Contour Integrals of Functions of a Complex Variable · Introduction 1.Complex functions of a...

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Contours Contour Integrals Examples

Contour Integrals of Functions of a ComplexVariable

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Contour Integrals of Functions of a Complex Variable

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Introduction

1. Complex functions of a complex variable are usually integratedalong parametric curves.

2. The integrals are ultimately reduced to integrals of complexfunctions of a real variable as introduced in the previouspresentation.



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Introduction1. Complex functions of a complex variable are usually integrated

along parametric curves.

2. The integrals are ultimately reduced to integrals of complexfunctions of a real variable as introduced in the previouspresentation.



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Introduction1. Complex functions of a complex variable are usually integrated

along parametric curves.2. The integrals are ultimately reduced to integrals of complex

functions of a real variable as introduced in the previouspresentation.



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Definitions

A set C of points (x,y) in the complex plane is called an arc if andonly if there are continuous functions x(t) and y(t) with a≤ t ≤ b sothat for every point (x,y) in C there is a t so that x = x(t) and y = y(t).

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DefinitionsA set C of points (x,y) in the complex plane is called an arc if andonly if there are continuous functions x(t) and y(t) with a≤ t ≤ b sothat for every point (x,y) in C there is a t so that x = x(t) and y = y(t).

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Definitions

With the usual switching between two-dimensional notation andcomplex notation, we also write the continuous function asz(t) = x(t)+ iy(t).



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DefinitionsWith the usual switching between two-dimensional notation andcomplex notation, we also write the continuous function asz(t) = x(t)+ iy(t).



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Definitions

An arc C is a Jordan arc or a simple arc if and only if for t1 6= t2 wehave z(t1) 6= z(t2).

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DefinitionsAn arc C is a Jordan arc or a simple arc if and only if for t1 6= t2 wehave z(t1) 6= z(t2).

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Definitions

An arc C is called a simple closed curve if, except for z(a) = z(b) wehave that t1 6= t2 implies z(t1) 6= z(t2).

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DefinitionsAn arc C is called a simple closed curve if, except for z(a) = z(b) wehave that t1 6= t2 implies z(t1) 6= z(t2).

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Definitions

A simple closed curve is called positively oriented if and only if it istraversed in the counterclockwise (mathematically positive) direction.

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DefinitionsA simple closed curve is called positively oriented if and only if it istraversed in the counterclockwise (mathematically positive) direction.

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Definitions

An arc C is called smooth if and only if the function z(t) that traversesC is differentiable with continuous derivative and z′(t) 6= 0 for all t.

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DefinitionsAn arc C is called smooth if and only if the function z(t) that traversesC is differentiable with continuous derivative and z′(t) 6= 0 for all t.

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Definitions

An arc is called a contour or a piecewise smooth arc if and only if itconsists of smooth arcs joined end-to-end. It is called a simple closedcontour if and only if there is no self-intersection except that theinitial point equals the final point.

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DefinitionsAn arc is called a contour or a piecewise smooth arc if and only if itconsists of smooth arcs joined end-to-end. It is called a simple closedcontour if and only if there is no self-intersection except that theinitial point equals the final point.

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U>U

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(x(a),y(a)

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r(x(b),y(b)

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O.k. for contours.



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Example.

With z(θ) = eiθ and 0≤ θ ≤ 2π , the unit circleC = {z ∈ C : |z|= 1} is a positively oriented simple closed curve.

-

6ℑ(z)

ℜ(z)

K



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Example. With z(θ) = eiθ and 0≤ θ ≤ 2π , the unit circleC = {z ∈ C : |z|= 1} is a positively oriented simple closed curve.

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6ℑ(z)

ℜ(z)

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Example.

With z(θ) = e−iθ and 0≤ θ ≤ 2π , the unit circleC = {z ∈ C : |z|= 1} is a simple closed curve, but it is not positivelyoriented.

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Example. With z(θ) = e−iθ and 0≤ θ ≤ 2π , the unit circleC = {z ∈ C : |z|= 1} is a simple closed curve, but it is not positivelyoriented.

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Definition.

The length of a contour C parametrized by z(t) is

L :=∫ b

a

∣∣z′(t)∣∣ dt.

Discussion.

L =∫ b

a

∣∣z′(t)∣∣ dt

=∫ b

a

√(x′(t)

)2 +(y′(t)

)2 dt

which is the length formula from multivariable calculus.



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Definition. The length of a contour C parametrized by z(t) is

L :=∫ b

a

∣∣z′(t)∣∣ dt.

Discussion.

L =∫ b

a

∣∣z′(t)∣∣ dt

=∫ b

a

√(x′(t)

)2 +(y′(t)

)2 dt

which is the length formula from multivariable calculus.



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Definition.

Let f be a continuous function of a complex variable andlet C be a contour with parametrization z(t). Then we define thecontour integral of f over C as∫

Cf (z) dz :=

∫ b

af (z(t))z′(t) dt.



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Definition. Let f be a continuous function of a complex variable andlet C be a contour with parametrization z(t).

Then we define thecontour integral of f over C as∫

Cf (z) dz :=

∫ b




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Definition. Let f be a continuous function of a complex variable andlet C be a contour with parametrization z(t). Then we define thecontour integral of f over C as∫

Cf (z) dz

:=∫ b




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Definition. Let f be a continuous function of a complex variable andlet C be a contour with parametrization z(t). Then we define thecontour integral of f over C as∫

Cf (z) dz :=

∫ b




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Note.

If φ is a differentiable function with continuous, nonzeroderivative that maps the interval [α,β ] to the interval [a,b], thenz(φ(τ)) is another parametrization of C. It just uses the parameter τ in[α,β ] rather than the parameter t in [a,b]. But the integral of f over Cis not affected by interchanging parametrizations in this fashion,because with ξ (τ) := z(φ(τ)) we have∫

β

α

f (ξ (τ))ξ ′(τ) dτ =∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt

Therefore, the definition of the contour integral is sensible, as it onlydepends on the shape of the contour, not on the way we parametrizeit. (Omitted proof that any two parametrizations “differ” by a φ asabove.)



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Note. If φ is a differentiable function with continuous, nonzeroderivative that maps the interval [α,β ] to the interval [a,b], thenz(φ(τ)) is another parametrization of C.

It just uses the parameter τ in[α,β ] rather than the parameter t in [a,b]. But the integral of f over Cis not affected by interchanging parametrizations in this fashion,because with ξ (τ) := z(φ(τ)) we have∫

β

α

f (ξ (τ))ξ ′(τ) dτ =∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt




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Note. If φ is a differentiable function with continuous, nonzeroderivative that maps the interval [α,β ] to the interval [a,b], thenz(φ(τ)) is another parametrization of C. It just uses the parameter τ in[α,β ] rather than the parameter t in [a,b].

But the integral of f over Cis not affected by interchanging parametrizations in this fashion,because with ξ (τ) := z(φ(τ)) we have∫

β

α

f (ξ (τ))ξ ′(τ) dτ =∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt




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Note. If φ is a differentiable function with continuous, nonzeroderivative that maps the interval [α,β ] to the interval [a,b], thenz(φ(τ)) is another parametrization of C. It just uses the parameter τ in[α,β ] rather than the parameter t in [a,b]. But the integral of f over Cis not affected by interchanging parametrizations in this fashion,because with ξ (τ) := z(φ(τ)) we have

∫β

α

f (ξ (τ))ξ ′(τ) dτ =∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt




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Note. If φ is a differentiable function with continuous, nonzeroderivative that maps the interval [α,β ] to the interval [a,b], thenz(φ(τ)) is another parametrization of C. It just uses the parameter τ in[α,β ] rather than the parameter t in [a,b]. But the integral of f over Cis not affected by interchanging parametrizations in this fashion,because with ξ (τ) := z(φ(τ)) we have∫

β

α

f (ξ (τ))ξ ′(τ) dτ

=∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt




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β

α

f (ξ (τ))ξ ′(τ) dτ =∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt




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β

α

f (ξ (τ))ξ ′(τ) dτ =∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt

Therefore, the definition of the contour integral is sensible, as it onlydepends on the shape of the contour, not on the way we parametrizeit.

(Omitted proof that any two parametrizations “differ” by a φ asabove.)



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β

α

f (ξ (τ))ξ ′(τ) dτ =∫

β

α

f (z(φ(τ)))z′(φ(τ))φ ′(τ) dτ

=∫ b

af (z(t))z′(t) dt




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Rules.

1. If w is a complex number, then∫

Cwf (z) dz = w

∫C

f (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.

3. Let −C denote the same contour as C, only traversed in theopposite direction. Then z(−τ) with −b≤ τ ≤−a is aparametrization and∫

−Cf (z) dz =

∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a

bf (z(t))z′(t) dt =−

∫C

f (z) dz

4. If the endpoint of C1 is the starting point of C2, then the union ofthe two contours in denoted C := C1 +C2 and we have∫

Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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Rules.1. If w is a complex number, then

∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.


−Cf (z) dz =

∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a


∫C

f (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.

3. Let −C denote the same contour as C, only traversed in theopposite direction.

Then z(−τ) with −b≤ τ ≤−a is aparametrization and∫

−Cf (z) dz =

∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a


∫C

f (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.

3. Let −C denote the same contour as C, only traversed in theopposite direction. Then z(−τ) with −b≤ τ ≤−a is aparametrization and

∫−C

f (z) dz =∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a


∫C

f (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.


−Cf (z) dz

=∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a


∫C

f (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.


−Cf (z) dz =

∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ

=∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a


∫C

f (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.


−Cf (z) dz =

∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a


∫C

f (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.


−Cf (z) dz =

∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a

bf (z(t))z′(t) dt

=−∫

Cf (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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∫C

wf (z) dz = w∫

Cf (z) dz.

2.∫

C(f +g)(z) dz =

∫C

f (z) dz+∫

Cg(z) dz.


−Cf (z) dz =

∫ −a

−bf (z(−τ))

ddτ

z(−τ) dτ =∫ −a

−bf (z(−τ))z′(−τ)(−1) dτ

=∫ a


∫C

f (z) dz


Cf (z) dz =

∫C1

f (z) dz+∫

C2

g(z) dz.



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Example.

Compute the contour integral of f (z) = |z| around theupper half of the positively oriented unit circle.∫

Cf (z) dz =

∫π

0

∣∣eit∣∣ ieit dt

=∫

π

0ieit dt

= eit∣∣∣π0

= eiπ − e0 =−2



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Example. Compute the contour integral of f (z) = |z| around theupper half of the positively oriented unit circle.

∫C

f (z) dz =∫

π

0


=∫

π

0ieit dt

= eit∣∣∣π0

= eiπ − e0 =−2



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Example. Compute the contour integral of f (z) = |z| around theupper half of the positively oriented unit circle.∫

Cf (z) dz

=∫

π

0


=∫

π

0ieit dt

= eit∣∣∣π0

= eiπ − e0 =−2



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Cf (z) dz =

∫π

0

∣∣eit∣∣

ieit dt

=∫

π

0ieit dt

= eit∣∣∣π0

= eiπ − e0 =−2



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Cf (z) dz =

∫π

0


=∫

π

0ieit dt

= eit∣∣∣π0

= eiπ − e0 =−2



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Cf (z) dz =

∫π

0


=∫

π

0ieit dt

= eit∣∣∣π0

= eiπ − e0

=−2



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Cf (z) dz =

∫π

0


=∫

π

0ieit dt

= eit∣∣∣π0

= eiπ − e0 =−2



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Theorem.

If the complex function f (z) has an antiderivative F(z),then the integral of f over the contour C parametrized with z(t),a≤ t ≤ b is equal to∫

Cf (z) dz = F(z(b))−F(z(a)).

Proof. ∫C

f (z) dz =∫ b

af(z(t))z′(t) dt

= F(z(t))∣∣∣b

a= F(z(b))−F(z(a))



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Theorem. If the complex function f (z) has an antiderivative F(z),then the integral of f over the contour C parametrized with z(t),a≤ t ≤ b is equal to∫

Cf (z) dz = F(z(b))−F(z(a)).

Proof. ∫C

f (z) dz =∫ b

af(z(t))z′(t) dt

= F(z(t))∣∣∣b

a= F(z(b))−F(z(a))



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Cf (z) dz = F(z(b))−F(z(a)).

Proof.

∫C

f (z) dz =∫ b

af(z(t))z′(t) dt

= F(z(t))∣∣∣b

a= F(z(b))−F(z(a))



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Cf (z) dz = F(z(b))−F(z(a)).

Proof. ∫C

f (z) dz =∫ b

af(z(t))z′(t) dt

= F(z(t))∣∣∣b

a= F(z(b))−F(z(a))



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Cf (z) dz = F(z(b))−F(z(a)).

Proof. ∫C

f (z) dz =∫ b

af(z(t))z′(t) dt

= F(z(t))∣∣∣b

a

= F(z(b))−F(z(a))



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Cf (z) dz = F(z(b))−F(z(a)).

Proof. ∫C

f (z) dz =∫ b

af(z(t))z′(t) dt

= F(z(t))∣∣∣b

a= F(z(b))−F(z(a))



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Example.

Let n ∈ C(!) Integrate the function f (z) = zn around thepositively oriented unit circle.∫

Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi

In particular, for n an integer not equal to −1, the integral is zero.Note that the computation is not affected by the fact that the contourcrosses a branch cut. We choose a branch and stay consistent with it.A value at a single point (where the power function is discontinuous)does not affect an integral.



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Example. Let n ∈ C

(!) Integrate the function f (z) = zn around thepositively oriented unit circle.∫

Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




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Example. Let n ∈ C(!)

Integrate the function f (z) = zn around thepositively oriented unit circle.∫

Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




logo1


Example. Let n ∈ C(!) Integrate the function f (z) = zn around thepositively oriented unit circle.

∫C

zn dz =∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




logo1


Example. Let n ∈ C(!) Integrate the function f (z) = zn around thepositively oriented unit circle.∫

Czn dz

=∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt

=∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0

(n 6=−1)∫C

z−1 dz =∫ 2π

0i dt = 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)

∫C

z−1 dz =∫ 2π

0i dt = 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz

=∫ 2π

0i dt = 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt

= 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi

In particular, for n an integer not equal to −1, the integral is zero.

Note that the computation is not affected by the fact that the contourcrosses a branch cut. We choose a branch and stay consistent with it.A value at a single point (where the power function is discontinuous)does not affect an integral.



logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi

In particular, for n an integer not equal to −1, the integral is zero.Note that the computation is not affected by the fact that the contourcrosses a branch cut.

We choose a branch and stay consistent with it.A value at a single point (where the power function is discontinuous)does not affect an integral.



logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi

In particular, for n an integer not equal to −1, the integral is zero.Note that the computation is not affected by the fact that the contourcrosses a branch cut. We choose a branch and stay consistent with it.

A value at a single point (where the power function is discontinuous)does not affect an integral.



logo1



Czn dz =

∫ 2π

0

(eit)n

ieit dt =∫ 2π

0iei(n+1)t dt

=1

n+1ei(n+1)t

∣∣∣2π

0(n 6=−1)∫

Cz−1 dz =

∫ 2π

0i dt = 2πi




Contour Integrals of Functions of a Complex Variable · Introduction 1.Complex functions of a...

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