Ð - Início — UNIVASF Universidade Federal do Vale do ...edson.araujo/Disciplinas/metmat/lists/2016.2.L... · 2 Lista de Exercic o - 2a Prova 192 Chapter 4 Elementary Functions

Universidade Federal do Vale do Sao FranciscoEngenharia Civil

Metodos Matematicos

Profo. Edson

2o SemestreLista de Exercicıo - 2a Prova 2016Data: 24 de Marco de 2017 Profo. Edson

4.1 Exponential and Logarithmic Functions 191

• Many properties of real logarithms apply to the complex log-arithm, such as ln (z1z2) = ln z1 + ln z2, but these propertiesdon’t always hold for the principal value Ln z.

(ii) Since the complex exponential function is not one-to-one, we canuse a Riemann surface, as described in the Remarks at the endof Section 2.4, to help visualize the mapping w = ez. The Rie-mann surface that we construct will also help us to visualize themultiple-valued function w = ln z. Consider the mapping w = ez

on the half-plane x ≤ 0. Each half-infinite strip Sn defined by(2n− 1)π < y ≤ (2n + 1)π, x ≤ 0, for n = 0, ±1, ±2, . . . is mappedonto the punctured unit disk 0 < |w| ≤ 1 shown in Figure 4.9(b)with the horizontal half-lines shown in color in Figure 4.9(a) map-ping onto the segment −1 ≤ u < 0 shown in black in Figure 4.9(b).Thus, w = ez describes an infinite-to-one covering of the puncturedunit disk. To visualize this covering, we imagine there being a dif-ferent image disk Bn for each half-infinite strip Sn. Now cut eachdisk Bn open along the segment −1 ≤ u < 0. We construct a Rie-mann surface for w = ez by attaching, for each n, the cut disk Bn

to the cut disk Bn+1 along the edge that represents the image of thehalf-infinite line y = (2n + 1)π. We place this surface in xyz -spaceso that for each z in the half-plane, the images . . . z−1, z0, z1, . . .of z in Bn−1, B0, B1, . . . , respectively, lie directly above the pointw = ez in the xy-plane. See Figure 4.10. By projecting the pointsof the Riemann surface vertically down onto the xy-plane we seethe infinite-to-one nature of the mapping w = ez. Conversely, themultiple-valued function F (z) = ln z may be visualized by consider-ing all points in the Riemann surface lying directly above a point inthe xy-plane. These infinitely many points on the Riemann surfacecorrespond to the infinitely many values of F (z) in the half-planeu ≤ 0.

x

y

u

v

(a) Collection of half-infinite strips Sn

(b) The image of each strip Sn is the punctured unit disk

π4π3

π2

–

π

π

–2π

–3π

–4π

S1

S–1

S0

Figure 4.9 The mapping w = ez

3π

2π

–2π

–3π

π

–π

–1

–1

1

1

0

0 0

Figure 4.10 A Riemann surface for

w = ez

EXERCISES 4.1 Answers to selected odd-numbered problems begin on page ANS-14.

4.1.1 Complex Exponential Function

In Problems 1–4, find the derivative f ′ of the given function f .

1. f(z) = z2ez+i 2. f(z) =3e2z − ie−z

z3 − 1 + i

3. f(z) = eiz − e−iz 4. f(z) = ie1/z

In Problems 5–8, write the given expression in terms of x and y.

5.∣∣∣ez2−z

∣∣∣ 6. arg(ez−i/z

)

7. arg(ei(z+z)

)8. iez + 1

192 Chapter 4 Elementary Functions

In Problems 9–12, express the given function f in the form f(z) = u(x, y)+iv(x, y).

9. f(z) = e−iz 10. f(z) = e2z+i

11. f(z) = ez2

12. f(z) = e1/z

13. Use the sufficient conditions for differentiability from page 155 in Section 3.2to determine where the function f(z) = e2z+i is differentiable.

14. Use the sufficient conditions for differentiability from page 155 in Section 3.2

to determine where the function f(z) = ez2

is differentiable.

In Problems 15–20, find the image of the given set under the exponential mapping.

15. The line y = −2.

16. The line x = 3.

17. The infinite strip 1 < x ≤ 2

18. The square with vertices at 0, 1, 1 + i, and i.

19. The rectangle 0 ≤ x ≤ loge 2, −π/4 ≤ y ≤ π/2.

20. The semi-infinite strip −∞ < x ≤ 0, 0 ≤ y ≤ π.

4.1.2 Complex Logarithmic Function

In Problems 21–26, find all complex values of the given logarithm.

21. ln (−5) 22. ln (−ei)23. ln (−2 + 2i) 24. ln (1 + i)

25. ln(√

2 +√

6i)

26. ln(−√

3 + i)

In Problems 27–32, write the principal value of the logarithm in the form a+ ib.

27. Ln (6 − 6i) 28. Ln(−e2

)

29. Ln (−12 + 5i) 30. Ln (3 − 4i)

31. Ln[(

1 +√

3i)5

]32. Ln

[(1 + i)4

]

In Problems 33–36, find all complex values of z satisfying the given equation.

33. ez = 4i 34. e1/z = −1

35. ez−1 = −ie3 36. e2z + ez + 1 = 0

In Problems 37–40, find a domain in which the given function f is differentiable;

then find the derivative f ′.

37. f(z) = 3z2 − e2iz + iLn z 38. f(z) = (z + 1)Ln z

39. f(z) =Ln(2z − i)

z2 + 140. f(z) = Ln

(z2 + 1

)

In Problems 41–46, find the image of the given set under the mapping w =Ln z.

41. The ray arg (z) = π/6.

42. The positive y-axis.

43. The circle |z| = 4.

44. The region in the first quadrant bounded by the circles |z| = 1 and |z| = e.

45. The annulus 3 ≤ |z| ≤ 5.



9. f(z) = e−iz 10. f(z) = e2z+i

11. f(z) = ez2

12. f(z) = e1/z




is differentiable.



16. The line x = 3.







21. ln (−5) 22. ln (−ei)23. ln (−2 + 2i) 24. ln (1 + i)

25. ln(√

2 +√

6i)

26. ln(−√

3 + i)


27. Ln (6 − 6i) 28. Ln(−e2

)

29. Ln (−12 + 5i) 30. Ln (3 − 4i)

31. Ln[(

1 +√

3i)5

]32. Ln

[(1 + i)4

]


33. ez = 4i 34. e1/z = −1

35. ez−1 = −ie3 36. e2z + ez + 1 = 0




39. f(z) =Ln(2z − i)

z2 + 140. f(z) = Ln

(z2 + 1

)






45. The annulus 3 ≤ |z| ≤ 5.



9. f(z) = e−iz 10. f(z) = e2z+i

11. f(z) = ez2

12. f(z) = e1/z




is differentiable.



16. The line x = 3.







21. ln (−5) 22. ln (−ei)23. ln (−2 + 2i) 24. ln (1 + i)

25. ln(√

2 +√

6i)

26. ln(−√

3 + i)


27. Ln (6 − 6i) 28. Ln(−e2

)

29. Ln (−12 + 5i) 30. Ln (3 − 4i)

31. Ln[(

1 +√

3i)5

]32. Ln

[(1 + i)4

]


33. ez = 4i 34. e1/z = −1

35. ez−1 = −ie3 36. e2z + ez + 1 = 0




39. f(z) =Ln(2z − i)

z2 + 140. f(z) = Ln

(z2 + 1

)






45. The annulus 3 ≤ |z| ≤ 5.

2 Lista de Exercicıo - 2a Prova



9. f(z) = e−iz 10. f(z) = e2z+i

11. f(z) = ez2

12. f(z) = e1/z




is differentiable.



16. The line x = 3.







21. ln (−5) 22. ln (−ei)23. ln (−2 + 2i) 24. ln (1 + i)

25. ln(√

2 +√

6i)

26. ln(−√

3 + i)


27. Ln (6 − 6i) 28. Ln(−e2

)

29. Ln (−12 + 5i) 30. Ln (3 − 4i)

31. Ln[(

1 +√

3i)5

]32. Ln

[(1 + i)4

]


33. ez = 4i 34. e1/z = −1

35. ez−1 = −ie3 36. e2z + ez + 1 = 0




39. f(z) =Ln(2z − i)

z2 + 140. f(z) = Ln

(z2 + 1

)






45. The annulus 3 ≤ |z| ≤ 5.



9. f(z) = e−iz 10. f(z) = e2z+i

11. f(z) = ez2

12. f(z) = e1/z




is differentiable.



16. The line x = 3.







21. ln (−5) 22. ln (−ei)23. ln (−2 + 2i) 24. ln (1 + i)

25. ln(√

2 +√

6i)

26. ln(−√

3 + i)


27. Ln (6 − 6i) 28. Ln(−e2

)

29. Ln (−12 + 5i) 30. Ln (3 − 4i)

31. Ln[(

1 +√

3i)5

]32. Ln

[(1 + i)4

]


33. ez = 4i 34. e1/z = −1

35. ez−1 = −ie3 36. e2z + ez + 1 = 0




39. f(z) =Ln(2z − i)

z2 + 140. f(z) = Ln

(z2 + 1

)






45. The annulus 3 ≤ |z| ≤ 5.



9. f(z) = e−iz 10. f(z) = e2z+i

11. f(z) = ez2

12. f(z) = e1/z




is differentiable.



16. The line x = 3.







21. ln (−5) 22. ln (−ei)23. ln (−2 + 2i) 24. ln (1 + i)

25. ln(√

2 +√

6i)

26. ln(−√

3 + i)


27. Ln (6 − 6i) 28. Ln(−e2

)

29. Ln (−12 + 5i) 30. Ln (3 − 4i)

31. Ln[(

1 +√

3i)5

]32. Ln

[(1 + i)4

]


33. ez = 4i 34. e1/z = −1

35. ez−1 = −ie3 36. e2z + ez + 1 = 0




39. f(z) =Ln(2z − i)

z2 + 140. f(z) = Ln

(z2 + 1

)






45. The annulus 3 ≤ |z| ≤ 5.

4.1 Exponential and Logarithmic Functions 193

46. The region outside the unit circle |z| = 1 and between the rays arg(z) = π/4and arg(z) = 3π/4.

Focus on Concepts

47. Use (1) to prove that ez1/ez2 = ez1−z2 .

48. Use (1) and de Moivre’s formula to prove that (ez1)n = enz1 , n an integer.

49. Determine where the complex function ez is analytic.

50. In this problem, we will show that the complex exponential function defined by(1) is the only complex entire function f that agrees with the real exponentialfunction ex when z is real and that has the property f ′(z) = f(z) for all z.

(a) Assume that f(z) = u(x, y) + iv(x, y) is an entire complex function forwhich f ′(z) = f(z). Explain why u and v satisfy the differential equations:

ux(x, y) = u(x, y) and vx(x, y) = v(x, y).

(b) Show that u(x, y) = a(y)ex and v(x, y) = b(y)ex are solutions to thedifferential equations in (a).

(c) Explain why the assumption that f(z) agrees with the real exponentialfunction for z real implies that a(0) = 1 and b(0) = 0.

(d) Explain why the functions a(y) and b(y) satisfy the system of differentialequations:

a(y) − b′(y) = 0

a′(y) + b(y) = 0.

(e) Solve the system of differential equations in (d) subject to the initial con-ditions a(0) = 1 and b(0) = 0.

(f) Use parts (a)–(e) to show that the complex exponential function definedby (1) is the only complex entire function f(z) that agrees with the realexponential function when z is real and that has the property f ′(z) = f(z)for all z.

51. Describe the image of the line y = x under the exponential function. [Hint:Find a polar expression r(θ) of the image.]

52. Prove that ez is a one-to-one function on the fundamental region −∞ < x <∞,−π < y ≤ π.

53. Prove that ln

(z1z2

)= ln z1 − ln z2 for all nonzero complex numbers z1 and z2.

54. Prove that ln zn1 = n ln z1 for all nonzero complex numbers z1 and all integersn.

55. (a) Find two complex numbers z1 and z2 so that Ln (z1z2) �= Ln z1+Ln z2.

(b) Find two complex numbers z1 and z2 so that Ln (z1z2) = Ln z1+ Ln z2.

(c) What must be true about z1 and z2 if Ln (z1z2) = Ln z1+ Ln z2?

56. Is Ln zn1 = nLn z1 for all integers n and complex numbers z1? Defend yourposition with a short proof or a counterexample.

4.2 Complex Powers 199


In Problems 1–6, find all values of the given complex power.

1. (−1)3i 2. 32i/π

3. (1 + i)1−i 4.(1 +

√3i

)i

5. (−i)i 6. (ei)√

2

In Problems 7–12, find the principal value of the given complex power.

7. (−1)3i 8. 32i/π

9. 24i 10. ii/π

11.(1 +

√3i

)3i12. (1 + i)2−i

13. Verify thatzα1

zα2= zα1−α2 for z �= 0.

14. (a) Verify that (zα)n = znα for z �= 0 and n an integer.

(b) Find an example that illustrates that for z �= 0 we can have(zα1)α2 �= zα1α2 .

Let zα represent the principal value of the complex power defined on the domain

|z| > 0, −π < arg(z) < π. In Problems 15–18, find the derivative of the given

function at the given point.

15. z3/2; z = 1 + i 16. z2i; z = i

17. z1+i; z = 1 +√

3i 18. z√

2; z = −i

Focus on Concepts

19. For any complex number z �= 0, evaluate z0.

20. If α = x+ iy where x = 0, ±1, ±2, ... , then what can you say about 1α?

21. Show that if α = 1/n where n is a positive integer, then the principal value ofzα is the same as the principal nth root of z.

22. (a) Show that if α is a rational number (that is, α = m/n where m and n areintegers with no common factor), then zα is finite-valued. That is, show thatthere are only finitely many values of zα.

(b) Show that if α is an irrational number (that is, not a rational number) ora complex number, then zα is infinite-valued.

23. Which of the identities listed in (5) hold for the principal value of zα?

24. A useful property of real numbers is xaya = (xy)a.

(a) Does the property zαwα = (zw)α hold for complex powers?

(b) Does the property zαwα = (zw)α hold for the principal value of a complexpower?

Computer Lab Assignments

Most CASs have a built in function to find the principal value of a complex power.

In Mathematica, the syntax (a + b I) (c + d I) is used to accomplish this.




1. (−1)3i 2. 32i/π

3. (1 + i)1−i 4.(1 +

√3i

)i

5. (−i)i 6. (ei)√

2


7. (−1)3i 8. 32i/π

9. 24i 10. ii/π

11.(1 +

√3i

)3i12. (1 + i)2−i

13. Verify thatzα1

zα2= zα1−α2 for z �= 0.






15. z3/2; z = 1 + i 16. z2i; z = i

17. z1+i; z = 1 +√

3i 18. z√

2; z = −i

Focus on Concepts

















1. (−1)3i 2. 32i/π

3. (1 + i)1−i 4.(1 +

√3i

)i

5. (−i)i 6. (ei)√

2


7. (−1)3i 8. 32i/π

9. 24i 10. ii/π

11.(1 +

√3i

)3i12. (1 + i)2−i

13. Verify thatzα1

zα2= zα1−α2 for z �= 0.






15. z3/2; z = 1 + i 16. z2i; z = i

17. z1+i; z = 1 +√

3i 18. z√

2; z = −i

Focus on Concepts
















1. (−1)3i 2. 32i/π

3. (1 + i)1−i 4.(1 +

√3i

)i

5. (−i)i 6. (ei)√

2


7. (−1)3i 8. 32i/π

9. 24i 10. ii/π

11.(1 +

√3i

)3i12. (1 + i)2−i

13. Verify thatzα1

zα2= zα1−α2 for z �= 0.






15. z3/2; z = 1 + i 16. z2i; z = i

17. z1+i; z = 1 +√

3i 18. z√

2; z = −i

Focus on Concepts
















1. (−1)3i 2. 32i/π

3. (1 + i)1−i 4.(1 +

√3i

)i

5. (−i)i 6. (ei)√

2


7. (−1)3i 8. 32i/π

9. 24i 10. ii/π

11.(1 +

√3i

)3i12. (1 + i)2−i

13. Verify thatzα1

zα2= zα1−α2 for z �= 0.






15. z3/2; z = 1 + i 16. z2i; z = i

17. z1+i; z = 1 +√

3i 18. z√

2; z = −i

Focus on Concepts














(iii) Since the complex sine function is periodic, the mapping w = sin zis not one-to-one on the complex plane. Constructing a Riemannsurface, for this function, as described in the Remarks at the endof Section 2.4 and Section 4.1, will help us visualize the complexmapping w = sin z. In order to construct a Riemann surface con-sider the mapping on the square S0 defined by −π/2 ≤ x ≤ π/2,−π/2 ≤ y ≤ π/2. From Example 3, we find that the square S0

shown in color in Figure 4.13(a) maps onto the elliptical region Eshown in gray in Figure 4.13(b). Similarly, the adjacent square S1

defined by π/2 ≤ x ≤ 3π/2, −π/2 ≤ y ≤ π/2, also maps onto E. ARiemann surface is constructed by starting with two copies of E, E0

and E1, representing the images of S0 and S1, respectively. We thencut E0 and E1 open along the line segments in the real axis from1 to cosh (π/2) and from −1 to − cosh (π/2). As shown in Figure4.14, the segment shown in color in the boundary of S0 is mappedonto the segment shown in black in the boundary of E0, while thedashed segment shown in color in the boundary of S0 is mappedonto the dashed segment shown in black in the boundary of E0. Ina similar manner, the segments shown in color in the boundary ofS1 are mapped onto the segments shown in black in the boundary ofE1. Part of the Riemann surface consists of the two elliptical regionsE0 and E1 with the segments shown in black glued together and thedashed segments glued together. To complete the Riemann surface,we take for every integer n an elliptical region En representing theimage of the square Sn defined by (2n− 1)π/2 ≤ x ≤ (2n + 1)π/2,−π/2 ≤ y ≤ π/2. Each region En is cut open, as E0 and E1 were,and En is glued to En+1 along their boundaries in a manner analo-gous to that used for E0 and E1. This Reimann surface, placed inxyz -space, is illustrated in Figure 4.15.

y

x

u

v

– π2

π2

π2

(a) The square S0

(b) The image E of S0

–2 –1

–1

1

2

–2

1 2

– π2

Figure 4.13 The mapping w = sin z

S0

E0

S1

E1

Figure 4.14 The cut elliptical regions E0

and E1

–2

–5–2

20

5

0

02

Figure 4.15 A Riemann surface for

w = sin z


4.3.1 Complex Trigonometric Functions

In Problems 1–8, express the value of the given trigonometric function in the form

a+ ib.

1. sin (4i) 2. cos (−3i)

3. cos (2 − 4i) 4. sin(π

4+ i

)

5. tan (2i) 6. cot (π + 2i)

7. sec(π

2− i

)8. csc (1 + i)

In Problems 9–12, find all complex values z satisfying the given equation.

9. sin z = i 10. cos z = 4

11. sin z = cos z 12. cos z = i sin z

4.3 Trigonometric and Hyperbolic Functions 213

In Problems 13–16, verify the given trigonometric identity.

13. sin (−z) = − sin z 14. cos (z1 + z2) = cos z1 cos z2 − sin z1 sin z2

15. cos z = cos z 16. sin(z − π

2

)= − cos z

In Problems 17–20, find the derivative of the given function.

17. sin(z2

)18. cos (iez)

19. z tan1

z20. sec

(z2 + (1 − i)z + i

)

4.3.2 Complex Hyperbolic Functions

In Problems 21–24, express the value of the given hyperbolic function in the form

a+ ib.

21. cosh (πi) 22. sinh(π

2i)

23. cosh(1 +

π

6i)

24. tanh (2 + 3i)


25. cosh z = i 26. sinh z = −1

27. sinh z = cosh z 28. sinh z = ez

In Problems 29–32, verify the given hyperbolic identity.

29. cosh2 z − sinh2 z = 1

30. sinh (z1 + z2) = sinh z1 cosh z2 + cosh z1 sinh z2

31. |sinh z|2 = sinh2 x+ sin2 y

32. Im (cosh z) = sinhx sin y


33. sin z sinh z 34. tanh z

35. tanh (iz − 2) 36. cosh(iz + eiz

)

Focus on Concepts

37. Recall that Euler’s formula states that eiθ = cos θ+ i sin θ for any real numberθ. Prove that, in fact, eiz = cos z + i sin z for any complex number z.

38. Solve the equation sin z = cosh 2 by equating real and imaginary parts.

39. If sin z = a with −1 ≤ a ≤ 1, then what can you say about z? Justify youranswer.

40. If |sin z| ≤ 1, then what can you say about z? Justify your answer.

41. Show that all the zeros of cos z are z = (2n+ 1)π/2 for n = 0, ±1, ±2, . . . .

42. Find all z such that |tan z| = 1.

43. Find the real and imaginary parts of the function sin z and use them to showthat this function is nowhere analytic.

44. Without calculating the partial derivatives, explain why sin x cosh y andcosx sinh y are harmonic functions in C.






2

)= − cos z


17. sin(z2

)18. cos (iez)

19. z tan1

z20. sec

(z2 + (1 − i)z + i

)



a+ ib.


2i)

23. cosh(1 +

π

6i)

24. tanh (2 + 3i)


25. cosh z = i 26. sinh z = −1










)

Focus on Concepts













2

)= − cos z


17. sin(z2

)18. cos (iez)

19. z tan1

z20. sec

(z2 + (1 − i)z + i

)



a+ ib.


2i)

23. cosh(1 +

π

6i)

24. tanh (2 + 3i)


25. cosh z = i 26. sinh z = −1










)

Focus on Concepts













2

)= − cos z


17. sin(z2

)18. cos (iez)

19. z tan1

z20. sec

(z2 + (1 − i)z + i

)



a+ ib.


2i)

23. cosh(1 +

π

6i)

24. tanh (2 + 3i)


25. cosh z = i 26. sinh z = −1










)

Focus on Concepts










5.1 Real Integrals 243

denotes the curve having the opposite orientation of C, then the analogue of(14) for line integrals is

∫

−C

P dx+Qdy = −∫

C

P dx+Qdy, (15)

or, equivalently

∫

−C

P dx+Qdy +

∫

C

P dx+Qdy = 0. (16)

For example, in part (a) of Example 1 we saw that∫Cxy2 dx = −64; we

conclude from (15) that∫−C

xy2 dx = 64.

�Note It is important to be aware that a line integral is independent of theparametrization of the curve C, provided C is given the same orientationby all sets of parametric equations defining the curve. See Problem 33 inExercises 5.1.


In Problems 1–10, evaluate the definite integral. If necessary, review the techniques

of integration in your calculus text.

1.

∫ 3

−1

x(x− 1)(x + 2) dx 2.

∫ 0

−1

t2 dt +

∫ 2

0

x2 dx +

∫ 3

2

u2 du

3.

∫ 1

1/2

sin 2πx dx 4.

∫ π/8

0

sec2 2x dx

5.

∫ 4

0

dx

2x + 16.

∫ ln 3

ln 2

e−x dx

7.

∫ 4

2

xe−x/2 dx 8.

∫ e

1

lnx dx

9.

∫ 4

2

dx

x2 − 6x + 510.

∫ 4

2

2x− 1

(x + 3)2dx

In Problems 11–14, evaluate the line integrals∫CG(x, y) dx,

∫CG(x, y) dy, and∫

CG(x, y) ds on the indicated curve C.

11. G(x, y) = 2xy; x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ π/4

12. G(x, y) = x3 + 2xy2 + 2x; x = 2t, y = t2, 0 ≤ t ≤ 1

13. G(x, y) = 3x2 + 6y2; y = 2x + 1, −1 ≤ x ≤ 0

14. G(x, y) = x2/y3; 2y = 3x3/2, 1 ≤ t ≤ 8

244 Chapter 5 Integration in the Complex Plane

In Problems 15–18, evaluate∫C

(2x + y) dx+ xy dy on the given curve from (−1, 2)

to (2, 5).

15. y = x + 3 16. y = x2 + 1

17. 18.y

x

(2, 2)

(2, 5)

(–1, 2)

Figure 5.9 Figure for Problem 17

y

x(2, 0)(–1, 0)

(2, 5)

(–1, 2)


In Problems 19–22, evaluate

∫

C

y dx+x dy on the given curve from (0, 0) to (1, 1).

19. y = x2 20. y = x

21. C consists of the line segments from (0, 0) to (0, 1) and from (0, 1) to (1, 1).


23. Evaluate

∫

C

(6x2 + 2y2) dx + 4xy dy, where C is given by x =

√t, y = t,

4 ≤ t ≤ 9.

24. Evaluate

∫

C

−y2 dx + xy dy, where C is given by x = 2t, y = t3, 0 ≤ t ≤ 2.

25. Evaluate

∫

C

2x3y dx + (3x + y) dy, where C is given by x = y2 from (1, −1)

to (1, 1).

26. Evaluate

∫

C

4x dx + 2y dy, where C is given by x = y3 + 1 from (0, −1)

to (9, 2).

In Problems 27 and 28, evaluate

∮

C

(x2 + y2) dx−2xy dy on the given closed curve.

27. 28.y

x

x2 + y2 = 4


y = x2

y = √x

y

x

(1, 1)




In Problems 15–18, evaluate∫C

(2x + y) dx+ xy dy on the given curve from (−1, 2)

to (2, 5).

15. y = x + 3 16. y = x2 + 1

17. 18.y

x

(2, 2)

(2, 5)

(–1, 2)


y

x(2, 0)(–1, 0)

(2, 5)

(–1, 2)


In Problems 19–22, evaluate

∫

C

y dx+x dy on the given curve from (0, 0) to (1, 1).

19. y = x2 20. y = x



23. Evaluate

∫

C

(6x2 + 2y2) dx + 4xy dy, where C is given by x =

√t, y = t,

4 ≤ t ≤ 9.

24. Evaluate

∫

C

−y2 dx + xy dy, where C is given by x = 2t, y = t3, 0 ≤ t ≤ 2.

25. Evaluate

∫

C

2x3y dx + (3x + y) dy, where C is given by x = y2 from (1, −1)

to (1, 1).

26. Evaluate

∫

C

4x dx + 2y dy, where C is given by x = y3 + 1 from (0, −1)

to (9, 2).

In Problems 27 and 28, evaluate

∮

C

(x2 + y2) dx−2xy dy on the given closed curve.

27. 28.y

x

x2 + y2 = 4


y = x2

y = √x

y

x

(1, 1)


5.2 Complex Integrals 245

In Problems 29 and 30, evaluate∮Cx2y3 dx− xy2 dy on the given closed curve.

29. 30.y

x

(1, 1)(–1, 1)

(1, –1)(–1, –1)


y

x

(2, 4)


31. Evaluate∮C

(x2 − y2

)ds, where C is given by x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ 2π.

32. Evaluate∫−C

y dx−x dy, where C is given by x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ π.

33. Verify that the line integral∫Cy2 dx + xy dy has the same value on C for each

of the following parametrizations:

C : x = 2t + 1, y = 4t + 2, 0 ≤ t ≤ 1

C : x = t2, y = 2t2, 1 ≤ t ≤√

3

C : x = ln t, y = 2 ln t, e ≤ t ≤ e3.

34. Consider the three curves between (0, 0) and (2, 4):

C : x = t, y = 2t, 0 ≤ t ≤ 2

C : x = t, y = t2, 0 ≤ t ≤ 2

C : x = 2t− 4, y = 4t− 8, 2 ≤ t ≤ 3.

Show that∫C1

xy ds =∫C3

xy ds, but∫C1

xy ds �=∫C2

xy ds. Explain.

35. If ρ(x, y) is the density of a wire (mass per unit length), then the mass ofthe wire is m =

∫Cρ(x, y) ds. Find the mass of a wire having the shape of

a semicircle x = 1 + cos t, y = sin t, 0 ≤ t ≤ π, if the density at a point P isdirectly proportional to the distance from the y-axis.

36. The coordinates of the center of mass of a wire with variable density are givenby x = My/m, y = Mx/m where

m =

∫

C

ρ(x, y) ds, Mx =

∫

C

yρ(x, y) ds, My =

∫

C

xρ(x, y) ds.

Find the center of mass of the wire in Problem 35.

5.2 Complex Integrals

5.2In the preceding section we reviewed two types of real integrals. We saw that the definitionof the definite integral starts with a real function y = f(x) that is defined on an interval onthe x-axis. Because a planar curve is the two-dimensional analogue of an interval, we then

generalized the definition of∫ b

af(x) dx to integrals of real functions of two variables defined

on a curve C in the Cartesian plane. We shall see in this section that a complex integral isdefined in a manner that is quite similar to that of a line integral in the Cartesian plane.

Since curves play a big part in the definition of a complex integral, we begin with abrief review of how curves are represented in the complex plane.

5.2 Complex Integrals 245

In Problems 29 and 30, evaluate∮Cx2y3 dx− xy2 dy on the given closed curve.

29. 30.y

x

(1, 1)(–1, 1)

(1, –1)(–1, –1)


y

x

(2, 4)


31. Evaluate∮C

(x2 − y2

)ds, where C is given by x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ 2π.

32. Evaluate∫−C

y dx−x dy, where C is given by x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ π.

33. Verify that the line integral∫Cy2 dx + xy dy has the same value on C for each

of the following parametrizations:

C : x = 2t + 1, y = 4t + 2, 0 ≤ t ≤ 1

C : x = t2, y = 2t2, 1 ≤ t ≤√

3

C : x = ln t, y = 2 ln t, e ≤ t ≤ e3.

34. Consider the three curves between (0, 0) and (2, 4):

C : x = t, y = 2t, 0 ≤ t ≤ 2

C : x = t, y = t2, 0 ≤ t ≤ 2

C : x = 2t− 4, y = 4t− 8, 2 ≤ t ≤ 3.

Show that∫C1

xy ds =∫C3

xy ds, but∫C1

xy ds �=∫C2

xy ds. Explain.

35. If ρ(x, y) is the density of a wire (mass per unit length), then the mass ofthe wire is m =

∫Cρ(x, y) ds. Find the mass of a wire having the shape of

a semicircle x = 1 + cos t, y = sin t, 0 ≤ t ≤ π, if the density at a point P isdirectly proportional to the distance from the y-axis.

36. The coordinates of the center of mass of a wire with variable density are givenby x = My/m, y = Mx/m where

m =

∫

C

ρ(x, y) ds, Mx =

∫

C

yρ(x, y) ds, My =

∫

C

xρ(x, y) ds.

Find the center of mass of the wire in Problem 35.

5.2 Complex Integrals

5.2In the preceding section we reviewed two types of real integrals. We saw that the definitionof the definite integral starts with a real function y = f(x) that is defined on an interval onthe x-axis. Because a planar curve is the two-dimensional analogue of an interval, we then

generalized the definition of∫ b

af(x) dx to integrals of real functions of two variables defined

on a curve C in the Cartesian plane. We shall see in this section that a complex integral isdefined in a manner that is quite similar to that of a line integral in the Cartesian plane.

Since curves play a big part in the definition of a complex integral, we begin with abrief review of how curves are represented in the complex plane.


∮

C

dz

z2 + 1=

1

2i

∮

C

[1

z − i −1

z + i

]dz.and

We now surround the points z = i and z = −i by circular contours C1 andC2, respectively, that lie entirely within C. Specifically, the choice |z − i| = 1

2for C1 and |z + i| = 1

2 for C2 will suffice. See Figure 5.32. From Theorem 5.5we can write

∮

C

dz

z2 + 1=

1

2i

∮

C1

[1

z − i −1

z + i

]dz +

1

2i

∮

C2

[1

z − i −1

z + i

]dz

=1

2i

∮

C1

1

z − idz −1

2i

∮

C1

1

z + idz +

1

2i

∮

C2

1

z − idz −1

2i

∮

C2

1

z + idz. (9)

C1

C2

C

y

i

–i

x

Figure 5.32 Contour for Example 5 Because 1/(z+i) is analytic on C1 and at each point in its interior and because1/(z− i) is analytic on C2 and at each point in its interior, it follows from (4)that the second and third integrals in (9) are zero. Moreover, it follows from(6), with n = 1, that

∮

C1

dz

z − i = 2πi and

∮

C2

dz

z + i= 2πi.

Thus (9) becomes

∮

C

dz

z2 + 1= π − π = 0.

C

D

Figure 5.33 Contour C is closed but

not simple.

Remarks

Throughout the foregoing discussion we assumed that C was a simpleclosed contour, in other words, C did not intersect itself. Although weshall not give the proof, it can be shown that the Cauchy-Goursat theoremis valid for any closed contour C in a simply connected domain D. Asshown in Figure 5.33, the contour C is closed but not simple. Nevertheless,if f is analytic in D, then

∮Cf(z) dz = 0. See Problem 23 in Exercises

5.3.


In Problems 1–8, show that∮Cf(z) dz = 0, where f is the given function and C is

the unit circle |z| = 1.

1. f(z) = z3 − 1 + 3i 2. f(z) = z2 +1

z − 4

3. f(z) =z

2z + 34. f(z) =

z − 3

z2 + 2z + 2

7 Lista de Exercicıo - 2a Prova5.3 Cauchy-Goursat Theorem 263

5. f(z) =sin z

(z2 − 25)(z2 + 9)6. f(z) =

ez

2z2 + 11z + 15

7. f(z) = tan z 8. f(z) =z2 − 9

cosh z

9. Evaluate

∮

C

1

zdz, where C is the contour shown in Figure 5.34.

10. Evaluate

∮

C

5

z + 1 + idz, where C is the contour shown in Figure 5.35.

C

x

y

2


C

x

y

x4 + y4 = 16


In Problems 11–22, use any of the results in this section to evaluate the given integral

along the indicated closed contour(s).

11.

∮

C

(z +

1

z

)dz; |z| = 2 12.

∮

C

(z +

1

z2

)dz; |z| = 2

13.

∮

C

z

z2 − π2dz; |z| = 3

14.

∮

C

10

(z + i)4dz; |z + i| = 1

15.

∮

C

2z + 1

z2 + zdz; (a) |z| = 1

2, (b) |z| = 2, (c) |z − 3i| = 1

16.

∮

C

2z

z2 + 3dz; (a) |z| = 1, (b) |z − 2i| = 1, (c) |z| = 4

17.

∮

C

−3z + 2

z2 − 8z + 12dz; (a) |z − 5| = 2, (b) |z| = 9

18.

∮

C

(3

z + 2− 1

z − 2i

)dz; (a) |z| = 5, (b) |z − 2i| = 1

2

19.

∮

C

z − 1

z(z − i)(z − 3i)dz; |z − i| = 1

2

20.

∮

C

1

z3 + 2iz2dz; |z| = 1

21.

∮

C

Ln(z + 10) dz; |z| = 2

22.

∮

C

[5

(z − 2)3+

3

(z − 2)2− 10

z − 2+ 7 csc z

]dz; |z − 2| = 1

2

23. Evaluate

∮

C

8z − 3

z2 − zdz, where C is the “figure-eight” contour shown in Figure

5.36. [Hint : Express C as the union of two closed curves C1 and C2.]

C

1x

y


24. Suppose z0 is any constant complex number interior to any simple closed curvecontour C. Show that for a positive integer n,

∮

C

dz

(z − z0)n=

2πi, n = 1

0, n > 1.

In Problems 25 and 26, evaluate the given contour integral by any means.

25.

∮

C

(ez

z + 3− 3z

)dz, where C is the unit circle |z| = 1

5.3 Cauchy-Goursat Theorem 263

5. f(z) =sin z

(z2 − 25)(z2 + 9)6. f(z) =

ez

2z2 + 11z + 15

7. f(z) = tan z 8. f(z) =z2 − 9

cosh z

9. Evaluate

∮

C

1


10. Evaluate

∮

C

5


C

x

y

2


C

x

y

x4 + y4 = 16




11.

∮

C

(z +

1

z

)dz; |z| = 2 12.

∮

C

(z +

1

z2

)dz; |z| = 2

13.

∮

C

z

z2 − π2dz; |z| = 3

14.

∮

C

10

(z + i)4dz; |z + i| = 1

15.

∮

C

2z + 1

z2 + zdz; (a) |z| = 1

2, (b) |z| = 2, (c) |z − 3i| = 1

16.

∮

C

2z

z2 + 3dz; (a) |z| = 1, (b) |z − 2i| = 1, (c) |z| = 4

17.

∮

C

−3z + 2

z2 − 8z + 12dz; (a) |z − 5| = 2, (b) |z| = 9

18.

∮

C

(3

z + 2− 1

z − 2i

)dz; (a) |z| = 5, (b) |z − 2i| = 1

2

19.

∮

C

z − 1

z(z − i)(z − 3i)dz; |z − i| = 1

2

20.

∮

C

1

z3 + 2iz2dz; |z| = 1

21.

∮

C

Ln(z + 10) dz; |z| = 2

22.

∮

C

[5

(z − 2)3+

3

(z − 2)2− 10

z − 2+ 7 csc z

]dz; |z − 2| = 1

2

23. Evaluate

∮

C

8z − 3



C

1x

y



∮

C

dz

(z − z0)n=

2πi, n = 1

0, n > 1.


25.

∮

C

(ez

z + 3− 3z



5. f(z) =sin z

(z2 − 25)(z2 + 9)6. f(z) =

ez

2z2 + 11z + 15

7. f(z) = tan z 8. f(z) =z2 − 9

cosh z

9. Evaluate

∮

C

1


10. Evaluate

∮

C

5


C

x

y

2


C

x

y

x4 + y4 = 16




11.

∮

C

(z +

1

z

)dz; |z| = 2 12.

∮

C

(z +

1

z2

)dz; |z| = 2

13.

∮

C

z

z2 − π2dz; |z| = 3

14.

∮

C

10

(z + i)4dz; |z + i| = 1

15.

∮

C

2z + 1

z2 + zdz; (a) |z| = 1

2, (b) |z| = 2, (c) |z − 3i| = 1

16.

∮

C

2z

z2 + 3dz; (a) |z| = 1, (b) |z − 2i| = 1, (c) |z| = 4

17.

∮

C

−3z + 2

z2 − 8z + 12dz; (a) |z − 5| = 2, (b) |z| = 9

18.

∮

C

(3

z + 2− 1

z − 2i

)dz; (a) |z| = 5, (b) |z − 2i| = 1

2

19.

∮

C

z − 1

z(z − i)(z − 3i)dz; |z − i| = 1

2

20.

∮

C

1

z3 + 2iz2dz; |z| = 1

21.

∮

C

Ln(z + 10) dz; |z| = 2

22.

∮

C

[5

(z − 2)3+

3

(z − 2)2− 10

z − 2+ 7 csc z

]dz; |z − 2| = 1

2

23. Evaluate

∮

C

8z − 3



C

1x

y



∮

C

dz

(z − z0)n=

2πi, n = 1

0, n > 1.


25.

∮

C

(ez

z + 3− 3z



5. f(z) =sin z

(z2 − 25)(z2 + 9)6. f(z) =

ez

2z2 + 11z + 15

7. f(z) = tan z 8. f(z) =z2 − 9

cosh z

9. Evaluate

∮

C

1


10. Evaluate

∮

C

5


C

x

y

2


C

x

y

x4 + y4 = 16




11.

∮

C

(z +

1

z

)dz; |z| = 2 12.

∮

C

(z +

1

z2

)dz; |z| = 2

13.

∮

C

z

z2 − π2dz; |z| = 3

14.

∮

C

10

(z + i)4dz; |z + i| = 1

15.

∮

C

2z + 1

z2 + zdz; (a) |z| = 1

2, (b) |z| = 2, (c) |z − 3i| = 1

16.

∮

C

2z

z2 + 3dz; (a) |z| = 1, (b) |z − 2i| = 1, (c) |z| = 4

17.

∮

C

−3z + 2

z2 − 8z + 12dz; (a) |z − 5| = 2, (b) |z| = 9

18.

∮

C

(3

z + 2− 1

z − 2i

)dz; (a) |z| = 5, (b) |z − 2i| = 1

2

19.

∮

C

z − 1

z(z − i)(z − 3i)dz; |z − i| = 1

2

20.

∮

C

1

z3 + 2iz2dz; |z| = 1

21.

∮

C

Ln(z + 10) dz; |z| = 2

22.

∮

C

[5

(z − 2)3+

3

(z − 2)2− 10

z − 2+ 7 csc z

]dz; |z − 2| = 1

2

23. Evaluate

∮

C

8z − 3



C

1x

y



∮

C

dz

(z − z0)n=

2πi, n = 1

0, n > 1.


25.

∮

C

(ez

z + 3− 3z



5. f(z) =sin z

(z2 − 25)(z2 + 9)6. f(z) =

ez

2z2 + 11z + 15

7. f(z) = tan z 8. f(z) =z2 − 9

cosh z

9. Evaluate

∮

C

1


10. Evaluate

∮

C

5


C

x

y

2


C

x

y

x4 + y4 = 16




11.

∮

C

(z +

1

z

)dz; |z| = 2 12.

∮

C

(z +

1

z2

)dz; |z| = 2

13.

∮

C

z

z2 − π2dz; |z| = 3

14.

∮

C

10

(z + i)4dz; |z + i| = 1

15.

∮

C

2z + 1

z2 + zdz; (a) |z| = 1

2, (b) |z| = 2, (c) |z − 3i| = 1

16.

∮

C

2z

z2 + 3dz; (a) |z| = 1, (b) |z − 2i| = 1, (c) |z| = 4

17.

∮

C

−3z + 2

z2 − 8z + 12dz; (a) |z − 5| = 2, (b) |z| = 9

18.

∮

C

(3

z + 2− 1

z − 2i

)dz; (a) |z| = 5, (b) |z − 2i| = 1

2

19.

∮

C

z − 1

z(z − i)(z − 3i)dz; |z − i| = 1

2

20.

∮

C

1

z3 + 2iz2dz; |z| = 1

21.

∮

C

Ln(z + 10) dz; |z| = 2

22.

∮

C

[5

(z − 2)3+

3

(z − 2)2− 10

z − 2+ 7 csc z

]dz; |z − 2| = 1

2

23. Evaluate

∮

C

8z − 3



C

1x

y



∮

C

dz

(z − z0)n=

2πi, n = 1

0, n > 1.


25.

∮

C

(ez

z + 3− 3z



5. f(z) =sin z

(z2 − 25)(z2 + 9)6. f(z) =

ez

2z2 + 11z + 15

7. f(z) = tan z 8. f(z) =z2 − 9

cosh z

9. Evaluate

∮

C

1


10. Evaluate

∮

C

5


C

x

y

2


C

x

y

x4 + y4 = 16




11.

∮

C

(z +

1

z

)dz; |z| = 2 12.

∮

C

(z +

1

z2

)dz; |z| = 2

13.

∮

C

z

z2 − π2dz; |z| = 3

14.

∮

C

10

(z + i)4dz; |z + i| = 1

15.

∮

C

2z + 1

z2 + zdz; (a) |z| = 1

2, (b) |z| = 2, (c) |z − 3i| = 1

16.

∮

C

2z

z2 + 3dz; (a) |z| = 1, (b) |z − 2i| = 1, (c) |z| = 4

17.

∮

C

−3z + 2

z2 − 8z + 12dz; (a) |z − 5| = 2, (b) |z| = 9

18.

∮

C

(3

z + 2− 1

z − 2i

)dz; (a) |z| = 5, (b) |z − 2i| = 1

2

19.

∮

C

z − 1

z(z − i)(z − 3i)dz; |z − i| = 1

2

20.

∮

C

1

z3 + 2iz2dz; |z| = 1

21.

∮

C

Ln(z + 10) dz; |z| = 2

22.

∮

C

[5

(z − 2)3+

3

(z − 2)2− 10

z − 2+ 7 csc z

]dz; |z − 2| = 1

2

23. Evaluate

∮

C

8z − 3



C

1x

y



∮

C

dz

(z − z0)n=

2πi, n = 1

0, n > 1.


25.

∮

C

(ez

z + 3− 3z


8 Lista de Exercicıo - 2a Prova264 Chapter 5 Integration in the Complex Plane

26.

∮

C

(z3 + z2 + Re(z)

)dz, where C is the triangle with vertices z = 0, z = 1+2i,

and z = 1

Focus on Concepts

27. Explain why

∮

C

f(z) dz = 0 for each of the following functions and C is any

simple closed contour in the complex plane.

(a) f(z) = (5iz4 − 4z2 + 2 − 6i)9 (b) f(z) = (z2 − 3iz)e5z

(c) f(z) =sin z

ez2 (d) f(z) = z cos2 z

28. Describe contours C for which we are guaranteed that

∮

C

f(z) dz = 0 for each

of the following functions.

(a) f(z) =1

z3 + z(b) f(z) = csc z

(c) f(z) =1

1 − ez(d) f(z) = Ln z

29. Explain why the integral in Problem 25 is the same as

∮

C

(ez

z + 3− 3

z

)dz

and why, in view of (6), this form makes the integral slightly easier to evaluate.

30. Evaluate∫Cez dz from z = 0 to z = 2 + 2i on the contour C shown in Figure

5.37 that consists of the line y = x and a circle tangent to the line at (1, 1).

31. From Example 1 we know the value of∮Cezdz for any simple closed contour C

in the complex plane. In particular, use |z| = 1 as C and the parametrizationz = eiθ, 0 ≤ θ ≤ 2π, to discover the values of the real integrals

∫ 2π

0

ecos θsin(θ + sin θ)dθ and

∫ 2π

0

ecos θcos(θ + sin θ)dθ.

C3

C1

C2 (2, 2)

(1, 1)

x

y


5.4 Independence of Path

5.4In Section 5.1 we saw that when a real function f possesses an elementary antiderivative,that is, a function F for which F ′(x) = f(x), a definite integral can be evaluated by thefundamental theorem of calculus:

∫ b

a

f(x) dx = F (b)− F (a). (1)

5.4 Independence of Path 271

(ii) If f is a real function continuous on the closed interval [a, b], thenthere exists a number c in the open interval (a, b) such that

∫ b

a

f(x) dx = f(c)(b− a). (14)

The result in (14) is known as the mean-value theorem for definiteintegrals. If f is a complex function analytic in a simply connecteddomain D, it is continuous at every point on a contour C in D withinitial point z0 and terminal point z1. One might expect a resultparallel to (14) for an integral

∫ z1z0f(z) dz. However, there is no such

complex counterpart.


In Problems 1 and 2, evaluate the given integral, where the contour C is given in

the figure, (a) by using an alternative path of integration and (b) by using Theorem

5.7.

1.

∫

C

(4z − 1) dz 2.

∫

C

ez dz

|z| =1

i

–i

x

y


3 + i

3 + 3i

x

y

0


In Problems 3 and 4, evaluate the given integral along the indicated contour C.

3.

∫

C

2z dz, where C is z(t) = 2t3 + i(t4 − 4t3 + 2), −1 ≤ t ≤ 1

4.

∫

C

2z dz, where C is z(t) = 2 cos3 πt− i sin2 π

4t, 0 ≤ t ≤ 2

In Problems 5-20, use Theorem 5.7 to evaluate the given integral. Write each answer

in the form a + ib.

5.

∫ 3+i

0

z2 dz 6.

∫ 1

−2i

(3z2 − 4z + 5i) dz

7.

∫ 1+i

1−i

z3 dz 8.

∫ 2i

−3i

(z3 − z) dz

9.

∫ 1−i

−i/2

(2z + 1)2 dz 10.

∫ i

1

(iz + 1)3 dz

11.

∫ i

i/2

eπz dz 12.

∫ 1+2i

1−i

zez2

dz

13.

∫ π+2i

π

sinz

2dz 14.

∫ πi

1−2i

cos z dz

5.4 Independence of Path 271

(ii) If f is a real function continuous on the closed interval [a, b], thenthere exists a number c in the open interval (a, b) such that

∫ b

a

f(x) dx = f(c)(b− a). (14)

The result in (14) is known as the mean-value theorem for definiteintegrals. If f is a complex function analytic in a simply connecteddomain D, it is continuous at every point on a contour C in D withinitial point z0 and terminal point z1. One might expect a resultparallel to (14) for an integral

∫ z1z0f(z) dz. However, there is no such

complex counterpart.


In Problems 1 and 2, evaluate the given integral, where the contour C is given in

the figure, (a) by using an alternative path of integration and (b) by using Theorem

5.7.

1.

∫

C

(4z − 1) dz 2.

∫

C

ez dz

|z| =1

i

–i

x

y


3 + i

3 + 3i

x

y

0


In Problems 3 and 4, evaluate the given integral along the indicated contour C.

3.

∫

C

2z dz, where C is z(t) = 2t3 + i(t4 − 4t3 + 2), −1 ≤ t ≤ 1

4.

∫

C

2z dz, where C is z(t) = 2 cos3 πt− i sin2 π

4t, 0 ≤ t ≤ 2

In Problems 5-20, use Theorem 5.7 to evaluate the given integral. Write each answer

in the form a + ib.

5.

∫ 3+i

0

z2 dz 6.

∫ 1

−2i

(3z2 − 4z + 5i) dz

7.

∫ 1+i

1−i

z3 dz 8.

∫ 2i

−3i

(z3 − z) dz

9.

∫ 1−i

−i/2

(2z + 1)2 dz 10.

∫ i

1

(iz + 1)3 dz

11.

∫ i

i/2

eπz dz 12.

∫ 1+2i

1−i

zez2

dz

13.

∫ π+2i

π

sinz

2dz 14.

∫ πi

1−2i

cos z dz


15.

∫ 2πi

πi

cosh z dz 16.

∫ 1+(π/2)i

i

sinh 3z dz

17.

∫

C

1

zdz, C is the arc of the circle z = 4eit, −π/2 ≤ t ≤ π/2

18.

∫

C

1

zdz, C is the line segment between 1 + i and 4 + 4i

19.

∫ 4i

−4i

1

z2dz, C is any contour not passing through the origin

20.

∫ 1+√

3i

1−i

(1

z+

1

z2

)dz, C is any contour in the right half-plane Re(z) > 0

In Problems 21–24, use integration by parts (13) to evaluate the given integral.

Write each answer in the form a + ib.

21.

∫ i

π

ez cos z dz 22.

∫ i

0

z sin z dz

23.

∫ 1+i

i

zez dz 24.

∫ πi

0

z2ez dz

In Problems 25 and 26, use Theorem 5.7 to evaluate the given integral. In each

integral z1/2 is the principal branch of the square root function. Write each answer

in the form a + ib.

25.

∫

C

1

4z1/2dz, C is the arc of the circle z = 4eit, −π/2 ≤ t ≤ π/2

26.

∫

C

3z1/2 dz, C is the line segment between z0 = 1 and z1 = 9i

Focus on Concepts

27. Find an antiderivative of f(z) = sin z2. Do not think profound thoughts.

28. Give a domain D over which f(z) = z(z + 1)1/2 is analytic. Then find anantiderivative of f in D.

5.5 Cauchy’s Integral Formulas andTheir Consequences

5.5In the last two sections we saw the importance of the Cauchy-Goursat theorem in theevaluation of contour integrals. In this section we are going to examine several more con-sequences of the Cauchy-Goursat theorem. Unquestionably, the most significant of these isthe following result:

The value of a analytic function f at any point z0 in a simply connected domaincan be represented by a contour integral.



15.

∫ 2πi

πi

cosh z dz 16.

∫ 1+(π/2)i

i

sinh 3z dz

17.

∫

C

1


18.

∫

C

1


19.

∫ 4i

−4i

1


20.

∫ 1+√

3i

1−i

(1

z+

1

z2




21.

∫ i

π

ez cos z dz 22.

∫ i

0

z sin z dz

23.

∫ 1+i

i

zez dz 24.

∫ πi

0

z2ez dz



in the form a + ib.

25.

∫

C

1


26.

∫

C


Focus on Concepts







15.

∫ 2πi

πi

cosh z dz 16.

∫ 1+(π/2)i

i

sinh 3z dz

17.

∫

C

1


18.

∫

C

1


19.

∫ 4i

−4i

1


20.

∫ 1+√

3i

1−i

(1

z+

1

z2




21.

∫ i

π

ez cos z dz 22.

∫ i

0

z sin z dz

23.

∫ 1+i

i

zez dz 24.

∫ πi

0

z2ez dz



in the form a + ib.

25.

∫

C

1


26.

∫

C


Focus on Concepts






5.5 Cauchy’s Integral Formulas and Their Consequences 281

Because f is a polynomial, it is analytic on the region defined by |z| ≤ 2. ByTheorem 5.16, max|z|≤2 | 2z + 5i | occurs on the boundary |z| = 2. Therefore,on |z| = 2, (9) yields

| 2z + 5i | =√

41 + 20 Im(z). (10)

The last expression attains its maximum when Im(z) attains its maximum on|z| = 2, namely, at the point z = 2i. Thus, max|z|≤2 | 2z + 5i | =

√81 = 9.

Note in Example 5 that f(z) = 0 only at z = − 52 i and that this point is

outside the region defined by |z| ≤ 2. Hence we can conclude that (10) attainsits minimum when Im(z) attains its minimum on |z| = 2 at z = −2i. As aresult, min|z|≤2 | 2z + 5i | =

√1 = 1.

Remarks Comparison with Real Analysis

In real analysis many functions are bounded, that is, |f(x)| ≤M for all x.For example, sin x and cos x are bounded since |sin x| ≤ 1 and |cosx| ≤ 1for all x, but we have seen in Section 4.3 that neither sin z nor cos z arebounded in absolute value.


5.5.1 Cauchy’s Integral Formulas

In Problems 1–22, use Theorems 5.9 and 5.10, when appropriate, to evaluate the

given integral along the indicated closed contour(s).

1.

∮

C

4

z − 3idz; |z| = 5 2.

∮

C

z2

(z − 3i)2dz; |z| = 5

3.

∮

C

ez

z − πidz; |z| = 4 4.

∮

C

1 + ez

zdz; |z| = 1

5.

∮

C

z2 − 3z + 4i

z + 2idz; |z| = 3 6.

∮

C

cos z

3z − πdz; |z| = 1.1

7.

∮

C

z2

z2 + 4dz; (a) |z − i| = 2, (b) |z + 2i| = 1

8.

∮

C

z2 + 3z + 2i

z2 + 3z − 4dz; (a) |z| = 2, (b) |z + 5| = 3

2

9.

∮

C

z2 + 4

z2 − 5iz − 4dz; |z − 3i| = 1.3 10.

∮

C

sin z

z2 + π2dz; |z − 2i| = 2

11.

∮

C

ez2

(z − i)3dz; |z − i| = 1 12.

∮

C

z

(z + i)4dz; |z| = 2

13.

∮

C

cos 2z

z5dz; |z| = 1 14.

∮

C

e−z sin z

z3dz; |z − 1| = 3

5.5 Cauchy’s Integral Formulas and Their Consequences 281

Because f is a polynomial, it is analytic on the region defined by |z| ≤ 2. ByTheorem 5.16, max|z|≤2 | 2z + 5i | occurs on the boundary |z| = 2. Therefore,on |z| = 2, (9) yields

| 2z + 5i | =√

41 + 20 Im(z). (10)

The last expression attains its maximum when Im(z) attains its maximum on|z| = 2, namely, at the point z = 2i. Thus, max|z|≤2 | 2z + 5i | =

√81 = 9.

Note in Example 5 that f(z) = 0 only at z = − 52 i and that this point is

outside the region defined by |z| ≤ 2. Hence we can conclude that (10) attainsits minimum when Im(z) attains its minimum on |z| = 2 at z = −2i. As aresult, min|z|≤2 | 2z + 5i | =

√1 = 1.

Remarks Comparison with Real Analysis

In real analysis many functions are bounded, that is, |f(x)| ≤M for all x.For example, sin x and cos x are bounded since |sin x| ≤ 1 and |cosx| ≤ 1for all x, but we have seen in Section 4.3 that neither sin z nor cos z arebounded in absolute value.


5.5.1 Cauchy’s Integral Formulas

In Problems 1–22, use Theorems 5.9 and 5.10, when appropriate, to evaluate the

given integral along the indicated closed contour(s).

1.

∮

C

4

z − 3idz; |z| = 5 2.

∮

C

z2

(z − 3i)2dz; |z| = 5

3.

∮

C

ez

z − πidz; |z| = 4 4.

∮

C

1 + ez

zdz; |z| = 1

5.

∮

C

z2 − 3z + 4i

z + 2idz; |z| = 3 6.

∮

C

cos z

3z − πdz; |z| = 1.1

7.

∮

C

z2

z2 + 4dz; (a) |z − i| = 2, (b) |z + 2i| = 1

8.

∮

C

z2 + 3z + 2i

z2 + 3z − 4dz; (a) |z| = 2, (b) |z + 5| = 3

2

9.

∮

C

z2 + 4

z2 − 5iz − 4dz; |z − 3i| = 1.3 10.

∮

C

sin z

z2 + π2dz; |z − 2i| = 2

11.

∮

C

ez2

(z − i)3dz; |z − i| = 1 12.

∮

C

z

(z + i)4dz; |z| = 2

13.

∮

C

cos 2z

z5dz; |z| = 1 14.

∮

C

e−z sin z

z3dz; |z − 1| = 3

10 Lista de Exercicıo - 2a Prova282 Chapter 5 Integration in the Complex Plane

15.

∮

C

2z + 5

z2 − 2zdz; (a) |z| = 1

2, (b) |z + 1| = 2 (c) |z − 3| = 2, (d) |z + 2i| = 1

16.

∮

C

z

(z − 1)(z − 2)dz; (a) |z| = 1

2, (b) |z + 1| = 1 (c) |z − 1| = 1

2, (d) |z| = 4

17.

∮

C

z + 2

z2(z − 1 − i)dz; (a) |z| = 1, (b) |z − 1 − i | = 1

18.

∮

C

1

z3(z − 4)dz; (a) |z| = 1, (b) |z − 2 | = 1

19.

∮

C

(e2iz

z4− z4

(z − i)3

)dz; |z| = 6

20.

∮

C

(cosh z

(z − π)3− sin2 z

(2z − π)3

)dz; |z| = 3

21.

∮

C

1

z3(z − 1)2dz; |z − 2| = 5

22.

∮

C

1

z2(z2 + 1)dz; |z − i| = 3

2

In Problems 23 and 24, evaluate the given integral, where C is the figure-eight

contour in the figure.

23.

∮

C

3z + 1

z(z − 2)2dz 24.

∮

C

eiz

(z2 + 1)2dz

C

0 2x

y


C

i

–i

x

y


5.5.2 Some Consequences of the Integral Formulas

In Problems 25 and 26, proceed as in Example 5 to find the maximum modulus of

the given function on indicated closed circular region.

25. f(z) = −iz + i; |z| ≤ 5 26. f(z) = z2 + 4z; |z| ≤ 1

27. Suppose the boundary C of the closed circular region R defined by |z| ≤ 1 isparametrized by x = cos t, y = sin t, 0 ≤ t ≤ 2π. By considering | f(z(t)) |,find the maximum modulus and the minimum modulus of the given analyticfunction f and the points z on C that give these values.

(a) f(z) = (iz + 3)2

Documents

Ð - Início — UNIVASF Universidade Federal do Vale do ...edson.araujo/Disciplinas/metmat/lists/2016.2.L... · 2 Lista de Exercic o - 2a Prova 192 Chapter 4 Elementary Functions