The function f(z) = z² is entire.

It has primitive F(z) = ¹/₃.z³.

So by our theorem,

along any contour joining 0 and 1 + i.

We recall that ¹/_z dz = – i,   ¹/_z dz = i, where C₁, C₂ are upper and lower semicircles of the unit circle. Now we can easily choose our domain D to avoid O and contain the given contour C_i.  Mysterious question! So why does the difference occur?

The answer is that for the different contours, we need different primitives!

For C₁, we might choose as primitive

log z = ln | z | + i arg z   (–/₂ < arg z < 3/₂).

Then               

But for C₂, our choice of primitive could be

log z = ln | z | + i arg z   (/₂ < arg z < 5/₂)

and            

Defintion A simply connected domain D is an open connected region such that every closed contour within it encloses only points of D. Otherwise the domain is multiply connected.

The Cauchy-Goursat Theorem has been stated for simply connected domains. i.e. if f(z) is analytic throughout a simply connected domain D, then for every closed contour C within D, f(z) dz = 0.

It will be useful to extend the Cauchy-Goursat Theorem to certain multiply-connected domains. Consider the illustrated (red) region D and suppose that f(z) is analytic over this (closed) region.

We assert that f(z) dz = 0, where B is the total directed boundary (C C₁ C₂), with all components traversed so that the region is on the left.

This is easy to prove. We insert the indicated green links partitioning D into two simply-connected domains. We apply the Cauchy-Goursat Theorem to the boundaries of the left and right regions, obtaining two line integrals having value 0. Putting the two circuits together, the integrals along the introduced links cancel, giving the required result.

Theorem 5.3 Let f be analytic everywhere within and on a closed contour C. If z₀ is any point interior to C, then

where the integral is taken in the positive sense around C.

Notes

(1) The formula is the Cauchy integral formula. It is remarkable because it gives the value of f at z₀ in terms of the values of f on the boundary. That is, for an analytic function, fixing the boundary values completely determines f at points inside C.

(2) The theorem can also be used to evaluate certain integrals.

Take f(z) = ^z/(9 – z²), z₀ = – i. Then we observe that f(z) is analytic in and on C. So using the Cauchy integral formula,

Let C₀ be a circle centre z₀, radius r₀ interior to C. Now the function ^f(z)/(z–z₀) is analytic at all points in and on C except at z = z₀, in particular in the region D lying between C and C₀. Hence as in Example 2 above

where both contours are traversed in the positive direction. Hence

or I = f(z₀) I₁ + I₂ say.

[Continued]

For z on C₀, we can write z – z₀ = r₀ cis = r₀ exp(i).

Hence dz = i r₀ exp(i) d, and

  for every r₀ > 0.

Also, f is continuous at z₀, so given > 0, there exists > 0: | z – z₀ | | f(z) – f(z₀) | < .

Take r₀ = . Then | z – z₀ | = , and

Hence I₂ can be made arbitrarily small by taking r₀ sufficiently small. Since I and I₁ are independent of r₀, I₂ must be too. Therefore I₂ = 0, and

I = f(z₀) I₁ = 2 i f(z₀).

Derivatives of Analytic Functions

Theorem 5.4 Suppose f is analytic inside and on a closed contour C, and z₀ lies inside C. Then

That is, we can take the Cauchy integral formula and formally differentiate with respect to z₀.

Proof  We omit this proof. It is not unlike the proof of the Cauchy integral formula.

Note  We can similarly prove:

One of most beautiful parts of course on Complex Functions is the development of
Taylor series. We look at this a little later, but for now, note the appearance here of the terms

f⁽ⁿ⁾(z₀)/_n!.

Example  Evaluate (a) , (b) where C is the unit circle | z | = 1 taken in the anti-clockwise direction.

(a) Take f(z) = cos z,  z₀ = 0 (within C). Since cos z is entire,

I_(a) = 2i .cos z₀ = 2i (the Cauchy integral formula).

(b) Take f(z) = sin z, z₀ = 0, f '(z) = cos z. Then

I_(b) = 2i f '(z₀) = 2i cos 0 = 2i (the Derivative formula).

Another Example

Let us take f(z) = 1 in Cauchy's integral formula and the Derivative formulae, and let C be any contour about z₀.

(a) By Cauchy's integral formula, we have

This can be rewritten as  

(b) By the Derivative formulae,

This can be rewritten as  

If a function f is analytic at a point, then by the Derivative formulae, its derivatives of all orders are also analytic functions at that point.

Now if f(z) = u + iv, then

f '(z) = u_x + iv_x = v_y – iu_y.

So if f '(z) is analytic, then u_x, v_x, u_y, v_y are all differentiable and so continuous.

In the same way, using f "(z) etc., we see that all partial derivatives of u, v of all orders are continuous at any point where f(z) is analytic. Thus we have:

Corollary If f = u + iv is analytic at a point, then all partial derivatives of u, v of all orders are continuous there.

Note  Cauchy's integral formula and the Derivative formulae easily extend to the boundaries of multiply connected domains.

Here is an interesting little result.

Lemma  If f is analytic, and | f | is constant, then f is constant.

Proof  Let f = u + iv. Then we are given that u² + v² = c.
So

2uu_x + 2vv_x = 0,   2uu_y + 2vv_y = 0,

leading to

u_x/u_y = v_x/v_y.

Also, by the Cauchy-Riemann equations, u_x = v_y,  u_y = –v_x.

So, eliminating the u terms, we get v_x² + v_y² = 0

and so v_x = 0 = v_y = u_x = u_y.

Hence f '(z)=0 and so f(z) is constant.

Let f be analytic and not constant on the open disk | z – z₀ | < r₀, centred at z₀. If C is any circle | z – z₀ | = r (0 < r < r₀) then by the Cauchy integral formula,

    (*)

Path C is in the positive sense and parametrizes: z() = z₀ + r exp(i ) (0 2).

So (*) becomes     (i.e. the value at the centre is the arithmetic mean of the values on the circle). Thus

   (0 r < r₀).  (+)

Now suppose | f(z₀) | is a maximum. Then | f(z) | | f(z₀) | for all z: | z – z₀| < r₀.

[Continued]

Maximum Modulus Principle

So,     (0 r < r₀).  (+)

Combining the two inequalities (+), we have

It follows that, | f(z) | = | f(z₀) | for all z : | z – z₀ | < r₀.

For, write the above equation as  

Since the integrand is non-negative, we deduce it must be 0 for all z : | z – z₀ | < r₀.

This shows that f(z) = f(z₀) for all z in the disk. So f is constant in the disk. This is a contradiction.

Theorem 5.5 (Maximum Modulus Principle)   If f is analytic and not constant in the interior of a region then | f(z) | has no maximum value in that interior.

We deduce that if a function f is continuous in a closed bounded region R and is analytic and not constant in the interior of R, then | f(z) | assumes its maximum value on the boundary of R and never in its interior.

Now let f be analytic in and on the circle C₀ defined by |z – z₀| = r₀, and traversed in a positive sense. Then by the Derivative formulae

,  n = 0, 1, 2, ... .

If | f(z) | M on C₀, then

| f ⁽ⁿ⁾(z₀) | ^{n! M} / r₀ⁿ  (n = 0, 1, 2, ...),

and for n = 1

| f '(z₀) | M / r₀.

The preceding observations lead to the following theorem.

Theorem 5.6 (Liouville)  If f is entire and bounded for all values of z in the complex plane, then f(z) is constant.

Proof  By assumption | f(z) | M for all z.

Therefore, as before, | f '(z₀) | M / r₀ for each z₀ in the plane, and for any positive r₀, no matter how large.

It follows that f '(z₀) = 0.

But z₀ is arbitrary. Thus for all z, f '(z) = 0, and so f(z) = constant.

Theorem 5.7 (Fundamental Theorem) Any polynomial

P(z) = a₀ + a₁z + ... + a_nzⁿ (a_n 0)

where n 1, has at least one zero. That is, there is at least one point z₁ : P(z₁) = 0.

It is curious that this important result has no easy algebraic solution.

Proof  Suppose P(z) 0 for any z. Then f(z) = ¹/_P(z) is entire.

In fact f(z) is also bounded. For f is continuous and so bounded in any closed disk centred at the origin. Further, if R is large and z is exterior to the disk | z | R, then

| f(z) | = ¹/_{| P(z) |} ¹/Rⁿ   (or worse!)

so f is bounded for all values of z in the plane. (We can tidy up this last argument, but the idea is that when | z | is large, so is | P(z) |.)

Now by Liouville's Theorem, f(z) and so P(z) is constant.
This contradiction shows that P(z) has at least one zero.

Let z₁ be the zero guaranteed by the Fundamental Theorem.

Then P(z) = (z – z₁)Q(z), where Q(z) is a polynomial of degree n – 1.

We deduce (by induction) that

P(z) = c(z – z₁)(z – z₂) ... (z – z_n).

Corollary Any polynomial of degree n, where n 1 can be expressed as a product of n linear factors. That is,

P(z) = c(z – z₁)(z – z₂) ... (z – z_n),

where c and the z_k are complex constants.

You might like to compare this result with what happens for polynomials over the reals R.

QUIZ 5.5A QUIZ 5.5B

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