lec34.ppt
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1. The document discusses analytic functions of complex variables through examples. It defines analytic functions as those whose derivatives of all orders exist in the region of analyticity. 2. The Cauchy-Riemann equations are derived and their implications are explored, including that they imply the Laplace equation and orthogonality of level curves. 3. Several examples are worked through to determine if functions are analytic by checking if they satisfy the Cauchy-Riemann equations. The Cauchy-Riemann equations are also derived in polar coordinates.
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lec34.ppt
- 1. NPTEL β Physics β Mathematical Physics - 1 Lecture 34 Analytic function of a complex variable (continued) Example 1. 1. The function π§3 is defined everywhere. So it is analytic also. (It may vanish, but should not blow up). It is analytic for all z. 2. π§ β3 is a triple valued function 1 ππ ππ π π β3, β3 π β3, ππ ππ 2ππ π π β3 π β3 4ππ It is not defined everywhere and hence not analytic. Example 2. π(π§) = π§3, π’ = π₯3 β 3π₯π¦2, π£ = 3π₯2π¦ β π¦3 π’π₯ = 3π₯2 β 3π¦2, π’π¦ = β6π₯π¦ π’π¦ = 3π₯2 β 3π¦2 π£π₯ = 6π₯π¦ Example 3. π(π§) = ππ§ is analytic in the entire finite z plane, whereas π(π§) = π§Μ is analytic nowhere. π§2 The function π(π§) = 1 is analytic for all nonzero z. (punctured z plane). Joint initiative of IITs and IISc β Funded by MHRD Page 16 of 66 Example 4. Determine whether π(π§) is analytic when π(π§) = (π₯ + πΌπ¦)2 + 2π(π₯ β πΌπ¦) for πΌ = real, constant π’(π₯, π¦) = (π₯ + πΌπ¦)2, π£(π₯, π¦) = 2(π₯ β πΌπ¦) π’π₯ = 2(π₯ + πΌπ¦), π£π¦ = β2πΌ π’π¦ = 2πΌ(π₯ + πΌπ¦), π£π₯ = 2 The CR conditions are satisfied for only πΌ2 = 1 and on the lines π₯ Β± π¦ = Β±1, because the derivative πβ²(π§) only exists there on these lines. So π(π§) is not analytic anywhere since it is not analytic in the neighbourhood of these lines. The word analytic is used in the same spirit as holomorphic.
- 2. NPTEL β Physics β Mathematical Physics - 1 A singular point π§0 is a point where f fails to be analytic. Thus π(π§) = π§2 has π§ = 0 as a singular point on the other hand π(π§) = π§Μ is analytic nowhere and has singular point everywhere in the complex plane. An analytic function, has derivatives of all orders, that is πβ², πβ²β², πβ²β²β² etc. in the region of analyticity and that the real and imaginary parts have continuous derivatives of all orders as well, that is all orders of derivatives of π’ and π£ exist. 1 Starting with π’π₯ = π£π¦, and π’π¦ = βπ£π₯ Taking second derivatives, π2π’ π2π£ π2π£ ππ₯2 = ππ₯ππ¦ , = β ππ₯ππ¦ ππ¦2 π2π’ Hence ββ 2π’ = π π’ + π π’ = 0 2 ππ₯2 ππ¦2 2 Similarly ββ 2π£ = β v + β v = 0 2 2 βx2 βy2 So π’ πππ π£ satisfy Laplaceβs equation ππ£ = ππ’ ππ£ = β ππ’ ππ¦ ππ₯ ππ₯ ππ¦ and A consequence of the CR condition is the two curves π’(π₯, π¦) = πΆ1 and π£(π₯, π¦) = πΆ2 are orthogonal ββ π’ = iΛ βπ’ + βπ₯ jΜ ππ’ ππ¦ ββ π£ = iΛ βπ£ + βπ₯ βπ¦ Λj β π£ ββ π’ . ββ π£ = βπ’ βπ£ + ππ’ βπ£ Joint initiative of IITs and IISc β Funded by MHRD Page 17 of 66 βπ₯ βπ₯ ππ¦ βy = βππ’ ππ’ ππ’ ππ’ ππ₯ ππ¦ + ππ¦ ππ₯ = 0 The gradients are perpendicular to each other. Hence the curves are also perpendicular to each other.
- 3. NPTEL β Physics β Mathematical Physics - 1 Example 5. Prove that π’ = πβπ₯(π₯π πππ¦ β π¦πππ π¦) is harmonic. Hence find v such that π(π§) = π’ + ππ£ is analytic Ans. π2π’ π2π’ ππ₯2 ππ¦ 2 + = 0 Use = = πβπ₯π πππ¦ β π₯πβπ₯π πππ¦ + π¦πβπ₯πππ π¦ ππ’ ππ¦ ππ₯ ππ’ (1) ππ£ = β ππ’ = πβπ₯πππ π¦ β π₯πβπ₯πππ π¦ β π¦πβπ₯π πππ¦ (2) ππ₯ ππ¦ Integrate (1) with respect to y keeping x constant π£ = βπβπ₯πππ π¦ + π₯πβπ₯πππ π¦ + πβπ₯(π¦π πππ¦ + πππ π¦) + πΉ(π₯) = π¦πβπ₯π πππ¦ + π₯πβπ₯πππ π¦ + πΉ(π₯) (3) πΉ(π₯) is an arbitrary real function of x. From (3) calculate and substitute in (2), ππ’ ππ₯ βπ¦πβπ₯π πππ¦ β π₯πβπ₯πππ π¦ + πβπ₯πππ π¦ + πΜ (π₯) = βπ¦πβπ₯π πππ¦ β π₯πβπ₯πππ π¦ β π¦πβπ₯π πππ¦ Thus, πΉβ²(π₯) = 0 So, πΉ(π₯) = πΆ, a constant π = πβπ₯(π¦π πππ¦ + π₯πππ π¦) + πΆ Example 6 πΉβ²(π§) = ππ₯ + π ππ₯ ππ’ ππ£ Using CR conditions = β ππ£ ππ₯ ππ’ ππ¦ πΉβ²(π§) = ππ’ β π ππ’ ππ₯ ππ¦ When u(x, y) is given, use this to calculate πΉβ²(π§). This is actually πΉβ²(π₯, 0). Integrate this to calculate πΉ(π§). Now separate the real and imaginary parts. Joint initiative of IITs and IISc β Funded by MHRD Page 18 of 66
- 4. NPTEL β Physics β Mathematical Physics - 1 Example 7. Prove CR conditions in terms of polar coordinates Proof: π₯ = ππππ π, π¦ = ππ πππ ; π = βπ₯2 + π¦2, π‘πππ = π¦ π₯ ππ’ = π’ = ππ’ ππ ππ’ π π ππ₯ π₯ + ππ ππ₯ ππ ππ₯ = ππ’ π πππ β ππ 1 ππ’ = ππ’ 1 ππ πππ ππ ππ πππ π ππ πππ π π β 1 ππ’ ---------------------- ------- (1) Similarly = π£π¦ = ππ£ ππ¦ ππ£ ππ ππ£ π π ππ ππ¦ ππ π π¦ + = 1 ππ£ 1 π πππ ππ ππππ π π π + ππ£ ------------------------------------------- (2) Now CR conditions in cartesian coordinates read, π’π₯ = π£π¦ ------------------------------------------------------ (3) Substituting (1) and (2) in (3) (ππ’ β 1 ππ£ ) π πππ = (ππ£ + 1 ππ£ ) πππ π ----------------------------------- (4) ππ π ππ ππ π ππ Using the other CR condition π’π¦ = βπ£π₯ (ππ’ β 1 ππ£ ) πππ π = β (ππ£ + 1 ππ’ ) π πππ -------------------------------- ------- (5) ππ π ππ ππ π ππ Multiplying (4) by cππ π and (5) by π πππ and adding, ----------------------------------------------------------------------- (6) Again multiplying (4) by π πππ and (5) by πππ π and substract, ---------------------- ------------- (7) (6) and (7) are CR conditions in the polar form. ππ’ = 1 π π£ ππ π π π ππ’ = 1 π π£ ππ π π π Joint initiative of IITs and IISc β Funded by MHRD Page 19 of 66