2 Cauchy’s integral formula

This is a generalization of the result in Key Point 2:

Key Point 3

Cauchy’s Integral Formula

If f ( z ) is analytic inside and on the boundary C of a simply-connected region then for any point z 0 inside C ,

C f ( z ) z z 0 d z = 2 π i f ( z 0 ) .

Example 13

Evaluate C z z 2 + 1 d z   where C is the path shown in Figure 15:

C 1 : z i = 1 2

Figure 15

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Solution

We note that z 2 + 1 ( z + i ) ( z i ) .

Let z z 2 + 1 = z ( z + i ) ( z i ) = z ( z + i ) z i .

The numerator z ( z + i ) is analytic inside and on the path C 1 so putting z 0 = i in the Cauchy integral formula (Key Point 3)

C 1 z z 2 + 1 d z = 2 π i i i + i = 2 π i . 1 2 = π i .

Task!

Evaluate C z z 2 + 1 d z   where C is the path (refer to the diagram)

  1. C 2 : | z + i | = 1 2
  2. C 3 : | z | = 2 .

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  1. Use the Cauchy integral formula to find an expression for C 2 z z 2 + 1 d z :

    z z 2 + 1 = z ( z i ) z + i . The numerator is analytic inside and on the path C 2 so putting z 0 = i in the Cauchy integral formula gives

    C 2 z z 2 + 1 d z = 2 π i i 2 i = π i .

  2. Now find C 3 z z 2 + 1 d z :

    By analogy with the previous part,

    C 3 z z 2 + 1 d z = C 1 z z 2 + 1 d z + C 2 z z 2 + 1 d z = π i + π i = 2 π i .

2.1 The derivative of an analytic function

If f ( z ) is analytic in a simply-connected region then at any interior point of the region, z 0 say, the derivatives of f ( z ) of any order exist and are themselves analytic (which illustrates what a powerful property analyticity is!). The derivatives at the point z 0 are given by Cauchy’s integral formula for derivatives:

f ( n ) ( z 0 ) = n ! 2 π i C f ( z ) ( z z 0 ) n + 1 d z

where C is any simple closed curve, in the region, which encloses z 0 .

Note the case n = 1 :

f ( z 0 ) = 1 2 π i C f ( z ) ( z z 0 ) 2 d z .

Example 14

Evaluate the contour integral

C z 3 ( z 1 ) 2 d z

where C is a contour which encloses the point z = 1 .

Solution

Since f ( z ) = z 3 ( z 1 ) 2 has a pole of order 2 at z = 1 then C f ( z ) d z = C z 3 ( z 1 ) 2 d z

where C is a circle centered at z = 1 .

If g ( z ) = z 3 then C f ( z ) d z = C g ( z ) ( z 1 ) 2 d z

Since g ( z ) is analytic within and on the circle C we use Cauchy’s integral formula for derivatives to show that

C z 3 ( z 1 ) 2 d z = 2 π i × 1 1 ! g ( z ) z = 1 = 2 π i 3 z 2 z = 1 = 6 π i .

Exercise

Evaluate  C z z 2 + 9 d z  where C is the path:

  1. C 1 : z 3 i = 1
  2. C 2 : z + 3 i = 1
  3. C 3 : z = 6.
  1. We will use the fact that   z z 2 + 9 = z ( z + 3 i ) ( z 3 i ) = z ( z + 3 i ) z 3 i

    The numerator   z z + 3 i   is analytic inside and on the path C 1 so putting z 0 = 3 i in Cauchy’s integral formula

    C 1 z z 2 + 9 d z = 2 π i 3 i 3 i + 3 i = 2 π i × 1 2 = π i .

  2. Here   z ( z 3 i ) z + 3 i   

    The numerator is analytic inside and on the path C 2 so putting z = 3 i in Cauchy’s integral formula:

    C 2 z z 2 + 9 d z = 2 π i 3 i 3 i 3 i = π i .

  3. The integral is the sum of the two previous integrals and has value 2 π i .