Contour Integrals Revisited

Contour integrals are integrals of complex functions f(z), i.e. of the complex variable z = x + iy. Such integrals are defined in terms of the values of f(z) along a given contour C extending from a point z = z1 to a point,  say, z = z2 in the complex plane.  It is therefore a line integral. This can be written as:

ò_C  f(z) dz

It may alternatively be written with limits, say from z1 to z2, usually when the value of the integral is independent of the choice of contour taken between the two end points.

Illustration: Suppose that the equation:  z = z(t)  such that (a <  t  < b)  represents a contour C extending from some point z1 = z(a) to a point z2 = z(b). let the function f(z) be piecewise continuous on C. That is, f(z(t))  is piecewise continuous on the interval a <  t  < b. We define the line integral or contour integral of f along C as:

ò f(z) dz   =  ò _a ^b  f[z(t)] z’(t) dz 

Since C is a contour then z’(t), i.e. dz/dt is piecewise continuous on the interval  a <  t  < b so the existence of the integral is assured.

Example (1)

This is by reference to the contour shown on  left side below:

Here C is the right hand half of the circle  êz ê  = 3.8 from z = -3.8i to z = 3.8i. Hence we want to find the integral of:

I = ò _C  z z* dz

For which z = 3.8 exp (i q)  (- p/2 <   q  <  p/2)

Then:    I=  ò _p/2 ^-p/2  (3.8 exp(i q)) (3.8 exp(- i q))  dq 

=  14.44i   ò _p/2 ^-p/2   dq

= 14.44 (p/2  - (-p/2)) = 14.44i   (p)  =  14.44 pi

Note that for such a point z on the circle êz ê  = 3.8, it follows that zz* = 14.44 or z* = 14.44/z. So that the result  14.44 pi can also be written: I = ò_C  dz/ z =  pi

Example(2):

The reference contour is shown in the graphic to the right in preceding diagram, along the path OAB. So  the integral can be solved as follows:

ò _C  f(z) dz  =   ò _OA  f(z) dz  +    ò _AB  f(z) dz  

Where f(z) =  y – x   -i3x²   (z = x + iy)

The segment OA can be represented parametrically as:  

z = 0 + iy (0 < y < 1)

Since x = 0 at all points on that segment the values of f there vary with the parameter y according to the equation: f(z) = y(0 < y < 1)

Therefore:

ò _OA  f(z) dz   =     ò ₀ ¹   y idy = i ò ₀ ¹   y dy = i/2

Meanwhile, on the segment AB, z = x + i(0 < x < 1) so that:

ò _AB  f(z) dz   =    =     ò ₀ ¹   (1 – x –i3x²) 1 dx =

ò ₀ ¹   (1 – x)dx  –  3i ò ₀ ¹    x²dx = ½ - i

Based on the original contour definition  (adding the integrals for the segments OA and AB):

ò _C1  f(z) dz  =   1 – i/2

If C2 denotes the segment OB of the line y = x then we have:

 z = x + ix (0 < x < 1)

And we can write:

ò _C2  f(z) dz  =    ò ₀ ¹   -i3x²( 1 +i) dx =

 3(1-i)  ò ₀ ¹    x²dx = 1 – i

  

We can see from this that the integrals of  f(z) have different values though the two paths C1 and C2 have the same initial and starting points. It follows from this that the integral of f(z) over the simple closed contour OABO or C1 – C2 is:

ò _C1  f(z) dz   - ò _C2  f(z) dz    = [½ - i] – (1 – i ) =    -1 +i / 2

Problem For Math Mavens:

Redo Example (1) except change the limits of the contour to have z = -2i to z = 2i.

More Difficult Contour Integrals

Time for some more contour integrals. We have seen earlier examples of contour line integrals and now we look at a more detailed example:

We want to integrate around the closed contour  for which f(z) = 3x + 2iy:

We check first to see if the function is analytic using the Cauchy –Riemann relations:

Recall that we require:  (see e.g.   http://brane-space.blogspot.com/2013/10/looking-at-harmonic-conjugates-and.html   )

i) ¶ u/ ¶x = ¶ v/ ¶y  and ii) ¶ u/ ¶y   =  - ¶ v/ ¶x

For f(z) = 3x + 2iy  =  u(x,y) + iv(x,y)

We have:  ¶ u/ ¶x =  3   and ¶ v/ ¶y    = 2,    so  ¶ u/ ¶x ¹ ¶ v/ ¶y 

And:

¶ u/ ¶y   = 0 =     - ¶ v/ ¶x

Since condition (i) is not fulfilled the function is not analytic, hence we cannot evaluate using Cauchy’ theorem , i.e. 

 ò  _C  f(z) dz = 0

So we must integrate line segment by line segment, viz.

I  =   ò _C  f(z) dz  =   ò _C  (u + iv) (dx + idy)  =     

 ò  _C  (3x + 2iy) (dx + idy)

=    ò  _C  (3x dx  -   2y dy)  +  i ò  _C  ( 2y dx + 3x dy) 

I =     å³ _{n = 1}     [ò  _C  (3x dx  -   2y dy)   + iò  _C  ( 2y dx + 3x dy) 

On C1:  0 < x  < 1, y = 3, dy = 0

ò  _C1  (3x dx  -   2y dy)  +  i ò  _C1  ( 2y dx + 3x dy)   

=  ∫ ₀ ¹    3x dx  +   i∫ ₀ ¹    6 dy  =    [3/2 x²] ₀ ¹       +   i[6y] ₀ ¹    

=  3/2 + 6i

On C2: 3 < y  < 5, x = 1, and dx = 0

ò  _C2  (3x dx  -   2y dy)  +  i ò  _C2  ( 2y dx + 3x dy)   

=   ò₃ ⁵   ( -2y ) dy  +   i ò₃ ⁵     3 dy  =      [- y²] ₃ ⁵      +   i[3y] ₃ ⁵

= - 16 + 6i

On C3: 1 < t  < 0, x = t, and dx = dt, y = 2t +3, dy = 2dt 

Then:

ò  _C3  (3x dx  -   2y dy)  +  i ò  _C3  ( 2y dx + 3x dy)   

=   ∫ ₁ ⁰   [ 3t  dt  - 2(2t + 3) 2 dt] + i  ∫ ₁ ⁰   [ 2(2t + 3) dt + 3t (2dt)]

= - ∫ ₁ ⁰    (5t  + 12)  dt   + i∫ ₁ ⁰    (10t  + 6)  dt  

=   -  [5t² / 2 + 12t] ₁ ⁰  + i[5t²  + 6t] ₁ ⁰     =  29/2 – 11i

Then the contour integral value is:

I =  ( 3/2 + 6i)  + (-16 + 6i) + (29/2 – 11i) =  0 + i

Problems  for Math Mavens:

1) For the closed path shown in the diagram below, evaluate the contour integral. I.

Let f(z) = z ²

   2) For the example shown in the blog post (Fig. 1) , what would the value of I be if:

f(z) = 3x + i3y?

The Residue Theorem and Complex Integrals

Recall that we saw the “residue theorem” (due to Cauchy)i.e.  Let f(z) be analytic on and inside a closed contour C (see diagram) except for a finite number of isolated singularities at z = a1, a2…..etc., which are enclosed by C.

ò_C  f(z)  dz =       2 pi    åⁿ _{k = 1}    Res f (a _k) 

We now want to elaborate this a bit more by reference to the diagram shown. In this case we consider the function f(z) is analytic inside and ON the simple closed curve C except at a finite number of specified points: a, b, c, etc.  at which there exist residues:   a _{- 1}  ,        b _{- 1} ,  c _{- 1}      , etc.

In which case we can write:

ò_C  f(z)  dz =   2 pi   [a _{- 1}        +  b _{- 1}          +  c _{- 1}        + …………………….]

That is, 2 pi    times the sum of the residues at all the singularities enclosed by C. To ensure this, one would respectively construct circles C1, C2, C3 etc. as I have done with respective centers at a, b, c etc. If we take care to do this properly then we can write:


ò _C  f(z)  dz =       ò _C1  f(z)  dz   + ò _C2  f(z)  dz  + ò _C3  f(z)  dz   +    ..........


Where:

ò_C1  f(z)  dz   =   2 pi   a _{- 1}       

ò_C2  f(z)  dz  =  2 pi   b _{- 1}       

ò_C3  f(z)  dz   =   2 pi   c _{- 1}       

So that:

ò_C  f(z)  dz =   2 pi   [a _{- 1}        +  b _{- 1}          +  c _{- 1}        + ..] =

2 pi   (sum of residues)

Example 1:

Evaluate the integral:  ò _C  cot (z)  dz

f(z) = cot (z)

For which: ò _C  f(z)  dz   =   2 pi   c _{- 1}       

Re-write: f(z) = cot (z) = 1/ tan z

For which singularities occur at tan z = 0

Or: o, + p, + 2p,+  3p  etc.

Then Res f(z) =   1/ sec² z ÷ _{z = + n p}     =    1/ (1/ cos² z)

= cos² z÷ _{z = + n p}     =    cos² (np)  

And :  cos² (np)    = 1   at z =  (2n + 1) p)/ 2

Therefore:    c _{- 1}        =  1, and

  ò_C  cot (z)  dz  =    2 pi   (1) = 2 pi   

Example 2:

Evaluate the integral:

ò _C  exp (z)   dz  /  (z – 1) (z + 3)²

 

Where C is given by  ÷ z  ÷    =   3/2 

Solution:

Take the residue at the simple pole (z = 1) such that:

lim _{z ® 1}   [ (z – 1)  exp (z)    / ( z  -  1) (z + 3)²  ] = 

 exp(1)/ 16 = e/ 16

The residue at the 2^nd order pole (z = -3) is:

lim _{z ® -3}  d/ dz  [(z + 3)²    exp (z)    / ( z  -  1) (z + 3)²  ] = 

lim _{z ® -3}   [ (z – 1)  exp (z)    - exp(z) / (z – 1 )²  ]

 

   = - 5 exp (-3) / 16

The integral is therefore:

ò _C  exp (z)   dz  /  (z – 1) (z + 3)²    =  2 pi   a _{- 1}   =   2 pi   (e/ 16)

(We do not add the 2^nd residue because it lies beyond the circle ÷ z  ÷    =   3/2  )

Problems for Math Mavens:

1) Evaluate the integral:   ò _C  (z + 1)   dz / (2z +  i)

2) Consider Example (2) and obtain the integral if we have ÷ z  ÷   =  10   instead of

 ÷ z  ÷    =   3/2 

3)  Evaluate the integral:   ò _C    z   dz / (z²  - 2z + 2)²

in the upper half z-plane