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
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
0dz)z(f
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
4z
dz
4z
1
)z(f


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
4z
dz
c


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
i2z
e
15
14


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
i2z
e
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
zdzsec
zcos
1
zsec)z(f 
,....
2
5
,
2
3
,
2
z






0zdzsec
c

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
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,
• 𝐹 𝑛
𝑎 =
𝑛!
2𝜋𝑖
𝐹(𝑥)
(𝑧−𝑎) 𝑛+1 . 𝑑𝑧
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
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
Ex – 2 :-
𝒅𝒛
(𝒛 𝟐−𝟕𝒛+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,
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
Ex – 3 :-
𝒔𝒊𝒏𝝅𝒛 𝟐+𝒄𝒐𝒔𝝅𝒛 𝟐
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
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
Cont...
• 𝐼 = −2𝜋𝑖 𝐹 1 +
2𝜋𝑖 𝐹(2)
• 𝐼 = −2𝜋𝑖 0 − 1 +
2𝜋𝑖 0 + 1
• 𝐼 = 2𝜋𝑖 + 2𝜋𝑖
• 𝐼 = 4𝜋𝑖
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,
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
Cauchy integral theorem & formula (complex variable & numerical method )

Cauchy integral theorem & formula (complex variable & numerical method )

  • 1.
    Subject :- ComplexVariable & 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 0dz)z(f
  • 3.
    Examples 1. Evaluate wherec 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 4z dz 4z 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 4z dz 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 i2z e 15 14  
  • 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 i2z 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 zdzsec zcos 1 zsec)z(f  ,.... 2 5 , 2 3 , 2 z       0zdzsec 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 FormulaFor 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, • 𝐹 𝑛 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 (𝒛 𝟐−𝟕𝒛+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:- 𝒔𝒊𝒏𝝅𝒛 𝟐+𝒄𝒐𝒔𝝅𝒛 𝟐 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, Cis 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