lec37.ppt
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1) The document discusses evaluating contour integrals using the residue theorem. It provides examples of calculating residues and evaluating integrals where the contour encloses poles. 2) The residue of a function f(z) at a pole z=a is the coefficient of the (z-a)^-1 term in the Laurent series expansion of f(z) about z=a. 3) According to the residue theorem, the value of a contour integral of a function along a closed loop is equal to 2i times the sum of the residues of the function enclosed by the contour.
![NPTEL Physics Mathematical Physics - 1
1
1 =
2
()
Since the last expression involves only the coefficient 1, we call 1 as the residue of
() at = .
To obtain the residue of a function () at = , it may appear that the Laurent
expansion of () about = must
be obtained. However, for
= , a pole of the order , there is a simple formula for 1 given by,
1 = 瑞
р ( 1)! ю鐘1
1 鐘1
[( )情()]
For the case of a simple pole,
1 = 瑞 ( )()
р
Proof of the above Residue Theorem
It can be shown easily,
() = () + () + ()
1 2 3
21 +
Hence the required result.
21 + 21
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![NPTEL Physics Mathematical Physics - 1
Example 1. If () =
respectively.
(р1)(+1)2 , = 1 and = 1 are poles of order one and two
Residue at = 1, 瑞 ( 1) {
р1
(р1)(+1)2 4
} = 1
р
Residue at = 1, 瑞
[( + 1)2
(р1)(+1)2
] = 1
4
Let () be single valued and analytic inside and on the simple closed curve C except the
singularity at , , ..
() = 2(1 + 1 + 1 + )
Exemple 2. If f(z) = then z = 0 is an essential singularity. From the known formula
for expansion,
1
1 = 1
1
+ +
22 63
1 1
The residue at = 0 is coefficient of 1
which is 1.
Exemple 3. Obtain the Laurent expansion of
1
(р2)2 around = 0 and = 1
() = (1 )2 = (1 + 2 + 32 + 43 + . . ) = + 2 + 3 + 42,
1 1
1
0 < || < 1
Put 1 = 4, () = 1
(1 + 4)1 = 1
[1 4 + 42 43 +
. ]
42 42
=
1 1
(р1)2 р1
+ 1 ( 1)
..
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lec37.ppt
- 1. NPTEL Physics Mathematical Physics - 1 Lecture 37 (2+2)2 Example: Evaluate where is a circle || = 4 The poles of (2+2)2 (р)2(+)2 = are at = 賊 (of order 2) Residue at = 瑞 р 1! 1 {( )2 ( )2( + )2 } = + 43 Similarly, Residue at = 1 = 43 (2+2)2 = 2 (Summing the residues) = 2 ( + + ) 43 43 = The Residue Theorem: Evaluation of the integrals: Let () be single valued and analytic inside and on a circle C except at the point. = chosen as the center of . Then () has a Laurent series about = given by, () = ( ) = = 0 + 1( ) + 2( )2 + + 1 2 р (р)2 + Where is given by, 1 () = 2 (р)+1 Page 31 of 66 Joint initiative of IITs and IISc Funded by MHRD = 0, 賊1, 賊2 . . In the special case when n = -1 we have,
- 2. NPTEL Physics Mathematical Physics - 1 1 1 = 2 () Since the last expression involves only the coefficient 1, we call 1 as the residue of () at = . To obtain the residue of a function () at = , it may appear that the Laurent expansion of () about = must be obtained. However, for = , a pole of the order , there is a simple formula for 1 given by, 1 = 瑞 р ( 1)! ю鐘1 1 鐘1 [( )情()] For the case of a simple pole, 1 = 瑞 ( )() р Proof of the above Residue Theorem It can be shown easily, () = () + () + () 1 2 3 21 + Hence the required result. 21 + 21 Page 32 of 66 Joint initiative of IITs and IISc Funded by MHRD
- 3. NPTEL Physics Mathematical Physics - 1 Example 1. If () = respectively. (р1)(+1)2 , = 1 and = 1 are poles of order one and two Residue at = 1, 瑞 ( 1) { р1 (р1)(+1)2 4 } = 1 р Residue at = 1, 瑞 [( + 1)2 (р1)(+1)2 ] = 1 4 Let () be single valued and analytic inside and on the simple closed curve C except the singularity at , , .. () = 2(1 + 1 + 1 + ) Exemple 2. If f(z) = then z = 0 is an essential singularity. From the known formula for expansion, 1 1 = 1 1 + + 22 63 1 1 The residue at = 0 is coefficient of 1 which is 1. Exemple 3. Obtain the Laurent expansion of 1 (р2)2 around = 0 and = 1 () = (1 )2 = (1 + 2 + 32 + 43 + . . ) = + 2 + 3 + 42, 1 1 1 0 < || < 1 Put 1 = 4, () = 1 (1 + 4)1 = 1 [1 4 + 42 43 + . ] 42 42 = 1 1 (р1)2 р1 + 1 ( 1) .. Page 33 of 66 Joint initiative of IITs and IISc Funded by MHRD
- 4. NPTEL Physics Mathematical Physics - 1 Example 4. Evaluate I = (216) Where C is the boundary of the annulus between the circles || = 1, || = 3 So, = 2 + (1) And in the region between 1 and 2 the function, () = (216) is analytic (its derivative exists and is continuous). The only non analytic points are at z = 0, 賊4 which are not included in the region. Hence = 0. Exemple 5. = 1 2 ( ) = 1, 2 The function () = then = 0. 1 (р) is analytic for all . Thus if does not include = , If includes a, then we have to deform it. Page 34 of 66 Joint initiative of IITs and IISc Funded by MHRD
- 5. NPTEL Physics Mathematical Physics - 1 The contour 駒 is small but finite circle of radius r centered at = () () = 0 Thus, () = () Let us evaluate this. The RHS is Let = = = 1 1 2 ( ) 2 = 1 1 2 0 = 1 2 (1)+1 = 2 0 ,1 = 1 for = 1 = 0 otherwise So = 0 even if the pole at a is inside C for 1 Page 35 of 66 Joint initiative of IITs and IISc Funded by MHRD
- 6. NPTEL Physics Mathematical Physics - 1 Example. 6. = +2 ( +1) where C is a circle. || = 2 + 2 ( + 1) + 1 = + + 2 = ( + 1) + 巨 Thus = 2, = 1 = (2 +1 1 ) = 2(2 1) = 2 Page 36 of 66 Joint initiative of IITs and IISc Funded by MHRD