5. COMPLEX
        INTEGRATION    

(TO PART B)

 Definite Integrals

Integrals are extremely important in the study of functions of a complex variable. The theory is elegant, and the proofs generally simple. The theory is put to much good use in applied mathematics.

We shall study line integrals of f(z). In order to do this we shall need some preliminary definitions.

Let F(t) = U(t) +iV(t) (a t b) where U,V are real-valued, piecewise-continuous functions of t on [a, b], i.e. continuous except for at most a finite number of jumps. The definite integral of F on the interval a t b is now defined by:

Definition              F(t) dt = U(t) dt + i V(t) dt.



Properties of the Definite Integral

1. Re ( F(t) dt) = U(t) dt = Re (F(t)) dt. [Real part property]

2. kF = k F ;(k C, const).      [Scalar multiple property]

3. (F + G)= F + G.       [Addition property]

4. F = – F.          [Interchange endpoints property]

5. | F | | F | (a b)     [Modulus property]

Property (1) follows immediately from the definition. The proofs of (2), (3), (4) are trivial, and follow from the properties of real integrals.
We see from (2) and (3) that the integral behaves in a linear way.



Proof of Property (5)

By definition, F dt is a complex number. So we can set r0 cis 0 = F dt.

By Property (2), r0 = F cis (–0) dt.

Each side is real, and when a complex number is real, it is the same as its real part.

So by Property (1),   r0 = Re (F cis (–0)) dt.

But Re (F cis (–0))  | F cis (–0) | = | F |.| cis (–0)| = | F |.

So r0 | F | dt,   providing a b.

Hence   | F dt | | F | dt.



Arcs

Definition A continuous arc is a set of points (x, y): x = (t), y = (t) (a t b), where , are real continuous functions of the real parameter t.

Notes

1. The definition gives a continuous mapping of [a, b] to the arc with a corresponding ordering of points.

2. If no two distinct values of t map to the same point (x, y), the arc is a simple arc or a Jordan arc.

Examples

Jordan arc     Not simple     Jordan curve  Non-simple closed
                                            curve      

3. If Note (2) holds except that (a) = (b) and (a) = (b), the arc is a simple closed curve or Jordan curve.



Contours

If functions , have continuous derivatives '(t),'(t) not simultaneously zero, we say the arc (x, y): x = (t), y = (t) (a t b) is smooth (has a continuously turning tangent).

Definition  A contour is a continuous chain of a finite number of smooth arcs joined end to end.

Examples

       Contour     Simple closed contour 



Length of an Arc

Definition  For a smooth arc, the length exists and is given by

, providing a b.

Now, where does this strange formula come from? From Pythagoras' Theorem, the length s of a small portion of the arc is given by

.

Dividing through by t gives

.

Integrating this expression with respect to t gives the required result.

Note It can be shown that this formula is independent of the choice of parameter used.






Line Integrals

Let C be a contour extending from z = to z = , and set z = x + iy. Thus for z on C, x = (t), y = (t) (a t b) say, where , are continuous and ',' are sectionally continuous. Also t = a when z = , and t = b when z = .

Let f(t) = u(t) + iv(t) be a (sectionally) continuous function on C (that is, the real functions u = u(t) and v = v(t) are sectionally continuous over a t b).

Definition We define

f(z) dz = f [(t) + i (t)].['(t) + i '(t)] dt.

Here, f(z) dz is a line integral, or contour integral.

Note that the line integral exists because the integrand on right is sectionally continuous.



Expanding the Line Integral

Suppose f(z) = u + iv = u((t), (t)) + iv((t),(t)).

Then substituting f(z) = u + iv in the defining expression

f(z) dz = f [(t) + i (t)].['(t) + i '(t)] dt.

gives

f(z) dz = (u' – v' )dt + i (u' + v' )dt,

or simply

f(z) dz = (u dx – v dy) + i (u dy + v dx).

Note  In summary,        f(z) dz = (u + iv)(dx + i dy).

Thus we have expressed the complex line integral in terms of two real line integrals.



Properties of the Line Integral

1.   (taken over the same contour).

2. kf(z)dz = k f(z)dz   (k  C , constant).

3. [f(z) + g(z)] dz = f(z) dz + g(z) dz

4. If C1 is a contour (with parameter t from to ), C2 a contour from to , and C = C1 C2 a contour from to , then .

5. If C has length L and |f(z)| M on C, then | f(z)dz | ML.

Properties (2) and (3) express the linearity of the integral.
Properties (1) to (4) follow easily from known properties of the real integral.



Proof of Property 5

Proof  We use Property 5 of the Definite Integral:

| F | | F |   (a b)      (*)

Now,

| f(z)dz | = | f ((t) + i (t)].['(t) + i '(t)) dt.|  (a b)

              | f ((t) + i (t)].['(t) + i '(t)) | dt (by inequality (*))
         M | '(t) + i '(t) | dt   ( since | f(z) | M)
         = M {('(t))2 + ('(t))2 }dt
         = ML.



Line Integral Example I

Find I1 = z2 dz, where C1 is the illustrated path OB.

Line OB has equation x = 2y. Taking y as parameter we obtain the set of points (2y, y) (0 y 1).

Now, z2 is continuous and on C1,

z2 = (x2 – y2) + 2ixy = 3y2 + 4y2i ,
dz = dx + i dy = (2 + i) dy

[strictly ('(y) + i'(y)) dy].

Hence

I1 = (3y2 + 4y2i)(2 + i)dy = (3 + 4i)(2 + i) y2 dy = 1/3.(2 + 11i).

Line Integral Example II

Find I2 = z2 dz, where C2 is the illustrated path OAB.

Now

I2 = OA z2 dz + AB z2 dz
 = x2 dx + [(4 – y2) + 4iy] i dy
 = 8/3 + [4 – 1/3 + 2i] = 1/3(2 + 11i).



Some Interesting Questions

Question 1  We see that in the previous examples, the integral from O to B is the same for both contours. Is this a coincidence? Or does it always happen? If it doesn't always happen, when does it?

We can express this in a different way. We see that

OABO = OAB + BO = I2 – I1 = 0.

So we ask: Is the integral around a closed contour always zero?

Question 2  We observe that if we forget the contour altogether, and simply integrate, then we obtain:

, again!

Does this always happen? This would say that C f is independent of the contour C.



Line Integral Example III

Find I3 = dz, where the contour C3 is shown in red.

Now C3 is the contour of points

(x, y) | x = cos , y = sin , 0

and       dz = (– sin + i cos ) d.        

Hence we have

I3 = dz

   = (cos – i sin)(– sin + i cos )d

  = i d = – i.



Line Integral Example IV


Find I4 = dz, where the contour C4 is shown in red.

In this case we obtain

I4 = dz = i d = i.

We observe that this is different from I3!
And if we set C = (– C3) C4, then dz = 2 i. (not 0) where the contour C is traversed in an anti-clockwise direction.

Note On the contour C, |z|2 = z = 1. That is, = 1/z. This suggests that there might be some special significance about the singular point z = 0 being inside the contour.



Line Integral Example V


Without evaluating the integral, find an upper bound for | dz/z4 | on the given contour C5.

We make use of Property 5:

| f(z)dz | ML.

In this case, the length L = 2 .
Also, the contour C has equation y = 1 – x (0 x 1).

Therefore for z on C

| z4 | = (x2 + y2)2 = [ x2 + (1 – x)2 ]2 = [2x2 – 2x + 1]2.

That is,         | z4 | = [2(x – 1/2)2 + 1/2]2 1/4.

(In retrospect, this is obvious from the figure! Why?)

Hence                | 1/z4 | 4 and | I | 42 .




Green’s Theorem

Let R be a closed region in the real plane made up of a closed contour C and all interior points.

If P(x, y), Q(x, y) are real continuous functions over R and have continuous first order partial derivatives, then Green’s Theorem says:

(P dx + Q dy) = R (Qx – Py) dxdy

where C is described in the anti-clockwise direction.

Outline Proof

Py dxdy = [ Py dy] dx

         = [ P(x, u(x)) – P(x, l(x)) ] dx

         = – P dx   etc.



Cauchy’s Theorem

Now consider a function f(z) = u(x, y) + iv(x, y) which is analytic at all points within and on the closed contour C, and is such that f '(z) is continuous there.

We show that f(z) dz = 0.

The given conditions tell us that u, v and their first order partial derivatives are continuous. Now

f(z) dz = (u dx – v dy) + i (u dy + v dx)

           = – R (vx + uy) dxdy + i R (ux – vy) dxdy

           = 0, using Green's Theorem and the Cauchy-Riemann equations.

This result was discovered by Cauchy.

Examples and the Cauchy-Goursat Theorem

As a consequence of Cauchy's Theorem we have the following:

Examples    dz = 0,   z dz = 0,   z2 dz = 0,

for all closed contours C, since these functions are analytic everywhere and their derivatives are everywhere continuous.

Goursat showed that Cauchy’s condition ‘f '(z) is continuous’ can be omitted. This discovery is important, because from it we can deduce that all derivatives of analytic functions are also analytic. But, proving this stronger result takes much more effort!

Theorem 5.1 (Cauchy-Goursat Theorem) If a function f is analytic at all points interior to and on a closed contour C, then f(z) dz = 0.


Cauchy-Goursat Theorem, Example I

Find where the contour is the unit circle | z | = 1.

Now the integrand z2/(z–3) is analytic everywhere except at the point z = 3. This point lies outside the circular disk | z | 1. Hence by the Cauchy-Goursat Theorem


Most of the useful applications of the Cauchy-Goursat Theorem are as yet
beyond us, but the next example gives us a glimpse of what can be done.


Cauchy-Goursat Theorem, Example II

Assuming Laplace’s Integral: show that

[We note that Laplace's integral can be evaluated by writing

and expressing this second integral in polar coordinates.]

We integrate f(z) = exp(–z2) around the rectangle C defined by | x | R, 0 y b. Later we let R .

Since exp(–z2) is entire, the Cauchy-Goursat Theorem applies, that is,

[Continued]

Example II, continued

Hence

Taking out the factor from the second and fourth integrals and letting R , we get a zero contribution (using the | dy | ML. formula). Hence letting R ,

Equating the real parts,

or

since the integrand is even.

Indefinite Integrals

Let f be analytic in D and z0, z D. Then for contours C1, C2 lying in D, joining points z0, z, and being traversed from z0 to z,

  by the Cauchy-Goursat Theorem.

Theorem 5.2 (‘Primitive’ Theorem)   For all such paths in the domain D

has the same value, and F'(z) = f(z).

Note  This will allow us to evaluate some line integrals by straight integration. We shall see that the name of the theorem comes from the fact that the function F satisfying F'(z) = f(z) is called a primitive of f.


Proof of the ‘Primitive’ Theorem

Proof Let z + z be a point in D. Then

F(z + z) – F(z) = ,

where since z is very small, we may take the path from z to z + z to be a line segment.

Now clearly on the segment,   So

.

Then

[Continued]

Proof of the ‘Primitive’ Theorem (Continued)

Now f is continuous at z. Hence for all > 0, there exists > 0:

| z | <      | f(z ') – f(z) | < .

Hence, when | z | < ,

(using Property 5).

That is,

and F(z) exists at each point of D, and F'(z) = f(z).


Notes on the ‘Primitive’ Theorem

We say that F is an indefinite integral or primitive (anti-derivative) of f and write F(z) = f(z) dz.

That is, F is an analytic function whose derivative is f(z).

Since

we can use this as a means of evaluating line integrals.
(All paths here are assumed to be in the domain D.)




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