1. CHAUDHARY BANSI Lal university
Cauchy’s Integral Formula And
Poisson Integral Formula
PRESENTED TO: PRESENTED BY:
DEPARTMENT OF RASHMI
MATHEMATICS M.Sc (MATHEMATICS)
2nd SEM
220000603038
2. CAUCHY’S INTEGRAL FORMULA
Statement:- If f(z) is analytic function within and on a closed contour C, and if 𝑧0 is any point inside C then
f(𝑧0) =
1
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz
Proof :- To prove the formula describe a circle V about the center 𝑧0 of small radius r such that this circle |z-
𝑧0|=r does not intersect the curve C.
The function
𝑓(𝑧)
𝑧−𝑧0
is analytic in the region bounded by C and V.
Then from Cauchy theorem , we have
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz =
𝑉
𝑓(𝑧)
(𝑧−𝑧0)
dz …………….(1)
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz =
𝑉
𝑓 𝑧 −𝑓(𝑧0)
(𝑧−𝑧0)
dz +
𝑉
𝑓(𝑧0)
(𝑧−𝑧0)
dz ……………..(2)
Since f(z) is analytic within C and so it is continuous at z = 𝑧0
So , that given 𝜀 >0 , there exits 𝛿 > 0 such that
|f(z) - f( 𝑧0)|= 𝜀 …( 3) for |z- 𝑧0|= 𝛿 ……………..(4)
3. Since r is at our choice and we can take r < 𝛿 so that (4) is satisfied for all z on the
circle V
For any point z on V , z-𝑧0 = r𝑒𝑖𝜃
hence dz = r𝑖𝑒𝑖𝜃
d𝜃
𝑉
𝑓(𝑧0)
(𝑧−𝑧0)
dz = 0
2𝜋 𝑓(𝑧0)
𝑟𝑒𝑖𝜃 r𝑖𝑒𝑖𝜃
d𝜃 = 2π𝑖f(𝑧0)
Hence by (2)
|
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz - 2π𝑖f(𝑧0) | = |
𝑉
𝑓 𝑧 −𝑓(𝑧0)
(𝑧−𝑧0)
dz | ≤
𝑉
|𝑓 𝑧 −𝑓(𝑧0)|
|(𝑧−𝑧0)|
|dz|
∈
𝑟 𝑉
|dz| =
∈
𝑟
2𝜋𝑟
= 2𝜋𝜖
Since , 𝜖 is arbitrary small making 𝜖 → 0
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz - 2π𝑖f(𝑧0) = 0
f(𝑧0) =
1
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz
Hence the proof of the theorem
4. Example:- Using Cauchy integral formula calculate
𝒄
𝒛
(𝟗−𝒛𝟐)(𝒛+𝒊)
dz , where C is
the circle |z|=2 described in the positive sense
Solution :- Let f(z) =
𝑧
9−𝑧2
Therefore z = 3 and 3 lies outside the circle hence the function is analytic
within and on the contour C
And z = - 𝑖 lies inside C
Hence by Cauchy integral formula , f(𝑧0) =
1
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz
f(- 𝑖) =
1
2𝜋𝑖
𝑐
𝑧
(𝟗−𝒛𝟐)(𝑧+1)
dz = 2𝜋𝑖 𝑓 −𝑖
= 2𝜋𝑖 [
−𝑖
9−(−𝑖)2]
= 2𝜋𝑖 [
−𝑖
10
]
=
2𝜋
10
5. Some More Extensions Of Cauchy Integral Formula
• Cauchy integral formula for multiply connected region:-
Statement :- if f(z) is analytic in a ring shaped bounded by two closed contours C₁
and C₂ and 𝑧0 is a point in the region between C₁ and C₂ . Then
f(𝑧0) =
1
2𝜋𝑖
𝐶1
𝑓(𝑧)
(𝑧−𝑧0)
dz -
1
2𝜋𝑖
𝑐2
𝑓(𝑧)
(𝑧−𝑧0)
dz where C₂ is the outer curve
• Cauchy integral formula for the derivative of an analytic function:-
Statement :- if f(z) is analytic within and on a closed contour C and 𝑧0 is any point
lying in it . Then
𝑓′(𝑧0) =
1
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−𝑧0)2 dz
• Cauchy integral formula for the higher order derivative of an analytic function:-
Statement :- if f(z) is analytic within and on a closed contour C and 𝑧0 is any point
within C . Then derivative of all orders are analytic and are given by
𝑓n(𝑧0) =
𝑛!
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−𝑧0)𝑛+1 dz
6. Example :- evaluate
𝒄
𝒆𝟐𝒛
(𝒛+𝟏)𝟒 dz where c is |z|=3
Solution :- we know that 𝑓n
(𝑧0) =
𝑛!
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−𝑧0)𝑛+1 dz ……………(1)
Let f(z) = 𝑒2𝑧
is analytic within and on C and z = -1 lies inside C . Taking n = 3
in (1)
𝑓3
(-1) =
3!
2𝜋𝑖
𝑐
𝑒2𝑧
(𝑧+1)4 dz …… (2)
f(z) = 𝑒2𝑧
, 𝑓1
(z) = 2𝑒2𝑧
,𝑓2
(z) = 4𝑒2𝑧
, 𝑓3
(z) = 8𝑒2𝑧
therefore 𝑓3
(-1) = 8𝑒−2
, put this value in (2)
8𝑒−2
=
3
𝜋𝑖
𝑐
𝑒2𝑧
(𝑧+1)4 dz
𝑐
𝑒2𝑧
(𝑧+1)4 dz =
8𝜋𝑖
3𝑒2
7. POISSON INTEGRAL FORMULA
STATEMENT :- let(z) be analytic in the region |z| ≤ R , Then for 0 < r < R we have
f(r𝑒𝑖𝜃
) =
1
2𝜋 0
2𝜋 𝑅2−𝑟2 𝑓(R𝑒𝑖∅)
𝑅2−2𝑟𝑅𝑐𝑜𝑠 𝜃−∅ +𝑟2 d∅
where ∅ is the value of 𝜃 on the circle |z|= R
PROOF :- Let C denote the circle |z|= R and let 𝑧0 = r𝑒𝑖𝜃
, 0 < r < R be any point inside C . Then by Cauchy’s
integral formula
f(𝑧0) =
1
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−𝑧0)
dz …………………(1)
The inverse of 𝑧0 w.r.t the circle |z|= R is
𝑅2
𝑧0
and lies outside the circle
So by Cauchy’s theorem , we have
0 =
1
2𝜋𝑖
𝑐
𝑓(𝑧)
(𝑧−
𝑅2
𝑧0
)
dz …………………………(2)
Subtracting (2) from (1) , we get
f(𝑧0) =
1
2𝜋𝑖
𝑐
𝑓(𝑧)(𝑧0−
𝑅2
𝑧0
)
(𝑧−𝑧0)(𝑧−
𝑅2
𝑧0
)
dz ……………………..(3)
8. Now any point on the circle C is expressible as z =R𝑒𝑖∅ and also 𝑧0= r𝑒𝑖𝜃 so
𝑧0 = r𝑒−𝑖𝜃
Therefore 𝑅2- 𝑧0𝑧0 = 𝑅2- 𝑟2 ……………..(4)
Now (𝑧 − 𝑧0)(𝑅2- 𝑧0𝑧0) = z𝑅2- 𝑧2𝑧0- 𝑧0𝑅2+z𝑧0𝑧0
= 𝑅3𝑒𝑖∅- 𝑅2𝑒2𝑖∅r𝑒−𝑖𝜃-r 𝑅2𝑒𝑖𝜃+R𝑟2𝑒𝑖∅
= R𝑒𝑖∅[𝑅2 - Rr𝑒𝑖∅𝑒−𝑖𝜃-r 𝑒𝑖𝜃 𝑒−𝑖∅R + 𝑟2]
= R𝑒𝑖∅
[𝑅2
- Rr(𝑒𝑖(𝜃−∅)
+ 𝑒𝑖(∅−𝜃)
) + 𝑟2
]
= R𝑒𝑖∅[𝑅2 - Rr(𝑒−𝑖(𝜃−∅)+ 𝑒𝑖(𝜃−∅)) + 𝑟2 ]
= R𝑒𝑖∅[𝑅2 - Rr(cos (𝜃 − ∅)- 𝑖sin(𝜃 − ∅)+cos(𝜃 − ∅)+ 𝑖sin (𝜃 − ∅))+𝑟2 ]
= R𝑒𝑖∅[𝑅2 - 2Rrcos (𝜃 − ∅)+ 𝑟2 ] ………..(5)
Thus equation (3) becomes
f(r𝑒𝑖𝜃) =
1
2𝜋𝑖 0
2𝜋 𝑅2−𝑟2 𝑓 R𝑒𝑖∅ 𝑅𝑒𝑖∅
R𝑒𝑖∅[𝑅2 − 2Rrcos (𝜃−∅)+ 𝑟2 ]
d∅
f(r𝑒𝑖𝜃) =
1
2𝜋 0
2𝜋 𝑅2−𝑟2 𝑓(R𝑒𝑖∅)
𝑅2−2𝑟𝑅𝑐𝑜𝑠 𝜃−∅ +𝑟2 d∅
Which is the required formula
9. REFRENCES
• H.A Priestly , introduction to complex analysis
• Real and Complex analysis by Walter Rudh
• Complex analysis by H.D Pathak
