lec36.ppt
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The document summarizes key results from complex analysis including: Cauchy's integral theorem which states that if f(z) is analytic in a simply connected region, the integral around any closed curve in that region is zero. Morera's theorem, the converse of Cauchy's theorem, which states that if the integral is zero around every closed curve, then the function is analytic. Cauchy's integral formula which expresses the value of an analytic function at a point inside a contour as a contour integral. The generalization of Cauchy's integral formula for functions with poles inside the contour.
![NPTEL – Physics – Mathematical Physics - 1
𝑓′(𝑎) = ∮ [
1 1 1 1
2𝜋𝑖 ℎ 𝑧 − (𝑎 + ℎ) 𝑧 − 𝑎
𝑐
− ] 𝑓(𝑧)𝑑𝑧
=
1 𝑓(𝑧)𝑑
𝑧
2𝜋𝑖 𝑐 (𝑧−𝑎−ℎ)(𝑧−𝑎)
∮
The result follows as ℎ → 0.
Important formula
If 𝑓(𝑧) is analytic inside and on a simple closed curve 𝐶 except for a pole of order m
at 𝑧 = 𝑎, inside 𝐶, prove that
1 1
2𝜋𝑖
∮ 𝑓(𝑧)𝑑𝑧 = 𝐿𝑡
𝑐
𝑧→𝑎 (𝑚 − 1)! 𝑑𝑧𝑚−1
𝑑𝑚
[(𝑧 − 𝑎)𝑚𝑓(𝑧)]
also for two different poles of order 𝑚1 and 𝑚2,
1 1
2𝜋𝑖
∮ 𝑓(𝑧)𝑑𝑧 = 𝐿𝑡
𝑧→𝑎1 (𝑚1 − 1)! 𝑑𝑧𝑚1−1
𝑐
𝑑𝑚1
[(𝑧 − 𝑎)𝑚1 𝑓(𝑧)]
+ 𝐿𝑡
𝑧→𝑎2 (𝑚2 − 1)! 𝑑𝑧𝑚2−1
Joint initiative of IITs and IISc – Funded by MHRD Page 30 of 66
1 𝑑𝑚2
[(𝑧 − 𝑎)𝑚2 𝑓(𝑧)]
A similar formula can be worked out for more poles.](https://image.slidesharecdn.com/lec36-231023100404-fe4db3bb/85/lec36-ppt-6-320.jpg)
lec36.ppt
- 1. NPTEL – Physics – Mathematical Physics - 1 Lecture 36 Caucly-Goursat theorem Let 𝑓(𝑧) the analytic in a region R and on it’s boundary C. then ∮ 𝑓(𝑧)𝑑𝑧 = 0 → Cauchy’s integral theorem Morera’s Theorem: Let 𝑓(𝑧) be continuous in a simply connected and suppose that ∮ 𝑓(𝑧)𝑑𝑧 = 0 around every simple closed curve C in R. Then 𝑓(𝑧) is analytic. This is converse of Cauchy’s theorem. Example. 1. Evaluate ∫ 𝑧∗ 𝑑𝑧 for the curve 𝐶 = 𝐶1 − 𝐶2 𝐶 ∫ 𝑧∗𝑑𝑧 = ∫ (𝑥 − 𝑖𝑦)(𝑑𝑥 + 𝑖𝑑𝑦) 𝐶1−𝐶2 𝐶1− 𝐶2 1 1 = ∫𝑥=0 𝑥𝑑𝑥 + ∫𝑦= (1 − 𝑖𝑦)(𝑖 𝑑𝑦) 0 = 1 + 𝑖 ∫ 𝑧∗𝑑𝑧 = ∫ 1 (𝑥 − 𝑖𝑥)(𝑑𝑥 + 𝑖𝑑𝑥) 𝑥=0 𝐶3 = 1 Since 𝑧∗ is not analytic, the results are not same. 𝑢𝑥 = 𝑣𝑦 and 𝑢𝑦 = −𝑣𝑥 are not satisfied for 𝑓(𝑧) = 𝑧 ∗. Then f (z) not exist. does Joint initiative of IITs and IISc – Funded by MHRD Page 25 of 66
- 2. NPTEL – Physics – Mathematical Physics - 1 Example.2. Evaluate ∫ 1 𝐶 𝑧 𝑑𝑧 for (a) a simple contour C not enclosing the origin. (b) a simple contour C enclosing the origin. (a) Because is analytic for all points (other then 𝑧 = 0), So 1 𝑧 ∫ 1 𝐶 𝑧 𝑑𝑧 = 0 (b) The contour can be deformed as above. In the limit of 𝑟 ⟶ 0 (of 𝐶1) and the lengths of 𝐿1 and 𝐿2 tending to zero, we have a closed curve. 𝐶 = 𝐶2 + 𝐿1 + 𝐿2 − 𝐶1 1 Now as earlier ∫ 𝑧 𝑑𝑧 = 0 Joint initiative of IITs and IISc – Funded by MHRD Page 26 of 66
- 3. NPTEL – Physics – Mathematical Physics - 1 as ∫𝐿1 + ∫𝐿2 = 0 (𝑧 = 𝑟𝑒𝑖𝜃 , 𝑑𝑧 = 𝑟𝑖𝑒𝑖𝜃 𝑑𝜃 ) 2𝜋 ∫ 𝑑𝑧 = ∫ 𝑑𝑧 = ∫ 𝑟−1 𝑒−𝑖𝜃𝑖𝑒𝑖𝜃𝑟𝑑𝜃 1 1 𝑧 𝐶2 𝐶1 𝑧 0 = 2𝜋𝑖 Thus the integral ∫ 1 𝑧 𝑑𝑧 around any closed curve that encloses the origin is zero. Consequences of Cauchy’s integral theorem 𝑏 1. If 𝑓(𝑧) is analytic in a simply connected region R, prove that ∫ 𝑓(𝑧)𝑑𝑧 is 𝑎 independent of the path in R joining any two points. A and B in R. 2. Let 𝑓(𝑧) be analytic in a region R bounded by two simple closed curves 𝐶1and 𝐶2 and also on 𝐶1, 𝐶2. Prove that ∮𝐶 𝑓(𝑧)𝑑𝑧 = ∮𝐶 𝑓(𝑧)𝑑𝑧 1 2 Since 𝑓(𝑧) is analytic everywhere in R, ∫ 𝐴𝐸𝐹𝐺𝐻𝐸𝐴𝐷𝐶𝐵𝐴 𝑓(𝑧)𝑑𝑧 = 0 ∫ 𝑓(𝑧)𝑑𝑧 + ∫ 𝑓(𝑧)𝑑𝑧 + ∫ 𝑓(𝑧)𝑑𝑧 + 𝐴𝐸 𝐸𝐹𝐺𝐻 𝐸𝐴 ∫ 𝑓(𝑧)𝑑𝑧 = 0 𝐴𝐷𝐶𝐵𝐴 Thus, ∫𝐸𝐹𝐺𝐻 𝑓(𝑧)𝑑𝑧 = − ∫𝐴𝐷𝐶𝐵𝐴 𝑓(𝑧)𝑑𝑧 ∮𝐶 𝑓(𝑧)𝑑𝑧 = ∮𝐶 𝑓(𝑧)𝑑𝑧 1 2 Joint initiative of IITs and IISc – Funded by MHRD Page 27 of 66
- 4. NPTEL – Physics – Mathematical Physics - 1 Prove Cauchy’s Theorem (Cauchy Goursat theorem) The theorem states that, ∮𝐶 𝑓(𝑧)𝑑𝑧 = 0, if f(z) is analytic with derivatives 𝑓′(𝑧) which is continuous at all points inside and on the simple closed curve C. 𝑓(𝑧) = 𝑢 + 𝑖𝑣 is analytic and has a continuous derivative. 𝑓́(𝑧) = 𝜕𝑢 + 𝑖 𝜕𝑣 = 𝜕𝑣 − 𝑖 𝜕𝑢 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 It follows that the partial derivatives, 𝜕𝑢 = 𝜕𝑣 , 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 and 𝜕𝑣 = − 𝜕𝑢 are continuous inside and on C. Thus, ∮ 𝑓(𝑧)𝑑𝑧 = ∮𝐶 (𝑢 + 𝑖𝑣)(𝑑𝑥 + 𝑖𝑑𝑦) = ∮(𝑢𝑑𝑥 − 𝑣𝑑𝑦) + 𝑖 ∮𝐶 (𝑣𝑑𝑥 + 𝑢𝑑𝑦) Using Green’s theorem, ∫ (− 𝜕𝑣 − 𝜕𝑢 ) 𝑑𝑥𝑑𝑦 + 𝑖 ∫ ( 𝜕𝑢 − 𝜕𝑣 ) 𝑑𝑥𝑑𝑦 𝑅 𝜕𝑥 𝜕𝑦 𝑅 𝜕𝑥 𝜕𝑦 = 0 (Using CR Condition) Cauchy’s Integral formula If 𝑓(𝑧) is analytic inside and on the boundary C of a simply connected region R, prove the following, 𝑓(𝑎) = 1 𝑓(𝑧) ∮ 𝑑𝑧 (Cauchy’s Integral formula) 2𝜋𝑖 𝐶 (𝑧−𝑎) The function is analytic inside R and on C excepting the point. 𝑧 = 𝑎 𝑓(𝑧) 𝑧−𝑎 ∮ 𝑓(𝑧) 𝐶 𝑧−𝑎 Γ 𝑧−𝑎 𝑑𝑧 = ∮ 𝑑𝑧 𝑓(𝑧) (1) The equation for Γ is |𝑧 − 𝑎| = 𝜖 Joint initiative of IITs and IISc – Funded by MHRD Page 28 of 66
- 5. NPTEL – Physics – Mathematical Physics - 1 so, 𝑧 − 𝑎 = 𝜖𝑒𝑖𝜃, 0 ≤ 𝜃 < 2𝜋 𝑧 − 𝑎 Substituting 𝑧 = 𝑎 + 𝜖𝑒𝑖𝜃, 𝑑𝑧 = 𝑖𝜖𝑒𝑖𝜃𝑑𝜃 2𝜋 ∮ Γ 𝑓(𝑧) 𝑑𝑧 = ∫ 0 𝑓(𝑎 + 𝜖𝑒𝑖𝜃 )𝑖𝜖𝑒𝑖 𝜃 𝜖𝑒𝑖 𝜃 𝑑𝜃 2𝜋 = i∫ 𝑓(𝑎 + 𝜖𝑒𝑖𝜃)𝑑𝜃 0 (2) Thus using (1) ∮ 𝑓(𝑧) 𝐶 𝑧−𝑎 0 𝑑𝑧 = 𝑖 ∫ 𝑓(𝑎 + 𝜖𝑒𝑖𝜃 )𝑑 𝜃 2𝜋 Taking limit 𝜖 → 0 (remember 𝑓(𝑧) is continuous). ∮ 𝐶 𝑓(𝑧)𝑑𝑧 𝑧 − 𝑎 = 𝐿𝑖𝑚 𝑖 ∫ 𝑓(𝑎 + 𝜖𝑒𝑖𝜃)𝑑𝜃 = 𝑖 ∫ 𝑓(𝑎)𝑑𝜃 𝜖→0 2𝜋 2𝜋 0 0 = 2𝜋𝑖𝑓(𝑎) Example: If 𝑓(𝑧) is analytic inside and on the boundary C of a simply connected region R, prove that, 𝑓′(𝑎) = ∮ 1 𝑓(𝑧) 2𝜋𝑖 (𝑧 − 𝑎)2 𝑐 𝑑𝑧 𝑓′(𝑎) = 𝐿𝑖𝑚 𝑓(𝑎+ℎ)−𝑓(𝑎) ℎ→0 ℎ Consider both the points 𝑎 and 𝑎 + ℎ are inside the circle of radius 𝜖. Joint initiative of IITs and IISc – Funded by MHRD Page 29 of 66
- 6. NPTEL – Physics – Mathematical Physics - 1 𝑓′(𝑎) = ∮ [ 1 1 1 1 2𝜋𝑖 ℎ 𝑧 − (𝑎 + ℎ) 𝑧 − 𝑎 𝑐 − ] 𝑓(𝑧)𝑑𝑧 = 1 𝑓(𝑧)𝑑 𝑧 2𝜋𝑖 𝑐 (𝑧−𝑎−ℎ)(𝑧−𝑎) ∮ The result follows as ℎ → 0. Important formula If 𝑓(𝑧) is analytic inside and on a simple closed curve 𝐶 except for a pole of order m at 𝑧 = 𝑎, inside 𝐶, prove that 1 1 2𝜋𝑖 ∮ 𝑓(𝑧)𝑑𝑧 = 𝐿𝑡 𝑐 𝑧→𝑎 (𝑚 − 1)! 𝑑𝑧𝑚−1 𝑑𝑚 [(𝑧 − 𝑎)𝑚𝑓(𝑧)] also for two different poles of order 𝑚1 and 𝑚2, 1 1 2𝜋𝑖 ∮ 𝑓(𝑧)𝑑𝑧 = 𝐿𝑡 𝑧→𝑎1 (𝑚1 − 1)! 𝑑𝑧𝑚1−1 𝑐 𝑑𝑚1 [(𝑧 − 𝑎)𝑚1 𝑓(𝑧)] + 𝐿𝑡 𝑧→𝑎2 (𝑚2 − 1)! 𝑑𝑧𝑚2−1 Joint initiative of IITs and IISc – Funded by MHRD Page 30 of 66 1 𝑑𝑚2 [(𝑧 − 𝑎)𝑚2 𝑓(𝑧)] A similar formula can be worked out for more poles.