what is cauchy's integral formula
Answers
Let 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 \}}} {\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.} {\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 the reciprocal of the denominator of the integrand in Cauchy's integral formula can be expanded as a power series in the variable (a − z0) — namely, when z0 = 0,
{\displaystyle {\frac {1+{\frac {a}{z}}+\left({\frac {a}{z}}\right)^{2}+\cdots }{z}}} {\displaystyle {\frac {1+{\frac {a}{z}}+\left({\frac {a}{z}}\right)^{2}+\cdots }{z}}}
— it follows that holomorphic functions are analytic. 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.} {\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.
Answer:
Let 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 \}}} {\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.} {\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 the reciprocal of the denominator of the integrand in Cauchy's integral formula can be expanded as a power series in the variable (a − z0) — namely, when z0 = 0,
{\displaystyle {\frac {1+{\frac {a}{z}}+\left({\frac {a}{z}}\right)^{2}+\cdots }{z}}} {\displaystyle {\frac {1+{\frac {a}{z}}+\left({\frac {a}{z}}\right)^{2}+\cdots }{z}}}
— it follows that holomorphic functions are analytic. 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.} {\displaystyle f^{(n)}(a)={\frac {n!}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{\left(z-a\right)^{n+1}}}\,dz.}
Step-by-step explanation: