = sin z , then f is analytic function.
We have
f (2)
(0) = 2! ∫ f (z)
dz
2πi 2 +1
γ (z − 0)
− sin(0) = 1
∫ sin
z
dz
π i γ z3
Therefore ∫ sin
z
dz = 0 .
γ z3
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Example: Evaluate the integral ∫ 1 dz where γ (t ) = a + reit , 0 ≤ t ≤ 2π .
γ z − a
Solution: Let f (z) = 1, then f is analytic function.
We have
f (0)
(a ) = 0! ∫ f (z)
dz
2πi (z 0 +1
γ − a)
1 = 1
∫ 1
dz
2πi z − a γ
Therefore, ∫ 1
dz = 2πi .
z − a γ
Example: Evaluate the integral ∫ e z + sin z dz where γ (t ) = re
it , 0 ≤ t ≤ 2π .
γ z
Solution: Let f (z ) = e
z + sin z , then f is analytic function.
We have
f (0)
(0) = 1!
∫ f (z) dz
(z 0 +1
2πi γ − 0)
e 0 + sin 0 =
0! ∫
e z + sin z
dz
2πi z γ
Therefore ∫ e z + sin z dz = 2πi .
γ z
Theorem Let f be analytic in B ( a , R) and suppose that
f ( z )
≤ M for all f in B ( a , R) .
Then f ( n)
(a) ≤ n
! M
R
n
Theorem Let f be analytic in the disk B ( a , R) and suppose that γ is a closed rectifiable
curve in B ( a , R) . Then ∫ f = 0 . γ
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i) Let G be connected set and f :G → be analytic function. If f(z) is real for all z in G,
then f is constant function.
ii) f(z) is imaginary number for all z and f(z) with constant modulus.
Definition:
An entire (integral) function is a function which is defined and analytic in the
whole complex plane .
Note : 1. f ( z ) = e z ,sin z ,cos z are entire functions.
2. All polynomials are entire functions.
∞
Theorem : If f is entire function then f has power series expansion f ( z ) = ∑a n zn with
n=0
infinite radius of convergence.
Theorem : If f is bounded and entire function then f is constant.
Proof. Since f is bounded and entire function,
f ( z )
≤ M ∀ z ∈ and for a∈ ,
R > 0 , f is analytic in B ( a , R) .
By Cauchy’s Estimate, we have f
( n)
(a) ≤ n ! M
Rn
Corollary : If f and g are analytic on a region G , then f ≡ g iff { z ∈ G : f ( z ) = g ( z ) }
has a limit point in G .
Corollary : If f is analytic on an open connected set G and f is not identically zero then
for each a in G with f ( a) = 0 there is an integer n ≥ 1 and an analytic function g : G →
such that g ( a) ≠ 0 and f ( z ) = (z − a ) n g ( z) for all z in G . That is each zero of f has finite
multiplicity.
∞
Corollary : If f : G → is analytic and not constant, a ∈G and f ( a) = 0 then there is an
R > 0 such that B ( a; R ) ⊂ G and f ( z) ≠ 0 for 0 < z − a < R . That is zeros of are isolated.
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Maximum Modulus Theorem:
Theorem : Let G is a region and f : G → is an analytic function such that there is a
point a in G with
f ( a )
≥
f ( z )
for all z in G , then f is constant.
Definition : If γ is a closed rectifiable curve in then for a∉{γ } . Then the integer
1 ∫ 1 dz is called the index of γ with respect to the point a . 2πi z − a
γ
Definition : A subset D of a metric space X is called component of X , if it is maximal
connected subset of X.
1) Let f : G C be analytic function defined on a region G with f(a) ≤ f(z) , for all z
in G. Show that either f ≡ 0 or f is constant function.
2) Let f and g be analytic functions defined on the region G. If f.g = 0 on G, then
either f ≡ 0 or g ≡ 0.
Lemma : Let γ be rectifiable curve and suppose φ is a function defined and continuous on
{γ } . For each m ≥ 1 let F ( z ) = φ(w) dw for z ∉{γ } . Then each F is analytic on ∫
γ ( w − z)m
−{γ } and Fm' ( z ) = mFm +1(z) .
Theorem: Let G be an open subset of the plane and γ
1 ,γ 2 ,...,γ m are closed rectifiable curves in G such that
f : G → be an analytic function. If
n (γ 1 ;w) + n (γ 2 ;w) + ...+ n (γ m;w) = 0
for all w in − G , then for a in G −{γ }
m m 1
f ( z)
f ( n)
(a ) ∑ n (γ k ;a ) = n ! ∑ dz 2πi γ
∫ k =1 k=1 z − a
k
Definition : A closed polygonal path having three sides is called triangular path.
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Definition A function f has an isolated singularity at z = a if there is R > 0 such that f
is analytic in B
a; R
,otherwise z = a is non-isolated singularity of f .
Ex.
1
,
z =
1
are isolated singularities and z = 0 is non-isolated singularity.
sin( π
z ) n
Ex. 1 , z = 1,2 are isolated singularities.
( z − 1)( z − 2)
There are three kinds of isolated singularities
A) Removable singularity
B) Pole
C) Essential Singularity
Definition: An isolated singularity at z = a of a function f is removable singularity if
there is R > 0 and an analytic function g : B ( a , R) → such that g ( z ) = f ( z ) in
0 <
z − a
< R .
Ex. f ( z ) = sin z has removable singularity at z = 0.
z
Ex. f ( z ) = z has removable singularity at z = 0.
ez
−1
Theorem : If f has an isolated singularity at z = a , then the point z = a is removable
singularity iff lim( z − a ) f ( z ) = 0. z →a
Corollary An isolated singularity of a function f|z| at z = a is removable singularity iff
f is bounded in the neighborhood of z = a .
Theorem If G is a region with a in G and if f is analytic on G − {a} with pole at z = a ,
then there is a positive integer m and an analytic function g : G → such that
f ( z ) = g ( z )
.
( z − a)m
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The Riemann Mapping Theorem
Definition:
A region G1 is conformally equivalent to the region G2 if there is analytic function f
: G1 ® £ such that f is one-one and f G1 G2 .
This is an equivalence relation.
Definition:
An open set G is simply connected if G is connected and every closed rectifiable
curve in G is homotopic to zero.
Theorem: Let G be an open connected subset of £ . Then the following are equivalent. (a) G is simply connected.
(b) n ; q 0 for every closed rectifiable curve in G and every point a in G - £ .
(c) £ - G is connected. (d) For any f Î H G such that f ( z ) ¹ 0 for all z in G , there is a function g Î H G such
that f z = g z 2 .
Theorem (Open Mapping Theorem)
Let G be a region and suppose that f is a non-constant analytic function on G. Then
for any open set U in G, f (U) is open.
Theorem (Identity Theorem)
Let G be a connected open set and let f : G £ be an analytic function. Then the
following are equivalent statements.
(a) f 0
(b) there is a point a in G such that f ( n)
(a ) 0 for each n ³ 0 .
(c) z Î G : f z = 0 has a limit point in G..
With this basics in our hand we prove the well known Riemann mapping theorem.
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Theorem (Riemann Mapping Theorem)
Let G be a simply connected region which is not the whole plane and let a ÎG . Then
there is a unique analytic function f : G £ having the properties :
(a) f a = 0 and f 'a > 0 ; (b) f is one-one.
(c) f (G ) = z : z < 1 .
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