1) The document discusses the Cauchy integral theorem and formula. The theorem states that if a function f(z) is analytic inside and on a closed curve C, the integral of f(z) over C is equal to 0. 2) The Cauchy integral formula gives a way to evaluate the value of an analytic function F(a) at a point a inside C using a line integral over C. 3) Several examples are provided to demonstrate evaluating integrals using the Cauchy integral theorem and formula, where the integrals are evaluated to be 0 or other values depending on whether the conditions of the theorem are satisfied.

1. Subject :- Complex Variable & Numerical Method
(2141905)
Topic :- Cauchy Integral Theorem & Formula
Branch :- Mechanical
Semester :- 4th
•Made By :-
-Divyangsinh Raj (150990119004)
-Naved Fruitwala (150990119006)
-Utkarsh Gandhi (150990119007)
Guided By :-
Dr. Purvi Naik

2. Cauchy Integral Theorem
• If f(z) is analytic function and f '(z) is
continuous at each point inside and on
closed curve c then,
 
c
0
dz
)
z
(
f

3. Examples
1. Evaluate where c is circle |z| = 2.
f (z) is not analytic at z = - 4. Here c is |z| = 2
which is a circle with center (0,0) and
radios 2.
 
c
4
z
dz
4
z
1
)
z
(
f



4. Here z = - 4 lies outside the curve c. Therefore
f'(z) is analytic inside and on c and also f '(z) is
continuous in z plane.
Therefore by Cauchy integral theorem,
0
4
z
d
z
c




5. 2. Evaluate ; |c| : |z-1|=1
c is circle with center (1,0) and radios 1.
f(z) is not analytic at z=-2i
For |z-1|= |-2i-1 |
 
c
z
i
2
z
e
1
5
1
4





6. z = - 2i lies outside to curve c. So f(z) is
analytic on and inside c. f '(z) is continuous on
and inside c. Therefore by Cauchy integral
theorem,
 

c
z
0
i
2
z
e

7. 3. Evaluate ; |c| : |z|=1
f (z) is not analytic at cos z = 0.
i.e.
Now c is circle with center (0,0) and radios 1.
These points lie outside c. So the f(z) is analytic and
f(z) is continuous. So by Cauchy integral theorem,

c
zdz
sec
z
cos
1
z
sec
)
z
(
f 

,....
2
5
,
2
3
,
2
z







0
zdz
sec
c



8. Cauchy Integral Formula
• If a function F(a) is analytic inside a closed
curve C and if a is any point inside C then,
• 𝐹 𝑎 =
1
2𝜋𝑖
𝐹(𝑥)
𝑧−𝑎
. 𝑑𝑧

9. Cauchy Integral Formula For
Derivative
• If a function F(x) is analytic in region R then its
derivative at any point z=a is also analytic in R
and it is given by
• 𝐹′ 𝑎 =
1
2𝜋𝑖
𝐹(𝑥)
(𝑧−𝑎)2 . 𝑑𝑧
• In general,
• 𝐹𝑛
𝑎 =
𝑛!
2𝜋𝑖
𝐹(𝑥)
(𝑧−𝑎)𝑛+1 . 𝑑𝑧

10. Ex – 1 :-
𝒛
𝒛−𝟐
. 𝒅𝒛 , where C is 𝒛 − 𝟐 =
𝟑
𝟐
• Solution :-
• Here I =
𝑧
𝑧−2
. 𝑑𝑧
• And C is a circle with center (2,0) and radius
3
2
• Here z=2 lies inside C, so let F(z) = z
• F(z) is analytic everywhere inside and on C,
• Now, by Cauchy Integral Formula,
• 𝐹 𝑎 =
1
2𝜋𝑖
𝐹(𝑍)
(𝑍−𝑎)
. 𝑑𝑧

11. Cont...
•
𝐹(𝑧)
(𝑧−𝑎)
. 𝑑𝑧 = 2𝜋𝑖 𝐹(𝑎)
•
𝑧
𝑧−2
. 𝑑𝑧 = 2𝜋𝑖 𝐹(2)
•
𝑧
𝑧−2
. 𝑑𝑧 = 2𝜋𝑖 2
•
𝑧
𝑧−2
. 𝑑𝑧 = 4𝜋𝑖

12. Ex – 2 :-
𝒅𝒛
(𝒛𝟐−𝟕𝒛+𝟏𝟐)
, C is |Z| = 3.5
• Let 𝐼 =
𝑑𝑧
(𝑧2−7𝑧+12)
• 𝐼 =
𝑑𝑧
(𝑧−4)(𝑧−3)
• 𝐼 =
1 (𝑧−4)
(𝑧−3)
. 𝑑𝑧
• Here, C is a circle with center (0,0) and radius 3.5
• Z=3 lies inside C, so let F(z) = 1 (𝑧 − 4)
• F(z) is analytic everywhere inside and on C,
• Now, by Cauchy Integral Formula,

13. Cont...
•
𝐹(𝑧)
(𝑧−𝑎)
. 𝑑𝑧 = 2𝜋𝑖 𝐹(𝑎)
•
𝑑𝑧
(𝑧2−7𝑧+12)
=
2𝜋𝑖 𝐹(3)
•
𝑑𝑧
(𝑧2−7𝑧+12)
=
2𝜋𝑖
1
3−4
•
𝑑𝑧
(𝑧2−7𝑧+12)
= −2𝜋𝑖

14. Ex – 3 :-
𝒔𝒊𝒏𝝅𝒛𝟐+𝒄𝒐𝒔𝝅𝒛𝟐
𝒛−𝟏 (𝒛−𝟐)
. 𝒅𝒛 where C is |z|
= 3
• Here, C is a circle with (0,0) and radius 3
• Now,
1
𝑧−1 (𝑧−2)
=
𝐴
𝑧−1
+
𝐵
(𝑧−2)
• By partial fraction,
• 1 = 𝐴 𝑧 − 2 + 𝐵 𝑧 − 1
• For 𝑧 = 2, 𝐵 = 1
• For 𝑧 = 1, 𝐴 = −1
•
𝑠𝑖𝑛𝜋𝑧2+𝑐𝑜𝑠𝜋𝑧2
𝑧−1 (𝑧−2)
. 𝑑𝑧 = −
𝑠𝑖𝑛𝜋𝑧2+𝑐𝑜𝑠𝜋𝑧2
𝑧−1
. 𝑑𝑧 +
𝑠𝑖𝑛𝜋𝑧2+𝑐𝑜𝑠𝜋𝑧2

15. Cont...
• Here, C is a circle with center (0,0) and radius
3
• Z=1 and 2 lies inside C, so let F(z) = 𝑠𝑖𝑛𝜋𝑧2
+
𝑐𝑜𝑠𝜋𝑧2
• F(z) is analytic everywhere inside and on C,
• Now, by Cauchy Integral Formula,
• 𝐹 𝑎 =
1
2𝜋𝑖
𝐹(𝑧)
(𝑧−𝑎)
. 𝑑𝑧

16. Cont...
• 𝐼 = −2𝜋𝑖 𝐹 1 +
2𝜋𝑖 𝐹(2)
• 𝐼 = −2𝜋𝑖 0 − 1 +
2𝜋𝑖 0 + 1
• 𝐼 = 2𝜋𝑖 + 2𝜋𝑖
• 𝐼 = 4𝜋𝑖

17. Ex – 4 :-
𝐳
(𝐳−𝟏)𝟑 . 𝐝𝐳, C is |z|=2
• Let 𝐼 =
z
(z−1)3 . dz
• 𝐼 =
z
(z−1)2+1 . dz
• Here F(z)=z and a=1 which is on C
• Here, C is a circle with center (0,0) and radius
2
• F(z) is analytic everywhere inside and on C,
• Now by Cauchy Integral Formula For
Derivative,

18. Cont...
•
z
(z−1)3 . dz = 𝐹′′ 1 ×
2𝜋𝑖
2!
• 𝐹 𝑧 = 𝑧
• 𝐹′ 𝑧 = 1
• 𝐹′′ 𝑧 = 0
•
z
(z−1)3 . dz = 0 × 𝜋𝑖
•
z
(z−1)3 . dz = 0