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:

or

)()( 0zzzf

??0

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:

2

Complex Variables

ConzforMzz

zfzf

0

0 )()(

Mrr

zfzf

zz

zfzf

)()()()( 0

0

0

00)()(when 0 rzfzf

!!0

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

y

x

i2

i

i

y

x

i2

i

i

CCU

CL

iz

CU CL

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

)cos(

2

cos

2

1

i

i

i

i

i

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: .

2

2

2

2

2

2

2

2

'

)("

)(

y

vi

y

u

x

vi

x

uzf

y

ui

y

v

x

vi

x

uzf