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1
Complex Variables 4.5 The Cauchy Integral Formula and Its
Extension
The values of an analytic function f(z) on a closed loop C dictate its values at
every point inside.
Cauchy Integral Formula
THEOREM 8 (Cauchy Integral Formula) Let f(z) be analytic on and in the interior of
a simple closed contour C. Let z0 be a point in the interior of C. Then
Consider a new circular contour C0 lying entirely inside C
centered at z0.
The function is analytic everywhere except z0
Principle of deformation:
Recall :
So:
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If we define M like
and on the circle: |z - z0| = r = L ; the function is continuous: |f(z)-f(z0)| < for |z – z0 | <
ML inequality provides:
or
So:
EXAMPLE 1 a) Find , where C is the triangular contour shown in Fig.
Solution. cos z – entire function on the contour, and z0=1
is inside so:
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Complex Variables
ConzforMzz
zfzf
0
0 )()(
Mrr
zfzf
zz
zfzf
)()()()( 0
0
0
00)()(when 0 rzfzf
!!0
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b) Find , where C is the same as in part (a). z0 = -1 is outside C,
cosz/(z+1) is analytic on the contour and its interior, so:
EXAMPLE 2 Find , where C is the circle |z - 2i| = 2.
Solution. z = i is inside the contour and z = -i is outside. Fist we factorize:
cosz/(z+i) is analytic on the contour and its interior, so:
Consider now:
Recollect:
3
Complex Variables
0
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x
i2
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CCU
CL
iz
CU CL
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4
Complex Variables
Then:
Extension
Indeed:
THEOREM 9 (Extension of Cauchy Integral Formula) If a function f(z) is analytic
within a domain, then it possesses derivatives of all orders in that domain. These
derivatives are themselves analytic functions in the domain. If f(z) is analytic on and
in the interior of a simple closed contour C and if z0 is inside C, then
iz
CU CL
02
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2
cos
2
1
i
i
i
i
i
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5
Complex Variables
THEOREM 10 A function that is harmonic in a domain will possess partial derivatives
of all orders in that domain.
Observe for instance:
EXAMPLE 3 Determine the value of , where C is the contour |z| = 2.
Solution. Use:
And for f(z) = x3 + 2z +1:
EXAMPLE 4 Find , where C is the circle |z - 4| = 2.
Solution. (z-1)3 is not zero in the rang of integration, but (z-5)2 gets zero at z =5 inside
C.
So we consider
as the first derivative of at z = 5: .
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