lec37.ppt
Download as PPT, PDF0 likes33 views
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 2πi times the sum of the residues of the function enclosed by the contour.
![NPTEL – Physics – Mathematical Physics - 1
1
𝑎−1 =
2𝜋𝑖
∮ 𝑓(𝑧)𝑑𝑧
Since the last expression involves only the coefficient 𝑎−1, we call 𝑎−1 as the residue of
𝑓(𝑧) at 𝑧 = 𝑎.
To obtain the residue of a function 𝑓(𝑧) at 𝑧 = 𝑎 , it may appear that the Laurent
expansion of 𝑓(𝑧) about 𝑧 = 𝑎 must
be obtained. However, for
𝑧 = 𝑎, a pole of the order 𝐾, there is a simple formula for 𝑎−1 given by,
𝑎−1 = 𝐿𝑖𝑚
𝑧→𝑎 (𝐾 − 1)! 𝑑𝑧𝐾−1
1 𝑑𝐾−1
[(𝑧 − 𝑎)𝐾𝑓(𝑧)]
For the case of a simple pole,
𝑎−1 = 𝐿𝑡 (𝑧 − 𝑎)𝑓(𝑧)
𝑧→𝑎
Proof of the above Residue Theorem
It can be shown easily,
∮𝑐 𝑓(𝑧) = ∮𝑐 𝑓(𝑧)𝑑𝑧 + ∮𝑐 𝑓(𝑧)𝑑𝑧 + ∮𝑐 𝑓(𝑧)𝑑𝑧
1 2 3
2𝜋𝑖𝑎−1 +
Hence the required result.
2𝜋𝑖𝑏−1 + 2𝜋𝑖𝑐−1
Page 32 of 66
Joint initiative of IITs and IISc – Funded by MHRD](https://image.slidesharecdn.com/lec37-231023101030-f1feee55/85/lec37-ppt-2-320.jpg)
![NPTEL – Physics – Mathematical Physics - 1
Example 1. If 𝑓(𝑧) =
respectively.
𝑧
(𝑧−1)(𝑧+1)2 , 𝑧 = 1 and 𝑧 = −1 are poles of order one and two
Residue at 𝑧 = 1, 𝐿𝑡 (𝑧 − 1) {
𝑧→1
𝑧
(𝑧−1)(𝑧+1)2 4
} = 1
𝑧→𝑎 𝑑𝑧
Residue at 𝑧 = −1, 𝐿𝑡 𝑑
[(𝑧 + 1)2 𝑧
(𝑧−1)(𝑧+1)2
] = − 1
4
Let 𝑓(𝑧) be single valued and analytic inside and on the simple closed curve C except the
singularity at 𝑎, 𝑏, 𝑐 … … ..
∮ 𝑓(𝑧)𝑑𝑧 = 2𝜋𝑖(𝑎−1 + 𝑏−1 + 𝑐−1 + … … … … )
Exemple 2. If f(z) = 𝑒 then z = 0 is an essential singularity. From the known formula
for expansion,
−1⁄𝑧
𝑒−1⁄𝑧 = 1 −
1
+ − + … … …
… 𝑧 2𝑧2 6𝑧3
1 1
The residue at 𝑧 = 0 is coefficient of 1
which is −1.
𝑧
Exemple 3. Obtain the Laurent expansion of
1
𝑧(𝑧−2)2 around 𝑧 = 0 and 𝑧 = 1
𝑓(𝑧) = (1 − 𝑧)−2 = (1 + 2𝑧 + 3𝑧2 + 4𝑧3 + … . . ) = + 2 + 3𝑧 + 4𝑧2,
1 1
𝑧 𝑧
1
𝑧
0 < |𝑧| < 1
Put 𝑧 − 1 = 4, 𝑓(𝑧) = 1
(1 + 4)−1 = 1
[1 − 4 + 42 − 43 + …
… … … . ]
42 42
=
1 1
(𝑧−1)2 𝑧−1
− + 1 − (𝑧 − 1) … … …
..
Page 33 of 66
Joint initiative of IITs and IISc – Funded by MHRD](https://image.slidesharecdn.com/lec37-231023101030-f1feee55/85/lec37-ppt-3-320.jpg)
lec37.ppt
- 1. NPTEL – Physics – Mathematical Physics - 1 Lecture 37 (𝑧2+𝜋2)2 Example: Evaluate ∮ 𝑐 𝑒𝑧 𝑑𝑧 where 𝐶 is a circle |𝑧| = 4 The poles of 𝑒𝑧 (𝑧2+𝜋2)2 (𝑧−𝜋𝑖)2(𝑧+𝜋𝑖)2 = 𝑒𝑧 are at 𝑧 = ±𝜋𝑖 (of order 2) Residue at 𝑧 = 𝜋𝑖 𝐿𝑖𝑚 𝑧→𝜋𝑖 1! 𝑑𝑧 1 𝑑 {(𝑧 − 𝜋𝑖)2 𝑒𝑧 (𝑧 − 𝜋𝑖)2(𝑧 + 𝜋𝑖)2 } = 𝜋+𝑖 4𝜋3 Similarly, Residue at 𝑧 = −𝜋𝑖 𝑎−1 = 𝜋−𝑖 4𝜋3 ∮ 𝑒𝑧 (𝑧2+𝜋2)2 𝑑𝑧 = 2𝜋𝑖 (Summing the residues) = 2𝜋𝑖 ( 𝜋+𝑖 + 𝜋−𝑖 ) 4𝜋3 4𝜋3 = 𝑖 𝜋 The Residue Theorem: Evaluation of the integrals: Let 𝑓(𝑧) be single valued and analytic inside and on a circle C except at the point. 𝑧 = 𝑎 chosen as the center of 𝐶. Then 𝑓(𝑧) has a Laurent series about 𝑧 = 𝑎 given by, 𝑓(𝑧) = ∑∞ 𝑎𝑛(𝑧 − 𝑎)𝑛 𝑛=−∞ = 𝑎0 + 𝑎1(𝑧 − 𝑎) + 𝑎2(𝑧 − 𝑎)2 + … … … + 𝑎−1 𝑎−2 𝑧−𝑎 (𝑧−𝑎)2 + … … … Where 𝑎𝑛 is given by, 𝑎 1 𝑓(𝑧) 𝑛 = 2𝜋𝑖 ∮𝐶 (𝑧−𝑎)𝑛+1 𝑑𝑧 Page 31 of 66 Joint initiative of IITs and IISc – Funded by MHRD 𝑛 = 0, ±1, ±2 … … … … . . In the special case when n = -1 we have,
- 2. NPTEL – Physics – Mathematical Physics - 1 1 𝑎−1 = 2𝜋𝑖 ∮ 𝑓(𝑧)𝑑𝑧 Since the last expression involves only the coefficient 𝑎−1, we call 𝑎−1 as the residue of 𝑓(𝑧) at 𝑧 = 𝑎. To obtain the residue of a function 𝑓(𝑧) at 𝑧 = 𝑎 , it may appear that the Laurent expansion of 𝑓(𝑧) about 𝑧 = 𝑎 must be obtained. However, for 𝑧 = 𝑎, a pole of the order 𝐾, there is a simple formula for 𝑎−1 given by, 𝑎−1 = 𝐿𝑖𝑚 𝑧→𝑎 (𝐾 − 1)! 𝑑𝑧𝐾−1 1 𝑑𝐾−1 [(𝑧 − 𝑎)𝐾𝑓(𝑧)] For the case of a simple pole, 𝑎−1 = 𝐿𝑡 (𝑧 − 𝑎)𝑓(𝑧) 𝑧→𝑎 Proof of the above Residue Theorem It can be shown easily, ∮𝑐 𝑓(𝑧) = ∮𝑐 𝑓(𝑧)𝑑𝑧 + ∮𝑐 𝑓(𝑧)𝑑𝑧 + ∮𝑐 𝑓(𝑧)𝑑𝑧 1 2 3 2𝜋𝑖𝑎−1 + Hence the required result. 2𝜋𝑖𝑏−1 + 2𝜋𝑖𝑐−1 Page 32 of 66 Joint initiative of IITs and IISc – Funded by MHRD
- 3. NPTEL – Physics – Mathematical Physics - 1 Example 1. If 𝑓(𝑧) = respectively. 𝑧 (𝑧−1)(𝑧+1)2 , 𝑧 = 1 and 𝑧 = −1 are poles of order one and two Residue at 𝑧 = 1, 𝐿𝑡 (𝑧 − 1) { 𝑧→1 𝑧 (𝑧−1)(𝑧+1)2 4 } = 1 𝑧→𝑎 𝑑𝑧 Residue at 𝑧 = −1, 𝐿𝑡 𝑑 [(𝑧 + 1)2 𝑧 (𝑧−1)(𝑧+1)2 ] = − 1 4 Let 𝑓(𝑧) be single valued and analytic inside and on the simple closed curve C except the singularity at 𝑎, 𝑏, 𝑐 … … .. ∮ 𝑓(𝑧)𝑑𝑧 = 2𝜋𝑖(𝑎−1 + 𝑏−1 + 𝑐−1 + … … … … ) Exemple 2. If f(z) = 𝑒 then z = 0 is an essential singularity. From the known formula for expansion, −1⁄𝑧 𝑒−1⁄𝑧 = 1 − 1 + − + … … … … 𝑧 2𝑧2 6𝑧3 1 1 The residue at 𝑧 = 0 is coefficient of 1 which is −1. 𝑧 Exemple 3. Obtain the Laurent expansion of 1 𝑧(𝑧−2)2 around 𝑧 = 0 and 𝑧 = 1 𝑓(𝑧) = (1 − 𝑧)−2 = (1 + 2𝑧 + 3𝑧2 + 4𝑧3 + … . . ) = + 2 + 3𝑧 + 4𝑧2, 1 1 𝑧 𝑧 1 𝑧 0 < |𝑧| < 1 Put 𝑧 − 1 = 4, 𝑓(𝑧) = 1 (1 + 4)−1 = 1 [1 − 4 + 42 − 43 + … … … … . ] 42 42 = 1 1 (𝑧−1)2 𝑧−1 − + 1 − (𝑧 − 1) … … … .. Page 33 of 66 Joint initiative of IITs and IISc – Funded by MHRD
- 4. NPTEL – Physics – Mathematical Physics - 1 Example 4. Evaluate I = ∮ 𝑒𝑧 𝑐 𝑧(𝑧2−16) 𝑑𝑧 Where C is the boundary of the annulus between the circles |𝑧| = 1, |𝑧| = 3 So, 𝐶 = 𝐶2 + (−𝐶1) And in the region between 𝐶1 and 𝐶2 the function, 𝑓(𝑧) = 𝑧(𝑧2−16) is analytic (its 𝑒𝑧 derivative exists and is continuous). The only non analytic points are at z = 0, ±4 which are not included in the region. Hence 𝐼 = 0. Exemple 5. 𝐼 = ∮ 1 𝑑𝑧 2𝜋𝑖 (𝑧 − 𝑎)𝑚 𝑐 𝑚 = 1, 2 … … … … … 𝑀 The function 𝑓(𝑧) = then 𝐼 = 0. 1 (𝑧−𝑎)𝑚 is analytic for all 𝑧 ≠ 𝑎. Thus if 𝐶 does not include 𝑧 = 𝑎, If 𝐶 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 𝐶𝑎 is small but finite circle of radius r centered at 𝑧 = 𝑎 ∮ 𝑓(𝑧)𝑑𝑧 − ∫ 𝑓(𝑧)𝑑𝑧 = 0 𝑐𝑎 𝑐 Thus, ∫𝑐 𝑓(𝑧)𝑑𝑧 = ∫𝑐 𝑓(𝑧)𝑑𝑧 𝑎 Let us evaluate this. The RHS is Let 𝑧 – 𝑎 = 𝑟𝑒𝑖𝜃 𝑑𝑧 = 𝑖𝑒𝑖𝜃𝑟𝑑𝜃 𝐼 = 1 1 2𝜋𝑖 (𝑧 − 𝑎)𝑚 2𝜋𝑖 𝑐𝑎 ∮ 𝑑𝑧 = ∫ 1 1 𝑟𝑚 𝑒𝑖𝑚 𝜃 2𝜋 0 𝑖𝑒𝑖𝜃 𝑟𝑑 𝜃 = 1 2𝜋 ∫ 𝑖𝑒−𝑖(𝑚−1)𝜃𝑟−𝑚+1𝑑𝜃 = 𝛿 2𝜋𝑖 0 𝑚,1 = 1 for 𝑚 = 1 = 0 otherwise So 𝐼 = 0 even if the pole at a is inside C for 𝑚 ≠ 1 Page 35 of 66 Joint initiative of IITs and IISc – Funded by MHRD
- 6. NPTEL – Physics – Mathematical Physics - 1 Example. 6. 𝐼 = ∮ 𝑧+2 𝑐 𝑧(𝑧 +1) 𝑑𝑧 where C is a circle. |𝑧| = 2 𝑧 + 2 𝐴 𝐵 𝑧(𝑧 + 1) 𝑧 𝑧 + 1 = + 𝑧 + 2 = 𝐴(𝑧 + 1) + 𝐵𝑧 Thus 𝐴 = 2, 𝐵 = −1 𝐼 = ∮ (2 − 𝑧 𝑧+1 1 ) 𝑑𝑧 = 2𝜋𝑖(2 − 1) = 2𝜋𝑖 Page 36 of 66 Joint initiative of IITs and IISc – Funded by MHRD 𝑐