This document presents Cauchy's integral formula and Poisson integral formula. Cauchy's integral formula expresses a analytic function inside a contour as a contour integral over the boundary. Poisson integral formula expresses the value of an analytic function inside a circle in terms of its values on the outer circle boundary. Examples are included to demonstrate evaluating integrals using these formulas. Higher order derivatives and extensions to multiply connected regions are also discussed.

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

10. THANK YOU
FOR YOUR ATTENTION