THEOREM 1 Indefinite Integration of Analytic Functions

Let f(z) be analytic in a simply connected domain D. Then there exists an indefinite integral of f(z) in the domain D, that is, an analytic function F(z) such that F′(z) = f(z) in D, and for all paths in D joining two points z₀ and z₁ in D we have

(Note that we can write z₀ and z₁ instead of C, since we get the same value for all those C from z₀ to z₁.)

EXAMPLE 1

EXAMPLE 2

EXAMPLE 3

since e^z is periodic with period 2πi.

EXAMPLE 4

Here D is the complex plane without 0 and the negative real axis (where Ln z is not analytic). Obviously, D is a simply connected domain.

THEOREM 2 Integration by the Use of the Path

Let C be a piecewise smooth path, represented by z = z(t), where a t b. Let f(z) be a continuous function on C. Then

PROOF

The left side of (10) is given by (8) in terms of real line integrals, and we show that the right side of (10) also equals (8). We have z = x + iy, hence We simply write u for u[x(t), y(t)] and v for v[x(t), y(t)]. We also have and Consequently, in (10)

EXAMPLE 5 A Basic Result: Integral of 1/zAround the Unit Circle

We show that by integrating 1/z counterclockwise around the unit circle (the circle of radius 1 and center 0; see Sec. 13.3) we obtain

This is a very important result that we shall need quite often.

Solution. (A) We may represent the unit circle C in Fig. 330 of Sec. 13.3 by

so that counterclockwise integration corresponds to an increase of t from 0 to 2π.

• (B) Differentiation gives (chain rule!).
• (C) By substitution, f(z(t)) = 1/z(t) = e^−it.
• (D) From (10) we thus obtain the result

Check this result by using z(t) = cos t + i sin t.

Simple connectedness is essential in Theorem 1. Equation (9) in Theorem 1 gives 0 for any closed path because then z₁ = z₀, so that F(z₁) − F(z₀) = 0. Now 1/z is not analytic at z = 0. But any simply connected domain containing the unit circle must contain z = 0, so that Theorem 1 does not apply—it is not enough that 1/z is analytic in an annulus, say, because an annulus is not simply connected!

EXAMPLE 6 Integral of 1/z^m with Integer Power m

Let f(z) = (z − z₀)^m where m is the integer and z₀ a constant. Integrate counterclockwise around the circle C of radius ρ with center at z₀ (Fig. 342).

Fig. 342. Path in Example 6

Solution. We may represent C in the form

Then we have

and obtain

By the Euler formula (5) in Sec. 13.6 the right side equals

If m = −1, we have ρ^{m + 1} = 1, cos 0 = 1, sin 0 = 0. We thus obtain 2πi. For integer m ≠ −1 each of the two integrals is zero because we integrate over an interval of length 2π, equal to a period of sine and cosine. Hence the result is

EXAMPLE 7 Integral of a Nonanalytic Function. Dependence on Path

Integrate f(z) = Re z = x from 0 to 1 + 2i (a) along C* in Fig. 343, (b) along C consisting of C₁ and C₂

Solution. (a) C* can be represented by z(t) = t + 2it (0 t 1). Hence and f[z(t)] = x(t) = t on C*. We now calculate

Fig. 343. Paths in Example 7

(b) We now have

Using (6) we calculate

Note that this result differs from the result in (a).

PROOF

Taking the absolute value in (2) and applying the generalized inequality (6*) in Sec. 13.2, we obtain

Now |Δz_m| is the length of the chord whose endpoints are z_m−1 and z_m (see Fig. 340). Hence the sum on the right represents the length L* of the broken line of chords whose endpoints are z₀, z₁, …, z_n(= Z). If n approaches infinity in such a way that the greatest |Δt_m| and thus |Δz_m| approach zero, then L* approaches the length L of the curve C, by the definition of the length of a curve. From this the inequality (13) follows.

EXAMPLE 8 Estimation of an Integral

Find an upper bound for the absolute value of the integral

Fig. 344. Path in Example 8

Solution. and |f(z)| = |z² | 2 on C gives by (13)

The absolute value of the integral is (see Example 1).

PROBLEM SET 14.1

1–10 FIND THE PATH and sketch it.

1. 
2. z(t) = 3 + i + (1 − i)t (0 t 3)
3. z(t) = t + 2it² (1 t 2)
4. z(t) = t + (1 − t)²i (−1 t 1)
5. 
6. z(t) = 1 + i + e^−πit (0 t 2)
7. z(t) = 2 + 4e^πit/2 (0 t 2)
8. z(t) = 5e^−it (0 t π/2)
9. z(t) = t + it³ (−2 t 2)
10. z(t) = 2 cos t + i sin t (0 t 2π)

11–20 FIND A PARAMETRIC REPRESENTATION

and sketch the path.

• 11. Segment from (−1, 1) to (1, 3)
• 12. From (0, 0) to (2, 1) along the axes
• 13. Upper half of |z − 2 + i| = 2 from (4, −1) to (0, −1)
• 14. Unit circle, clockwise
• 15. x² − 4y² = 4, the branch through (2, 0)
• 16. Ellipse 4x² + 9y² = 36, counterclockwise
• 17. |z + a + ib| = r, clockwise
• 18. y = 1/x from (1, 1) to
• 19. Parabola
• 20. 4(x − 2)² + 5(y + 1)² = 20

21–30 INTEGRATION

Integrate by the first method or state why it does not apply and use the second method. Show the details.

• 21. ∫_C Re z dz, C the shortest path from 1 + i to 3 + 3i
• 22. ∫_C Re z dz, C the parabola from 1 + i to 3 + 3i
• 23. ∫_C e^z dz, C the shortest path from πi to 2πi
• 24. ∫_C cos 2z dz, C the semicircle |z| = π, x 0 from −πi to πi
• 25. ∫_C z exp (z²) dz, C from 1 along the axes to i
• 26. ∫_C (z + z⁻¹)dz, C the unit circle, counterclockwise
• 27. ∫_C sec²z dz, any path from π/4 to πi/4
• 28. C the circle |z − 2i| = 4, clockwise
• 29. ∫_C Im z²dz counterclockwise around the triangle with vertices 0, 1, i
• 30. ∫_C Re z²dz clockwise around the boundary of the square with vertices 0, i, 1 + i, 1
• 31. CAS PROJECT. Integration. Write programs for the two integration methods. Apply them to problems of your choice. Could you make them into a joint program that also decides which of the two methods to use in a given case?
• 32. Sense reversal. Verify (5) for f(z) = z², where C is the segment from −1 − i to 1 + i.
• 33. Path partitioning. Verify (6) for f(z) = 1/z and C₁ and C₂ the upper and lower halves of the unit circle.
• 34. TEAM EXPERIMENT. Integration. (a) Comparison. First write a short report comparing the essential points of the two integration methods.
  • (b) Comparison. Evaluate ∫_C f(z)dz by Theorem 1 and check the result by Theorem 2, where:
    1. f(z) = z⁴ and C is the semicircle |z| = 2 from −2i to 2i in the right half-plane,
    2. f(z) = e^2z and C is the shortest path from 0 to 1 + 2i.
  • (c) Continuous deformation of path. Experiment with a family of paths with common endpoints, say, z(t) = t + ia sin t, 0 t π, with real parameter a. Integrate nonanalytic functions (Re z, Re(z²), etc.) and explore how the result depends on a. Then take analytic functions of your choice. (Show the details of your work.) Compare and comment.
  • (d) Continuous deformation of path. Choose another family, for example, semi-ellipses z(t) = a cos t + i sin t, −π/2 t π/2, and experiment as in (c).
• 35. ML-inequality. Find an upper bound of the absolute value of the integral in Prob. 21.

THEOREM 1 Cauchy's Integral Theorem

If f(z) is analytic in a simply connected domain D, then for every simple closed path C in D,

EXAMPLE 1 Entire Functions

for any closed path, since these functions are entire (analytic for all z).

EXAMPLE 2 Points Outside the Contour Where f(x) is Not Analytic

where C is the unit circle, sec z = 1/cos z is not analytic at z = ±π/2, ±3π/2, …, but all these points lie outside C; none lies on C or inside C. Similarly for the second integral, whose integrand is not analytic at z = ±2i outside C.

EXAMPLE 3 Nonanalytic Function

where C: z(t) = e^it is the unit circle. This does not contradict Cauchy's theorem because is not analytic.

EXAMPLE 4 Analyticity Sufficient, Not Necessary

where C is the unit circle. This result does not follow from Cauchy's theorem, because f(z) = 1/z² is not analytic at z = 0. Hence the condition that f be analytic in D is sufficient rather than necessary for (1) to be true.

EXAMPLE 5 Simple Connectedness Essential

for counterclockwise integration around the unit circle (see Sec. 14.1). C lies in the annulus where 1/z is analytic, but this domain is not simply connected, so that Cauchy's theorem cannot be applied. Hence the condition that the domain D be simply connected is essential.

In other words, by Cauchy's theorem, if f(z) is analytic on a simple closed path C and everywhere inside C, with no exception, not even a single point, then (1) holds. The point that causes trouble here is z = 0 where 1/z is not analytic.

PROOF

Cauchy proved his integral theorem under the additional assumption that the derivative f′(z) is continuous (which is true, but would need an extra proof). His proof proceeds as follows. From (8) in Sec. 14.1 we have

Since f(z) is analytic in D, its derivative f′(z) exists in D. Since f′(z) is assumed to be continuous, (4) and (5) in Sec. 13.4 imply that u and v have continuous partial derivatives in D. Hence Green's theorem (Sec. 10.4) (with u and −v instead of F₁ and F₂) is applicable and gives

where R is the region bounded by C. The second Cauchy–Riemann equation (Sec. 13.4) shows that the integrand on the right is identically zero. Hence the integral on the left is zero. In the same fashion it follows by the use of the first Cauchy–Riemann equation that the last integral in the above formula is zero. This completes Cauchy's proof.

THEOREM 2 Independence of Path

If f(z) is analytic in a simply connected domain D, then the integral of f(z) is independent of path in D.

PROOF

Let z₁ and z₂ be any points in D. Consider two paths C₁ and C₂ in D from z₁ to z₂ without further common points, as in Fig. 348. Denote by the path C₂ with the orientation reversed (Fig. 349). Integrate from z₁ over C₁ to z₂ and over back to z₁. This is a simple closed path, and Cauchy's theorem applies under our assumptions of the present theorem and gives zero:

EXAMPLE 6 A Basic Result: Integral of Integer Powers

From Example 6 in Sec. 14.1 and the principle of deformation of path it follows that

for counterclockwise integration around any simple closed path containing z₀ in its interior.

Indeed, the circle |z − z₀| = ρ in Example 6 of Sec. 14.1 can be continuously deformed in two steps into a path as just indicated, namely, by first deforming, say, one semicircle and then the other one. (Make a sketch).

THEOREM 3 Existence of Indefinite Integral

If f(z) is analytic in a simply connected domain D, then there exists an indefinite integral F(z) of f(z) in D—thus, F′(z) = f(z)—which is analytic in D, and for all paths in D joining any two points z₀ and z₁ in D, the integral of f(z) from z₀ to z₁ can be evaluated by formula (9) in Sec. 14.1.

PROOF

The conditions of Cauchy's integral theorem are satisfied. Hence the line integral of f(z) from any z₀ in D to any z in D is independent of path in D. We keep z₀ fixed. Then this integral becomes a function of z, call if F(z),

which is uniquely determined. We show that this F(z) is analytic in D and F′(z) = f(z). The idea of doing this is as follows. Using (4) we form the difference quotient

We now subtract f(z) from (5) and show that the resulting expression approaches zero as Δz → 0. The details are as follows.

We keep z fixed. Then we choose z + Δz in D so that the whole segment with endpoints z and z + Δz is in D (Fig. 352). This can be done because D is a domain, hence it contains a neighborhood of z. We use this segment as the path of integration in (5). Now we subtract f(z). This is a constant because z is kept fixed. Hence we can write

By this trick and from (5) we get a single integral:

Since f(z) is analytic, it is continuous (see Team Project (24d) in Sec. 13.3). An > 0 being given, we can thus find a δ > 0 such that |f(z*) − f(z)| < when |z* − z| < δ. Hence, letting |Δz| < δ, we see that the ML-inequality (Sec. 14.1) yields

By the definition of limit and derivative, this proves that

Since z is any point in D, this implies that F(z) is analytic in D and is an indefinite integral or antiderivative of f(z) in D, written

Fig. 352. Path of integration

Also, if G′(z) = f(z), then F′(z) − G′(z) ≡ 0 in D; hence F(z) − G(z) is constant in D (see Team Project 30 in Problem Set 13.4). That is, two indefinite integrals of f(z) can differ only by a constant. The latter drops out in (9) of Sec. 14.1, so that we can use any indefinite integral of f(z). This proves Theorem 3.

PROOF

By two cuts and (Fig. 354) we cut D into two simply connected domains D₁ and D₂ in which and on whose boundaries f(z) is analytic. By Cauchy's integral theorem the integral over the entire boundary of D₁ (taken in the sense of the arrows in Fig. 354) is zero, and so is the integral over the boundary of D₂, and thus their sum. In this sum the integrals over the cuts and cancel because we integrate over them in both directions—this is the key—and we are left with the integrals over C₁ (counterclockwise) and C₂ (clockwise; see Fig. 354); hence by reversing the integration over C₂ (to counterclockwise) we have

and (6) follows.

PROBLEM SET 14.2

1–8 COMMENTS ON TEXT AND EXAMPLES

1. Cauchy's Integral Theorem. Verify Theorem 1 for the integral of z² over the boundary of the square with vertices ±1 ± i. Hint. Use deformation.
2. For what contours C will it follow from Theorem 1 that
   1. 
   2. 
3. Deformation principle. Can we conclude from Example 4 that the integral is also zero over the contour in Prob. 1?
4. If the integral of a function over the unit circle equals 2 and over the circle of radius 3 equals 6, can the function be analytic everywhere in the annulus 1 < |z| < 3?
5. Connectedness. What is the connectedness of the domain in which (cos z²)/(z⁴ + 1) is analytic?
6. Path independence. Verify Theorem 2 for the integral of e^z from 0 to 1 + i (a) over the shortest path and (b) over the x-axis to 1 and then straight up to 1 + i
7. Deformation. Can we conclude in Example 2 that the integral of 1/(z² + 4) over (a) |z − 2| and (b) |z − 2| = 3 is zero?
8. TEAM EXPERIMENT. Cauchy's Integral Theorem.

   (a) Main Aspects. Each of the problems in Examples 1–5 explains a basic fact in connection with Cauchy's theorem. Find five examples of your own, more complicated ones if possible, each illustrating one of those facts.

   (b) Partial fractions. Write f(z) in terms of partial fractions and integrate it counterclockwise over the unit circle, where

   • (i)
   • (ii)

   (c) Deformation of path. Review (c) and (d) of Team Project 34, Sec. 14.1, in the light of the principle of deformation of path. Then consider another family of paths with common endpoints, say, z(t) = t + ia (t − t²), 0 t 1, a a real constant, and experiment with the integration of analytic and nonanalytic functions of your choice over these paths (e.g., z, Im z, z², Re z², Im z², etc.).

9–19 CAUCHY'S THEOREM APPLICABLE?

Integrate f(z) counterclockwise around the unit circle. Indicate whether Cauchy's integral theorem applies. Show the details.

• 9. f(z) = exp(−z²)
• 10.
• 11. f(z) = 1/(2z − 1)
• 12. f(z) = ³
• 13. f(z) = 1/(z⁴ − 1.1)
• 14. f(z) = 1/
• 15. f(z) = Im z
• 16. f(z) = 1/(πz − 1)
• 17. f(z) = 1/|z|²
• 18. f(z) = 1/(4z − 3)
• 19. f(z) = z³ cot z

20–30 FURTHER CONTOUR INTEGRALS

Evaluate the integral. Does Cauchy's theorem apply? Show details.

• 20. C the boundary of the parallelogram with vertices ±i, ±(1 + i)
• 21. C the circle |z| = π counterclockwise.
• 22.
• 23. Use partial fractions.
• 24. Use partial fractions.
• 25. C consists of |z| = 2 counterclockwise and |z| = 1 clockwise.
• 26. C the circle clockwise.
• 27. C consists of |z| = 1 counterclockwise and |z| = 3 clockwise.
• 28. C the boundary of the square with vertices ±1, ±i clockwise.
• 29. C: |z − 4 − 2i| = 5.5 clockwise.
• 30. C: |z − 2| = 4 clockwise. Use partial fractions.

THEOREM 1 Cauchy's Integral Formula

Let f(z) be analytic in a simply connected domain D. Then for any point z₀ in D and any simple closed path C in D that encloses z₀ (Fig. 356),

the integration being taken counterclockwise. Alternatively (for representing f(z) by a contour integral, divide (1) by 2πi),

PROOF

By addition and subtraction, f(z) = f(z₀) + [f(z) − f(z₀)]. Inserting this into (1) on the left and taking the constant factor f(z₀) out from under the integral sign, we have

The first term on the right equals f(z₀) · 2πi, which follows from Example 6 in Sec. 14.2 with m = −1. If we can show that the second integral on the right is zero, then it would prove the theorem. Indeed, we can. The integrand of the second integral is analytic, except at z₀. Hence, by (6) in Sec. 14.2, we can replace C by a small circle K of radius ρ and center z₀ (Fig. 357), without altering the value of the integral. Since f(z) is analytic, it is continuous (Team Project 24, Sec. 13.3). Hence, an > 0 being given, we can find a δ > 0 such that |f(z) − f(z₀)| < for all z in the disk |z − z₀| < δ. Choosing the radius ρ of K smaller than δ, we thus have the inequality

Fig. 356. Cauchy's integral formula

Fig. 357. Proof of Cauchy's integral formula

at each point of K. The length of K is 2πρ. Hence, by the ML-inequality in Sec. 14.1,

Since (> 0) can be chosen arbitrarily small, it follows that the last integral in (2) must have the value zero, and the theorem is proved.

EXAMPLE 1 Cauchy's Integral Formula

for any contour enclosing z₀ = 2 (since e^z is entire), and zero for any contour for which z₀ = 2 lies outside (by Cauchy's integral theorem).

EXAMPLE 2 Cauchy's Integral Formula

EXAMPLE 3 Integration Around Different Contours

Integrate

counterclockwise around each of the four circles in Fig. 358.

Solution. g(z) is not analytic at −1 and 1. These are the points we have to watch for. We consider each circle separately.

(a) The circle |z − 1| = 1 encloses the point z₀ = 1 where g(z) is not analytic. Hence in (1) we have to write

thus

and (1) gives

(b) gives the same as (a) by the principle of deformation of path.

(c) The function g(z) is as before, but f(z) changes because we must take z₀ = −1 (instead of 1). This gives a factor z − z₀ = z + 1 in (1). Hence we must write

thus

Compare this for a minute with the previous expression and then go on:

(d) gives 0. Why?

PROBLEM SET 14.3

1–4 CONTOUR INTEGRATION

Integrate z²/(z² − 1) by Cauchy's formula counterclockwise around the circle.

1. |z + 1| = 1
2. |z − 1 − i| = π/2
3. |z + i| = 1.4
4. |z + 5 − 5i| = 7

5–8 Integrate the given function around the unit circle.

• 5. (cos 3z)/(6z)
• 6. e^2z/(πz − i)
• 7. z³/(2z − i)
• 8. (z² sin z)/(4z − 1)
• 9. CAS EXPERIMENT. Experiment to find out to what extent your CAS can do contour integration. For this, use (a) the second method in Sec. 14.1 and (b) Cauchy's integral formula.
• 10. TEAM PROJECT. Cauchy's Integral Theorem. Gain additional insight into the proof of Cauchy's integral theorem by producing (2) with a contour enclosing z₀ (as in Fig. 356) and taking the limit as in the text. Choose
  1. 
  2. 

  and (c) another example of your choice.

11–19 FURTHER CONTOUR INTEGRALS

Integrate counterclockwise or as indicated. Show the details.

• 11. C: 4x² + (y − 2)² = 4
• 12. C the circle with center −1 and radius 2
• 13. C: |z − 1| = 2
• 14. C: |z| = 0.6
• 15. C the boundary of the square with vertices ±2, ±2, ±4i.
• 16. C the boundary of the triangle with vertices 0 and ±1 + 2i.
• 17. C: |z − i| = 1.4
• 18. C consists of the boundaries of the squares with vertices ±3, ±3i counterclockwise and ±1, ±i clockwise (see figure).

  

  Problem 18

• 19. C consists of |z| = 2 counterclockwise and |z| = 1 clockwise.
• 20. Show that dz = 0 for a simple closed path C enclosing z₁ and z², which are arbitrary.

THEOREM 1 Derivatives of an Analytic Function

If f(z) is analytic in a domain D, then it has derivatives of all orders in D, which are then also analytic functions in D. The values of these derivatives at a point z₀ in D are given by the formulas

and in general

here C is any simple closed path in D that encloses z₀ and whose full interior belongs to D; and we integrate counterclockwise around C (Fig. 360).

PROOF

We prove (1′), starting from the definition of the derivative

On the right we represent f(z₀ + Δz) and f(z₀) by Cauchy's integral formula:

We now write the two integrals as a single integral. Taking the common denominator gives the numerator f(z){z − z₀ − [z − (z₀ + Δz)]} = f(z) Δz, so that a factor Δz drops out and we get

Clearly, we can now establish (1′) by showing that, as Δz → 0, the integral on the right approaches the integral in (1′). To do this, we consider the difference between these two integrals. We can write this difference as a single integral by taking the common denominator and simplifying the numerator (as just before). This gives

We show by the ML-inequality (Sec. 14.1) that the integral on the right approaches zero as Δz →0.

Being analytic, the function f(z) is continuous on C, hence bounded in absolute value, say, |f(z)| K. Let d be the smallest distance from z₀ to the points of C (see Fig. 360). Then for all z on C,

Furthermore, by the triangle inequality for all z on C we then also have

We now subtract |Δz| on both sides and let |Δz| d/2, so that −|Δz| −d/2. Then

Let L be the length of C. If |Δz| d/2, then by the ML-inequality

This approaches zero as Δz →0. Formula (1′) is proved.

Note that we used Cauchy's integral formula (1*) Sec. 14.3, but if all we had known about f(z₀) is the fact that it can be represented by (1*) Sec. 14.3, our argument would have established the existence of the derivative f′(z₀) of f(z). This is essential to the continuation and completion of this proof, because it implies that (1″) can be proved by a similar argument, with f replaced by f′, and that the general formula (1) follows by induction.

EXAMPLE 1 Evaluation of Line Integrals

From (1′), for any contour enclosing the point πi (counterclockwise)

EXAMPLE 2

From (1″), for any contour enclosing the point −i we obtain by counterclockwise integration

EXAMPLE 3

By (1′), for any contour for which 1 lies inside and ±2i lie outside (counterclockwise),

THEOREM 2 Liouville's Theorem

If an entire function is bounded in absolute value in the whole complex plane, then this function must be a constant.

PROOF

By assumption, |f(z)| is bounded, say, |f(z)| < K for all z. Using (2), we see that |f′(z₀)| < K/r. Since f(z) is entire, this holds for every r, so that we can take r as large as we please and conclude that f′(z₀) = 0. Since z₀ is arbitrary, f′(z) = u_x + iv_x = 0 for all z (see (4) in Sec. 13.4), hence u_x = v_x = 0, and u_y = v_y = 0 by the Cauchy–Riemann equations. Thus u = const, v = const, and f = u + iv = const for all z. This completes the proof.

THEOREM 3 Morera's² Theorem (Converse of Cauchy's Integral Theorem)

If f(z) is continuous in a simply connected domain D and if

for every closed path in D, then f(z) is analytic in D.

PROOF

In Sec. 14.2 we showed that if f(z) is analytic in a simply connected domain D, then

is analytic in D and F′(z) = f(z). In the proof we used only the continuity of f(z) and the property that its integral around every closed path in D is zero; from these assumptions we concluded that F(z) is analytic. By Theorem 1, the derivative of F(z) is analytic, that is, f(z) is analytic in D, and Morera's theorem is proved.

PROBLEM SET 14.4

1–7 CONTOUR INTEGRATION. UNIT CIRCLE

Integrate counterclockwise around the unit circle.

8–19 INTEGRATION. DIFFERENT CONTOURS

Integrate. Show the details. Hint. Begin by sketching the contour. Why?

• 8. C the boundary of the square with vertices ±2, ±2i counterclockwise.
• 9. C the ellipse 16x² + y² = 1 clockwise.
• 10. C consists of |z| = 3 counterclockwise and |z| = 1 clockwise.

• 11. C: |z − i| = 2 counterclockwise.
• 12. C: z − 3i| = 2 clockwise.
• 13. C: |z − 3| = 2 counterclockwise.
• 14. C the boundary of the square with vertices ±1.5, ±1.5i, counterclockwise.
• 15. C consists of |z| = 6 counterclockwise |z − 3| = 2 and clockwise.
• 16. C consists of |z −i| = 3 counterclockwise and |z| = 1 clockwise.
• 17. C consists of |z| = 5 counterclockwise and clockwise.
• 18. C: |z| = 1 counterclockwise, n integer.
• 19. C: |z| = 1, counterclockwise.
• 20. TEAM PROJECT. Theory on Growth
  1. Growth of entire functions. If f(z) is not a constant and is analytic for all (finite) z, and R and M are any positive real numbers (no matter how large), show that there exist values of z for which |z| > R and |f(z)| > M. Hint. Use Liouville's theorem.
  2. Growth of polynomials. If f(z) is a polynomial of degree n > 0 and M is an arbitrary positive real number (no matter how large), show that there exists a positive real number R such that for all |f(z)| > M for all |z| > R.
  3. Exponential function. Show that f(z) = e^x has the property characterized in (a) but does not have that characterized in (b).
  4. Fundamental theorem of algebra.If f(z) is a polynomial in z, not a constant, then f(z) = 0 for at least one value of z. Prove this. Hint. Use (a).

The complex line integral of a function f(z) taken over a path C is denoted by

If f(z) is analytic in a simply connected domain D, then we can evaluate (1) as in calculus by indefinite integration and substitution of limits, that is,

for every path C in D from a point z₀ to a point z₁ (see Sec. 14.1). These assumptions imply independence of path, that is, (2) depends only on z₀ and z₁ (and on f(z), of course) but not on the choice of C (Sec. 14.2). The existence of an F(z) such that F′(z) = f(z) is proved in Sec. 14.2 by Cauchy's integral theorem (see below).

A general method of integration, not restricted to analytic functions, uses the equation z = z(t) of C, where a t b,

Cauchy's integral theorem is the most important theorem in this chapter. It states that if f(z) is analytic in a simply connected domain D, then for every closed path C in D (Sec. 14.2),

Under the same assumptions and for any z₀ in D and closed path C in D containing z₀ in its interior we also have Cauchy's integral formula

Furthermore, under these assumptions f(z) has derivatives of all orders in D that are themselves analytic functions in D and (Sec. 14.4)

This implies Morera's theorem (the converse of Cauchy's integral theorem) and Cauchy's inequality (Sec. 14.4), which in turn implies Liouville's theorem that an entire function that is bounded in the whole complex plane must be constant.