The question is about the integral of complex functions.
[ c ]is a circle with center 0 and radius 2
Use the method shown in the photo
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no surprise that the value of an analytic function at a boundary (contour)
determines the function at all points inside the boundary.
19.1.4 Derivatives as Integrals
The CIF is a very powerful tool for working with analytic functions. One
of the applications of this formula is in evaluating the derivatives of such
functions. It is convenient to change the dummy integration variable to § and
write the CIF as
f(2) = f C) d£
2πί Jc ξ
1 [ f(E)
(19.9)
where C is a simple closed contour in the E-plane and z is a point within C.
By carrying the derivative inside the integral, we get
df
d [f(E) d£]
f(E) d£
2mi Jc (E – 2)²*
1
d
1
1
%3D
2ni de f LE) de
dz
dz
- Z
2ni
By repeated differentiation, we can generalize this formula to the nth deriva-
tive, and obtain
Theorem 19.1.11. The derivatives of all orders of an analytic function f(z)
exist in the domain of analyticity of the function and are themselves analytic
in that domain. The nth derivative of f(z) is given by
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With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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