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Complex Integration& VariablesAvradip Ghosh
University of Houston
Complex Integration
We often interpret real integrals in terms of area; now we define complex integrals in terms ofline integrals over paths in the complex plane.
Reference: Mathematical Methods for Physics and Engineering. K.F. Riley ,M.P. Hobson, S.J. Bence, Cambridge University Press
Reference: Mathematical Methods for Physics and Engineering. K.F. Riley ,M.P. Hobson, S.J. Bence, Cambridge University Press
Reference: Mathematical Methods for Physics and Engineering. K.F. Riley ,M.P. Hobson, S.J. Bence, Cambridge University Press
An Important Result
M is an upper bound on thevalue of |f(z)| on the path, i.e. |f(z)| β€ M on C
L is the length of the path C
What Is an Analytic Function
A function that is single-valued and differentiable at all points of a domain Ris said to be analytic (or regular) in R.
A function may be analytic in a domain except at a finite number of points (or an infinite number if the domain isinfinite);
Such Finite points are called
Singularities
Cauchy-Riemann Relations (to check for analyticity)
real imaginary
If we have a function that is
divided into its real and imaginary parts
to check for analyticity (differentiability)
Necessary and sufficient condition is that the four partial derivativesexist, are continuous and satisfy the CauchyβRiemann relations
Reference: Mathematical Methods for Physics and Engineering. K.F. Riley ,M.P. Hobson, S.J. Bence, Cambridge University Press
Power Series and Convergence
Convergence meaning:
The sum of the series either goes
to 0 or a finite value as the
terms goes to β
substituting
Converges if: Is also convergent
Radius and Circle of Convergence
R
|z|<R⦠Convergent|z|>R⦠Divergent
|z|=R⦠Inconclusive
Laurent Series
In mathematics, the Laurent series of a complex function f(z) is a representation of that function as a power series which includes even
terms of negative degree. It may be used to express complex functions in cases where a Taylor series expansion cannot be applied.
π(π§) =
π=ββ
β
ππ π§ β π π
ππ =1
2ππ
πΎ
)π(π§
π§ β π π+1 ππ§Nth Coefficient :
π(π§) =)π0(π§
π§ β π§0π=
πππ§ β π§0
π+. . . . . . . .
πβ1π§ β π§0
+ π0 + π1(π§ β π§0) + π1 π§ β π§02+. . . .
Singularities, Zeros, and Poles
the point z=Ξ± is called a singular point, or singularity of the complex function f(z) if f is not analytic at z=Ξ± but every neighbourhood of it
If a function f(z) is not analytic at z=Ξ±but analytic everywhere in the entire
plane the point z=Ξ± is an Isolated Singularity of the function f(z)
Isolated Singularity:
Types of Singularities
π(π§) =1
π§ β 1 2
Pole of order 2 at z=1
π(π§) = π Ξ€1 π₯ = 1 +1
π₯+
1
2π₯2+
1
6π₯3
Removable Essential Pole
Zeros of a Complex Function
Reference: Mathematical Methods for Physics and Engineering. K.F. Riley ,M.P. Hobson, S.J. Bence, Cambridge University Press
Cauchyβs Theorem
Note: For Proof Refer to Riley Hobson
Cauchyβs Residue Theorem
If we have a closed contour Χ―πΆ and there
are poles inside the contour then we can close the contour around the poles
(only poles that are inside the contour)
Take the residue at the selected poles
Select the poles according to the boundary conditions and
the specific problem
π(π§) =1
π§ + π§1)(π§ β π§2
Z1,Z2>0
Most Important
Calculating Residueβs
For a pole of Order 1
π(π§) =)π0(π§
π§ β π§0=
πβ1π§ β π§0
+ π0 + π1(π§ β π§0) + π1 π§ β π§02+. . . .
Residue
Reπ πππ’π|ππππ=π§0 = limπ§βπ§0
(π§ β π§0)π(π§).
πΆ
π(π§) = 2ππ
πππππ
Reπ πππ’ππ Final
Answer
Laurent Series
OR
Calculating Residueβs
For a pole of Order βnβ
Residue
πΆ
π(π§) = 2ππ
πππππ
Reπ πππ’ππ Final
Answer
π(π§) =)π0(π§
π§ β π§0π=
πππ§ β π§0
π+. . . . . . . .
πβ1π§ β π§0
+ π0 + π1(π§ β π§0) + π1 π§ β π§02+. . . .
Reπ πππ’π|ππππ=π§0 =1
(π β 1)!limπ§βπ§0
αππβ1
ππ§πβ1π§ β π§0
ππ(π§
Laurent Series
OR
Taking contour only in the upper half plane
Proof of Cauchyβs Residue theorem by
Laurent Series
π(π§) =)π0(π§
π§ β π§0π=
πππ§ β π§0
π+. . . . . . . .
πβ1π§ β π§0
+ π0 + π1(π§ β π§0) + π1 π§ β π§02+. . . .Laurent Series
Nth Coefficient : ππ =1
2ππ
πΆ
)π(π§
π§ β π§0π+1 ππ§
We are interested in
this
So we take n=-1,Thus π§ β π§0
0 = 1
πβ1 =1
2ππΰΆ»
πΆ
)π(π§ ππ§
ΰΆ»
πΆ
)π(π§ ππ§ = (2ππ)πβ1
Thus we are so interested in the
Residue: coefficient of the
power π§ β π§0
β1
Must See Videos
β’ https://youtu.be/T647CGsuOVU
β’ (You think you know about Complex Numbers: Decide after watching the above video: max 2 per day)
Other Videos (For Contour Integration)
https://youtu.be/b5VUnapu-qs
https://youtu.be/_3p_E9jZOU8
ReferencesMathematical Methods for
Physics and Engineering. K.F. Riley ,M.P. Hobson, S.J.
Bence, Cambridge University Press
Mathematical Methods for Physicists
A Comprehensive Guide,George B. Arfken, Hans J.
Weber and Frank E. Harris, Elsevier
Thank You