problem on Cauchy Integral Formula | Engineering mathematics скачать в хорошем качестве

problem on Cauchy Integral Formula | Engineering mathematics

cauchy Integral Theorem. application of Cauchy integral theorem. engineering mathematics. .  U be an open subset of the complex plane C, and suppose the closed disk D defined as {\displaystyle D={\bigl \{}z:|z-z_{0}|\leq r{\bigr \}}} is completely contained in U. Let f : U → C be a holomorphic function, and let γ be the circle, oriented counterclockwise, forming the boundary of D. Then for every a in the interior of D, {\displaystyle f(a)={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{z-a}}\,dz.} The proof of this statement uses the Cauchy integral theorem and like that theorem, it only requires f to be complex differentiable. Since {\displaystyle 1/(z-a)} can be expanded as a power series in the variable {\displaystyle a}: {\displaystyle {\frac {1}{z-a}}={\frac {1+{\frac {a}{z}}+\left({\frac {a}{z}}\right)^{2}+\cdots }{z}}} — it follows that holomorphic functions are analytic, i.e. they can be expanded as convergent power series. In particular f is actually infinitely differentiable, with {\displaystyle f^{(n)}(a)={\frac {n!}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{\left(z-a\right)^{n+1}}}\,dz.} This formula is sometimes referred to as Cauchy's differentiation formula. The theorem stated above can be generalized. The circle γ can be replaced by any closed rectifiable curve in U which has winding number one about a. Moreover, as for the Cauchy integral theorem, it is sufficient to require that f be holomorphic in the open region enclosed by the path and continuous on its closure. Note that not every continuous function on the boundary can be used to produce a function inside the boundary that fits the given boundary function. For instance, if we put the function f (z) = 1/z, defined for |z| = 1,into the Cauchy integral formula, we get zero for all points inside the circle. In fact, giving just the real part on the boundary of a holomorphic function is enough to determine the function up to an imaginary constant — there is only one imaginary part on the boundary that corresponds to the given real part, up to addition of a constant. We can use a combination of a Möbius transformation and the Stieltjes inversion formula to construct the holomorphic function from the real part on the boundary. For example, the function f (z) = i − iz has real part Re f (z) = Im z. On the unit circle this can be written i/z − iz/2. Using the Möbius transformation and the Stieltjes formula we construct the function inside the circle. The i/z term makes no contribution, and we find the function −iz. This has the correct real part on the boundary, and also gives us the corresponding imaginary part, but off by a constant, namely i. . . Smooth functionsEdit A version of Cauchy's integral formula is the Cauchy–Pompeiu formula,[2] and holds for smooth functions as well, as it is based on Stokes' theorem. Let D be a disc in C and suppose that f is a complex-valued C1 function on the closure of D. Then[3] (Hörmander 1966, Theorem 1.2.1) {\displaystyle f(\zeta )={\frac {1}{2\pi i}}\int _{\partial D}{\frac {f(z)\,dz}{z-\zeta }}-{\frac {1}{\pi }}\iint _{D}{\frac {\partial f}{\partial {\bar {z}}}}(z){\frac {dx\wedge dy}{z-\zeta }}.} One may use this representation formula to solve the inhomogeneous Cauchy–Riemann equations in D. Indeed, if φ is a function in D, then a particular solution f of the equation is a holomorphic function outside the support of μ. Moreover, if in an open set D, {\displaystyle d\mu ={\frac {1}{2\pi i}}\varphi \,dz\wedge d{\bar {z}}} for some φ ∈ Ck(D) (where k ≥ 1), then f (ζ, ζ) is also in Ck(D) and satisfies the equation {\displaystyle {\frac {\partial f}{\partial {\bar {z}}}}=\varphi (z,{\bar {z}}).} The first conclusion is, succinctly, that the convolution μ ∗ k(z) of a compactly supported measure with the Cauchy kernel {\displaystyle k(z)=\operatorname {p.v.} {\frac {1}{z}}} is a holomorphic function off the support of μ. Here p.v. denotes the principal value. The second conclusion asserts that the Cauchy kernel is a fundamental solution of the Cauchy–Riemann equations. Note that for smooth complex-valued functions f of compact support on C the generalized Cauchy integral formula simplifies to {\displaystyle f(\zeta )={\frac {1}{2\pi i}}\iint {\frac {\partial f}{\partial {\bar {z}}}}{\frac {dz\wedge d{\bar {z}}}{z-\zeta }},} and is a restatement of the fact that, considered as a distribution, (πz)−1 is a fundamental solution of the Cauchy–Riemann operator ∂/∂z̄.[4] The generalized Cauchy integral formula can be deduced for any bounded open region X with C1 boundary ∂X from this result and the formula for the distributional derivative of the characteristic function χX of X: {\displaystyle {\frac {\partial \chi _{X}}{\partial {\bar {z}}}}={\frac {i}{2}}\oint _{\partial X}\,dz,} where the distribution on the right hand side denotes contour integration al . . #CauchyIntegralformula #cauchytheorem