Section 1.1 Complex Numbers 1 Solutions to Exercises 1.1 1. We have So a 0 and b 1. 5. We have So a 3 and b 4. 9. We have i 0+ 1i. i +i because i i +i 1 {}}{ 4+4i + i 3+4i. 1 + i 3 7 i 1 3 3 + i 14 1 1 7 {}}{ i i 7 5 8 i 7 So a 5 8 and b 7. 13. Multiplying and dividing by the conjugate of the denominator, i.e. by i +i we get 14 + 13i i So a 3 and b 8. 14 + 13i + i i + i 8 + 14i +6i 13 5 17. Applying the quadratic formula we get 14 +14 i +13i +13i 4+1 15 5 + 40 5 i 3+8i x 0 ± 4 0± 4 0± i 6 1. Put u x. Then we get u +u +10. By the quadratic formula u ± 4 4 1. Therefore we get 1 x and the solution is x ±i.
Chapter 1 A Preview of Applications and Techniques 5. Suppose w can be written in the form w c + id. Then from the identity z + w 0 we have a + ib + c + id 0, or a + c+ib + d 0. It follows that a + c 0 and b + d 0. Hence c a and d b. And, this means w a ib. b Let e s represent a complex number such that z + e s z for all complex z. Show that e s 0; that is, Re e s 0 and Im e s 0. Thus e s 0 is the unique additive identity for complex numbers. Solution. Let us put z 0 into z + e s z. This gives 0 + e s 0, or if e s a + ib we get a + ib 0+i0. Since two numbers are equal if and only if their real and imaginary parts are the same we have a 0 and b 0. So, e s 0. 9. a We have a n z n + a n 1 z n 1 + + a 1 z + a 0 a n z n + a n 1 z n 1 + + a 1 z + a 0 a n z n + a n 1 z n 1 + + a 1 z + a 0 [Since a 0,a 1,,a n 1,a n are all real] a n z n + a n 1 z n 1 + + a 1 z + a 0 [by using the property that z n z n ] a n z n + a n 1 z n 1 + + a 1 z+a 0 b Suppose that z 0 is a root of the polynomial pz a n z n + a n 1 z n 1 + + a 1 z + a 0. This means that a n z0 n + a n 1 z0 n 1 + + a 1 z 0 + a 0 0. By using the part a we have a n z 0 n + a n 1 z 0 n 1 + + a 1 z 0 + a 0 a n z0 n + a n 1 z0 n 1 + + a 1 z 0 + a 0 0. Hence, the number z 0 is also a root of the same polynomial. 33. Project Problem: The cubic equation. a Using the given change of variables x y a 3 we get x 3 + ax + bx + c y a 3 + a y a + b y a + c 3 3 3 y 3 3y a a a 3 3 +3y + ay ay a a 3 3 3 3 + a +by b a 3 + c y 3 + y a + a+y a 3 a a3 + b+ 3 7 + a3 9 ab 3 + c y 3 + yb a 3 +a3 7 ab 3 + c y 3 + py + q 0
Section 1.1 Complex Numbers 3 if we put p b a a3 3 and q 3 + c. b Now we substitute y u + v into the equation we received in the previous step: 7 ab y 3 + py + q u + v 3 + pu + v+q u 3 +3u v +3uv + v 3 + pu + v+q u 3 + v 3 +3uvu + v+pu + v+q u 3 + v 3 +3uv + pu + v+q 0. c We suppose that u and v satisfy the desired property 3uv + p 0. Then we have uv p/3 and if we cube both sides we get u 3 v 3 p/3 3. And, also using the property that 3uv + p 0 and the equation from the part b we have u 3 + v 3 +3uv + pu + v+q u 3 + v 3 + q 0. So, it immediately follows that u 3 + v 3 q. d Let us show that U is a solutions of this quadratic equation. Because of the symmetry then V is also a solution! Using the equation U+V β we get V β U. We substitute it into the equation UV γ. We get 0UV γ U β U γ U βu γ, or if we rewrite it and multiply both sides by 1 we have U + βu + γ 0. And, this means that U is a solution of the given quadratic equation. e Straightforward we can put β q and γ p3 7 and use the part d to conclude that u 3 and v 3 are the solutions of the quadratic equation X + qx p3 7 0. By the quadratic formula we get that u 3 q + q +4p 3 /7 and v 3 q q +4p 3 /7, or u 3 q q + + p and v 3 q q 3 + p 3. Now if we get the cubic roots of both sides of the equations we get that u 3 q q p 3 + + and v 3 q q p 3. 3 + 3 f Substituting back the found values we have x y a 3 u + v a 3 3 q q p 3 + 3 + + q q p 3 3 a + 3 3.
4 Chapter 1 Complex Numbers and Functions Solutions to Exercises 1. 1. z has coordinates 1, 1, z has coordinates 1, 1, and z has coordinates 1, 1. 5. Since z 1 i 1 i 1+i we have that z has coordinates 1, 1, z has coordinates 1, 1, and z has coordinates 1, 1. 9. We use the property that ab a b twice below: {}}{{}}{{}}{ 1 + i1 i1 + 3i 1+i 1 i1 + 3i 1+i 1 i 1+3i 10. 13. Using the properties z1 z z 1 z and z z we have i i i i i i 1 5 5 5. 10 17. The equation z i 1 does not have any solution because z is a distance from the point z to the origin. And clearly a distance is always non-negative. 1. Let z x+iy satisfies the equation z + z 1 4. Since z x+iy and z 1 x 1+iy we have x +y + x 1 +y 4, or x +y 4 x 1 +y. If we square both sides of the equation we get x +y 16 8 x 1 +y +x 1 +y, or x +y 16 8 x 1 +y +x 4x +1+y. And if we subtract x +y from both sides of the equation and add 8 x 1 +y we get 8 x 1 +y 17 4x. If we square both sides of the equation one more time we get the equation of the ellipse we are looking for: 64x 1 +y 17 4x,or 64x 1 + 176y 17 4x. 5. a Let us recall that a b is the distance between a and b. So, z i 1 is the distance between z and 1 + i. Also Rez is the distance between z and 1 + i. Also Rez + 1 is the signed distance from z to the line Rez 1. Hence, if z satisfies the equation z 1 i Rez+1it follows that the distances from the line Rez 1 toz and from 1 + i toz are equal. So, this equation is a parabola equation which directrix is the line Rez 1 and the focus is 1 + i.
Section 1. The Complex Plane 5 b Similar z z 0 is the distance from z to z 0, and Rez a is the distance from z to the line whose equation is Rez a. So, it is also a parabola with the directrix whose equation is Rez a and the focus is z 0. 9. Let us concentrate on the circle which center is the origin and the radius is r z 1 z. Obviously the absolute value of any complex number lying outside of the disc bounded by this circle will be greater than r. Therefore, since from the picture it follows that the number z 1 z which is the tip of the vector z on the picture lies outside or on the circle of the disc described above then the absolute value which is z 1 z is greater than or equal to r. And, if we write this algebraically we get the needed inequality: 33. We get the estimate z 1 z z 1 z. cos θ + i sin θ cos θ + i sin θ simply by the triangle inequality. Now we notice that So, i sinθ i sin θ 1 sin θ sin θ. cos θ + i sin θ cos θ + sin θ. We know from algebra that sin θ 1, and cos θ 1 Therefore we have cos θ + sin θ 1+1, and this justifies the last step of the estimation above. b By definition of the absolute value we know that if z x + iy then z x + y. Therefore if z cos θ + i sin θ then z cos θ + sin θ. But from trigonometry we know the formula cos θ + sin θ 1 is true for any number θ. Therefore we have cos θ + i sin θ cos θ + sin θ 1. 1 37. We want to know something about z 4 given some information about z 1. Notice that an 1 upper bound of the quantity z 4 is given by the reciprocal of any lower bound for z 4. So, first we can find some information about z 4. For this we use some ideas from the example 8 in this section. We notice that z 4z 1 3. Therefore by the inequality z 1 z z 1 z we have z 4 z 1 3 z 1 3 Now since we know z 1 1 we get that z 1 3 1 3. Hence z 4. Finally taking reciprocals of the sides of the inequality we will reverse the sign of the inequality and get the needed upper estimate: 1 z 4 1. 41. a By triangle inequality we have v j w j v j w j.
6 Chapter 1 Complex Numbers and Functions And now we use the properties z 1 z z 1 z and z z and get v j w j v j w j v j w j. Using this identity in the triangle inequality above and recalling the assumption that we already proved?? we have v j w j v j w j n v j w j v j w j. b Using the hint we start from the obvious inequality. 0 v j w j Expanding the right hand side of this inequality we have n 0 v j w j v j v j w j + w j Therefore v j w j 1 v j + w j v j w j. v j + w j 1 n v j + w j n v j w j. 1 1 + 1 1 1 1 c Using the hint we can consider v v 1,v,..., v n and w w 1,w,..., w n and look at n them as at vectors. If we define v v n j and w w j. Then it turns out we need to prove the following inequality: v j w j v w. In order to do so we define new vectors U 1 u u and W 1 w w. So, the coordinates of these vectors are correspondingly U j uj u and W j wj w. Then we have U j u j u 1 u u j 1 u u 1, and similar we get n W j 1. Therefore we can apply part b to the vectors V and W.We have n V j W j V j W j.
Section 1. The Complex Plane 7 The left side is equal to v j w j v w 1 v w v j w j. The right side is equal to v j w j v w 1 v v j 1 w w j 1 n v j v w w j 1 u v 1. v w Hence we can rewrite the inequality we received in the form: 1 v j w j 1. v w So, if we multiply both sides by v w we get the needed inequality.
8 Chapter 1 Complex Numbers and Functions Solutions to Exercises 1.3 1. We need to present the number given in its polar form in the form with the real and imaginary parts z x + iy. We have z 3 cos 7π 7π + i sin 3 cos 7π 7π + i 3 sin 1 1 1 1 So, the number can be presented in the Cartesian coordinates as the point 3 cos 7π 7π 1, 3 sin 1. 5. Let z 3 3 i. Then we have r 3 + 3 9 3. Also, we can find the argument by evaluating cos θ x r 3 3 and sin θ y r 3 3. From the Table 1 in the Section 1.3 we see that θ 5π 4. Thus, arg z 5π 4 +kπ. Since 5π 4 is not from the interval π, π] we can subtract π and get that Arg z 5π 4 π 3π 4. So, the polar representation is 3 3 i 3 cos 3π + i sin 3π. 4 4 9. Again we can denote z i. Then we have r 1 1. We can evaluate cos θ x r 0 1/ 0 and sin θ y r 1/ 1/ 1. And, from the Table 1 in the Section 1.3 we find θ 3π. Also we could plot the complex number i as a point in the complex plane and find the angle from the picture. Therefore arg z 3π + πk. Since 3π is not from the interval π, π] we can subtract π and get that Arg z 3π π π. So, the polar representation is i 1 cos π + i sin π. 13. We have z x + iy 13+i. Since x>0 we compute Arg z tan 1 y x tan 1 13 0.153. Hence we express arg z 0.153 + πk for all integer k. 17. First, we need to express the number z 3+i in the polar form. We compute r 3+1. And since 3 < 0, and 1 > 0 we see that Arg z tan 1 1 tan 1 3 3 3 5π 6. Since we need to find the cube of the number z we can use the De Moivre s Identity to get the polar representation: 3 5π 5π 3+i 3 cos + i sin 6 6 3 5π 5π cos 3 + i sin 6 6 15π 15π 8 cos + i sin 6 6 3π 3π 8 cos + i sin. 6 6
Section 1.3 Polar form 9 Where in the last identity we used the fact that Arg 3+i 3 15π 6 π 3π 6 interval π, π]. should be in the 1. First, we find the modulus and the argument of the number z 1+i. We have r 1 +1. And, since for z x + iy 1+i1 we get x>0 we compute Arg z tan 1 y x tan 1 1 1 tan 1 1 π 4. Hence z can be expressed in the polar form 1+i π π cos + i sin. 4 4 After this we can use De Moivre s identity to get π cos 4 1 + i 30 30π cos 30 4 15 0 i 1 15 i. Thus Re 1 + i 30 0, and Im 1 + i 30 15. 5. By De Moivre s Identity we have π 30 + i sin 4 + i sin 30π i n cosπ/ + i sinπ/ n cosnπ/ + i sinnπ/. Now since the functions cosθ, and sinθ both have period π we notice that the values cosnπ/, and sinnπ/ depend only on what remainder gives n when is divided by 4. Therefore, we conclude that i n 1, if n 4k, for some integer k, i, if n 4k +1, for some integer k, 1, if n 4k +, for some integer k, i, if n 4k +3, for some integer k. 9. We know that for any complex numbers z 1 r 1 cos θ 1 + i sin θ 1, and z r cos θ + i sin θ we have z 1 z r 1 [cosθ 1 +i sinθ 1 ]r [cosθ +i sinθ ] r 1 r [cosθ 1 + θ +isinθ 1 + θ ]. We can use this property several times to get the identity above step by step. 4 z 1 z z n z 1 z z 3 z n r 1 cosθ 1 +i sinθ 1 r cosθ +i sinθ z 3 z n r 1 r cosθ 1 + θ +isinθ 1 + θ z 3 z 4 z n r 1 r cosθ 1 + θ +isinθ 1 + θ r 3 cosθ 3 +i sinθ 3 z 4 z n r 1 r r 3 cosθ 1 + θ + θ 3 +isinθ 1 + θ + θ 3 z 4 z 5 z n r 1 r r n 1 cosθ 1 + θ + + θ n 1 +isinθ 1 + θ + + θ n 1 z n r 1 r r n 1 cosθ 1 + θ + + θ n 1 +isinθ 1 + θ + + θ n 1 r n cosθ n +i sinθ n r 1 r r n 1 r n cosθ 1 + θ + + θ n 1 + θ n +isinθ 1 + θ + + θ n 1 + θ n
10 Chapter 1 Complex Numbers and Functions 33. First, we express the number i in the polar form: i 1cosπ/ + i sinπ/. Now the solutions of the equation z 4 i are the 4th roots of i. So, by the formula for the nth roots with n 4 we find the solutions z 1 cosπ/8 + i sinπ/8, z 3 cos9π/8 + i sin9π/8, z cos5π/8 + i sin5π/8, and z 4 cos13π/8 + i sin13π/8. 37. In order to solve this equation we need to find the 7-th root of 1. Thus we express 1 in the polar form. 1 cosπ+i sinπ. Hence by the formula for the n-th root with n 7 we have that the solutions are z 1 cosπ/7 + i sinπ/7, z cos3π/7 + i sin3π/7, z 3 cos5π/7 + i sin5π/7, z 4 cosπ+i sinπ/7, z 5 cos9π/7 + i sin9π/7, z 6 cos11π/7 + i sin11π/7, and z 7 cos13π/7 + i sin13π/7. 41. First, we can make a substitution and put w z 5. Then we need to solve the equation w 3 15 15cosπ+i sinπ. The solution set is given by the n-th root formula: w 1 3 15cosπ/3 + i sinπ/3 51/+i 3/, w 3 15cosπ+i sinπ 5 1+i 0, and w 3 3 15cos5π/3 + i sin5π/3 51/ i 3/. Recalling that z w + 5 we conclude that the solutions for the original problem are z 1 5+w 1 15 + i 5 3, z 5+w 0, and z 3 5+w 3 15 i 5 3. 45. We can use the formula for the solution of the quadratic equation az + bz + c 0 given by z b ± b 4ac. a We have in our problem a 1,b 1, and c i 4. Therefore the solution is z 1 ± 1 i. Therefore we need to find all possible square roots of 1 i. So, we express 1 i in the polar form: 1 i 1/ cos Hence two square roots of this number are π 4 + i sin π 4 w 1 1/4 cos π + i sin π, and w cos 1/4 8 8. 7π + i sin 8 7π. 8
Section 1.3 Polar form 11 And, of course w 1 w. Thus the solution for the original problem is z 1 1+w 1, and z 1 w 1. Also since π/ < π < 0 by the half-angle formulas we find 8 cos π 1 + cosπ/4 1 8 + 4 and sin π 1 + sinπ/4 1 8 + 4. Finally we can write down the solutions for the original equation in the form with radicals: z 1 1 + 1/4 z 1 1/4 1 + 4 i 1/4 49. By De Moivre s Identity with n 4wehave 1 + 4, and 1 + 4 + i 1/4 1 + 4. cos θ + i sinθ 4 cos 4θ + i sin 4θ. Therefore the real part of the number z cos θ + i sinθ 4 is equal to cos 4θ. So, we can directly evaluate the number z: cos θ + i sinθ 4 cos θ + i sinθ cos θ +i cos θ sin θ + i sin θ cos θ +i cos θ sin θ sin θ cos θ +i cos θ sin θ + sin θ + cos θi cos θ sin θ +i cos θ sin θ sin θ+ sin θ cos θ In the last equality we used the formula a + b + c a + b + c +ab +bc +ca which is easy to verify by direct computation similar to a + b a +ab + b. cos 4 θ 4 cos θ sin θ + sin 4 θ +4i cos 3 θ sin θ 4i cos θ sin 3 θ sin θ cos θ cos 4 θ 6 cos θ sin θ + sin 4 θ + i cos 3 θ sin θ 4 cos θ sin 3 θ. Finally, we see that the real part of the number z is equal to cos 4 θ 6 cos θ sin θ + sin 4 θ and is equal to cos 4θ by De Moivre s Identity. That is why these two trigonometric expressions are equal to each other. 53. a We compute ω 0 ω j n ω0 n ωj n 1 11, since both numbers ω 0 and ω j are nth roots of unity. This verifies that z ω 0 ω j satisfies z n 1 and thus is an nth root of 1. b We see that ω 0 ω j ω 0 ω k ω 0 ω j ω k 0 because ω 0 0 and ω j ω k since they are different nth roots of unity. This verifies that ω 0 ω j ω 0 ω k.
1 Chapter 1 Complex Numbers and Functions Also since all the complex numbers ω 0 ω j for 1 j n are different, are nth roots of unity by the part a, and there are exactly nnth roots of unity we conclude that the set ω 0 ω j n is the same as the set of all n roots of unity. c First, we choose some nth root of unity ω 0 which is different from 1 such one exists because all roots are different and there are at least of them since n. Since by the part b the set of all nth roots of unity ω j n is the same as the set ω 0 ω j n we see that the sum of all the numbers from the first set will be the same as the sum of all the numbers from the second one. That is ω 1 + ω + + ω n ω 0 ω 1 + ω 0 ω + + ω 0 ω n, or ω 1 + ω + + ω n ω 0 ω 1 + ω + + ω n. Now if we subtract the right hand side of the equation from both sides of it and factor out the sum of all nth roots of unity we find that 1 ω 0 ω 1 + ω + + ω n 0. This means that either 1 ω 0 0,orω 1 + ω + + ω n 0. But the first equation cannot be true because ω 0 is different from 1. Therefore we conclude that ω 1 + ω + + ω n 0. d Let us call z 1+ω 0 + ω0 + + ωn 1 0.Ifzis not zero then the number z1 ω 0 is also not zero because 1 ω 0 0 we remember that ω 0 is different from 1. So, we evaluate z1 ω 0 1+ω 0 + ω0 + + ωn 1 0 1 ω 0 11 ω 0 +ω 0 1 ω 0 + + ω0 n 1 ω 0 +ω0 n 1 1 ω 0 0 0 {}}{{}}{ 1 ω 0 + ω 0 1 ω0 n 0, since ω 0 is an nth root of unity. Hence we have ω 0 + ω 0 ω 3 0 + + ω n 0 z1 ω 0 0, and we divide by 1 ω 0 0 to show that z 0. 57. a We can use a formula from the problem 56. For n 1,,..., 0 {}}{ ω0 n 1 + ω0 n 1 ω0 n [ n ] n cos nθ k k0 cos θ n k 1 k sin θ k Since for the trigonometric functions there is an identity sin θ 1 cos θ we can use it in the formula above to get cos nθ [ n ] n k k0 [ n ] n k k0 cos θ n k 1 k sin θ k cos θ n k 1 k 1 cos θ k
Section 1.3 Polar form 13 Now if we substitute x cos θ into the formula above we find T n x T n cos θ cos nθ [ n ] n k k0 [ n ] n k k0 cos θ n k 1 k 1 cos θ k x n k 1 k 1 x k. b We can use the general formula for Chebyshev polynomials proved above for particular n. We have [ 0 ] 0 T 0 x 1 k k x 0 k 1 x k 0! 0!0! x0 1 x 0 1 1 1 1 11. k0 For n 1 we have [ 1 ] 1 T 1 x k For n we have For n 3 we get For n 4 we get k0 T x T 4 x Finally for n 5wehave T 5 x 1 k x 1 k 1 x k 1! 0!1! x1 0 1 x 0 1 1 1 x 1x. [ ] k k0 1 k x k 1 x k! x 0 1 x 0 + 1 1! x 1 1 x!0! 0!! x 1 x x 1. T 3 x [ 4 ] 4 k k0 [ 3 ] 3 k k0 1 k x 3 k 1 x k 3! 3!0! x3 1 x 0 3! 1!! x1 1 x 1 x 3 3x1 x 4x 3 3x. 1 k x 4 k 1 x k 4! 4!0! x4 1 x 0 4!!! x 1 x 1 + 4! 0!4! x0 1 x x 4 6x 1 x +1 x + x 4 8x 4 8x +1. [ 5 ] 5 k k0 1 k x 5 k 1 x k 5! 5!0! x5 1 x 0 5! 3!! x3 1 x 1 + 5! 1!4! x1 1 x x 5 10x 3 1 x +x1 x + x 4 16x 5 0x 3 +5x.
14 Chapter 1 Complex Numbers and Functions Solutions to Exercises 1.4 1. We compute f3 + 4i 3+4i+i 3+5i. 5. In order to find the principal root we need to express the value of the expression z + for z i + 1 in the polar form z +i +1 +i +i +1 +i + π π 8 cos + i sin. 4 4 Then using the formula for the principal root, that is rcos θ + i sin θ 1/ r 1/ cosθ/ + i sinθ/ we find the needed value: π π 1/ f1 + i 8 cos + i sin 8 1/ π π cos + i sin. 4 4 8 8 The values of cos π 8 and sin π 8 can be found by the half-angle formulas: cos θ 1 + cos θ We can use them for θ π/4. and sin θ 1 cos θ. cos π 8 1+ and sin θ 1. Finally we substitute these values to the expression found above to get f1 + i 8 1/4 1 + 4 + i 1. 4 9. We compute for z x + iy fz z 3 x + iy 3 x 3 +3x iy+3xiy +iy 3 x 3 3xy + i3x y y 3. Hence we find ux, y x 3 3xy and vx, y 3x y y 3. 13. We know that for z rcos θ + i sin θ 0, we have 1 z 1 cos θ +i sin θ. r So, we see that the formula makes sense for any complex number z except when z 0. Hence the largest subset of C where this function makes is all complex plane C except z 0. 17. We again use the formula for the principal root. For the complex number z rcos θ + i sin θ we have z 1 1 θ θ r cos + i sin. Since r 0 and the trigonometric functions cos θ and sin θ make sense for any θ we see that the principal root is defined for any z. And it follows that fz is defined for all complex numbers z.
Section 1.4 Complex Functions 15 1. We see that S is the disc of radius 1 with center at the origin. Now if we multiply any number z from S by 4 we get a point inside the disc of radius 4 centered at the origin which is S 1 {z : z < 4}. This means that the image fs is contained in S 1. On the other hand if z 1 is in S 1 then z1 4 is in the original disc S and f z 14 4 z 4 z. And this means that S 1 is in the image fs. So, S 1 is contained in fs and contains fs. This is only possible if fs S 1 {z : z < 4}. 5. a We compute fgz agz+b acz + d+b acz + ad + b acz +ad + b. This means that fgz is also linear. b First, we express the number a in the polar form a rcos θ + i sin θ. If we substitute this value to the function fz we obtain fz rcos θ + i sin θz + b cos θ + i sin θrz + b. So, if we take g 1 z z + b, g z cos θ + i sin θz, and g 3 z rz we can find a representation for fz: fz g 1 cos θ + i sin θrz g 1 g rz g 1 g g 3 z. And we notice that g 1 is a translation, g is a rotation, and g 3 is a dilation. 9. If a real number z is positive then we can express it in the polar form z r rcos0+i sin0 for r>0. And it follows that Argz 0. If a real number z is negative then we can express it in the polar form z r rcosπ+i sinπ for r>0. And it follows that Argz π. So, it follows that the image of the set S is two numbers 0 and π. 33. If we write z rcos θ + i sin θ, then fz 1 z 1 r cos θ +i sin θ. Hence the polar coordinates of w fz ρcos φ+i sin φ are 1 3 <ρ<, and π 3 Arg w π 3.Asrincreases from 0 to 3, ρ decreases from to 1 3 ; and as θ goes from π 3 up to π 3, φ decreases from π 3 to π 3. 37. a From the equation for fz we get fz z 1. So, since z 0 and 1 0 it follows that fz 0 otherwise the product of fz and z would be 0, not 1. b We know from the part a that for any z from S we have fz 0. Since f[s] is simply the union of all possible values of fz when z is from S it follows that f[s] cannot include zero. c We compute ffz 1 fz 1 1 z 1 z 1 z z z 1 z. We know that f[f[s]] is the set of all complex numbers w such that w fu when u is from d f[s]. But then u fz for some complex number z from S. And, if we use the identity from the part c we get w fu ffz z. So, it follows that f[f[s]] is contained in S. On the other hand if z is in S then fz is in f[s] and ffz is in f[f[s]]. So, it follows that S is contained in f[f[s]]. Combining these two facts we conclude that f[f[s]] S. 41. We need to solve the equation 1 z z which is equivalent to z 1. The solutions of the last equation are the square roots of 1 which are +1 and 1. Hence the fixed points are ±1. 45. a We know that w is in f[l]. This means that there is z m + in with integers m and n such that z w. By the way we can compute w z m + in m +imn+in m n +imn
16 Chapter 1 Complex Numbers and Functions Now take m 1 n and n 1 m. Then for z 1 m 1 + in 1 we compute Hence w is also from f[l]. z1 m 1 + in 1 m 1 n 1 +im 1 n 1 n m +i mn m n +imn w. If we take z z m in then we find Hence w is from f[l]. z z z w. Finally, we can compute Re w + i Im w Rew i Im w w. Since we already proved u w is from f[l]. And, for u from f[l] we have u is from f[l] as well. We can conclude that Re w + i Im w u is from f[l]. b This part follows from the formula given in the proof of the part a. That is for z m + in w z m + in m +imn+in m n +i mn. We notice that m n is integer and mn is integer. So, w z is also from L. c By the part b it follows that w is in L. Hence the function f maps the number w to fw which is already in f[l]. And this is what we needed to prove.
Section 1.5 The Complex Exponential 17 Solutions to Exercises 1.5 1. We compute e iπ e 0+iπ e 0 cos π + i sin π 1 1+i 0 1. 5. We compute e i 3π 4 e 0+i 3π 4 e 0 1 + i cos 3π 4 + i sin 3π 4 + i. 9. We evaluate 3e 3+i π 3e 3 cos π + i sin π 3e 3 0 + i 13e 3 i. 13. For z 1 3 i we have that the radius r z 1 + 3 4. 3 Since z is in the third quadrant, we have Arg z tan 1 + π 4π 3. 1 by dividing and multi- 17. First, we can find the standard representation for the number z 1+i 1 i plying by the conjugate of the denominator 1+i 1 i 1 + i1 + i 1 i1 i 1+i + i 1 +1 i i. Then we find z 0 +1 1. And, since the number z i is pure imaginary we find that Arg z π. Therefore z i 1 cos π + i sin π 1 e π. 1. We can notice that we need to find only the radius of the given complex number z e i π 1. And, if we write it in the polar form we have z e i π 1 e 0 i π 1 1 cos π 1 i sin π. 1 From this it follows that e i π 1 z 1. 5. a For z x + iy we have Similar we have Re e 3z Ree 3x+iy Re e 3x cos 3y + i sin 3y e 3x cos 3y. Im e 3z Im e 3x cos 3y + i sin 3y e 3x sin 3y.
18 Chapter 1 Complex Numbers and Functions b For z x + iy we first compute z x + iy x +ixy+iy x y +xy i. Then we find Re e z Ree x y +xy i Re e x y cos xy + i sin xy e x y cos xy. Similar we have Im e z Im e x y cos xy + i sin xy e x y sin xy. c For z x + iy we compute Re e z Ree x iy Ree x cos y +i sin y e x cos y. Similar we compute Im e z Ime x iy Ime x cos y +i sin y e x sin y. d For z x + iy we compute Re e iz Ree ix y Ree y cosx+i sinx e y cosx. Similar we compute Im e iz Ime ix y Ime y cosx+i sinx e y sinx. 9. Any complex number z inside the square S can be expressed in the form z x + iy where x 1 <x<x and α 1 <y<α. We can compute fz e z e x+iy e x cosy +i siny This means that in the polar form the radius of the number fz ise x. And, since x is a real number between x 1 and x then e x is any number between e x1 and e x. The argument of the complex number fz is equal to y. And, y is any number between α 1 and α. Therefore, f[s] is a sector of all complex numbers having in the polar form the radius between e x1 and e x and the argument between α 1 and α. 33.a Since we know that sin θ eiθ e iθ i
Section 1.5 The Complex Exponential 19 we compute e sin 4 iθ e iθ 4 θ i 1 e 4iθ 4e 3iθ e iθ +6e iθ e iθ 4e iθ e 3iθ + e 4iθ 16 1 e 4iθ + e 4iθ 4 eiθ + e iθ +6 8 1 cos 4θ 4 cos θ +6. 8 b We compute the integral using the linearization we received in a. 1 sin 4 θdθ 8 cos 4θ 4 cos θ +6dθ 1 3 sin 4θ 1 4 sin θ + 3 4 θ + C. 37. For a complex number z x + iy we can compute e z and e z separately. We have Also we have e z e x+iy e x e iy e x e x. e z e x+iy e x +y Then we use a property of the exponential function. For real numbers x 1 x we have e x1 e x. Now combining with the fact that x x x + y we conclude e z e x e x +y e z and the equality happens only if x x + y. And this is true only if y 0 and x 0. Or in other words when z is a non-negative real number.
Complex Numbers and Functions Solutions to Exercises 1.6 1. We have z i. Then we evaluate using the definition of cosz and sinz cosz eiz + e iz which verifies and 15; and ei + e i e 1 + e 1 sinz eiz e iz ei e i i i e 1 e i i i which verifies 3 and 16. 1 e + e cosh 1, e 1 e 1 i ie i i sinh 1, e 5. a For z 1+i we evaluate using the definition of cosz cosz eiz + e iz ei1+i + e i1+i ei 1 + e 1 i e 1 cos1 + i sin1 + e 1 cos 1 + i sin 1 e 1 cos1 + i sin1 + e 1 cos1 i sin1 cos1 e 1 + e 1 i sin1 e1 e 1 cos1 cosh1 i sin1 sinh1. Similar we evaluate sinz sinz eiz e iz i and tanz ei1+i e i1+i i ei 1 e 1 i i e 1 cos1 + i sin1 e 1 cos 1 + i sin 1 i 1 i e 1 cos1 + i sin1 e 1 cos1 i sin1 i cos1 e 1 e 1 + i sin1 e1 + e 1 i sin1 sinh1 + i cos1 cosh1, tan1 + i tanz sinz cosz sin1 sinh1 + i cos1 cosh1 cos1 cosh1 i sin1 sinh1 sin1 sinh1 + i cos1 cosh1 cos1 cosh1 + i sin1 sinh1 cos1 cosh1 i sin1 sinh1 cos1 cosh1 + i sin1 sinh1 i cos 1 cosh 1 + sin 1 sinh 1 cos 1 cosh 1 + sin 1 sinh 1
Section 1.6 Trigonometric and Hyperbolic Functions 1 Where we used the computed values of cos1 + i and sin1 + i for evaluating tan1 + i. b In this part we use the formulas REF17 and 18 for the values of cos z and sin z. We have for z x + iy with x 1 and y 1 cos1 + i cos z cos x + sinh y cos 1 + sinh 1, and sin1 + i cos z sin x + sinh y sin 1 + sinh 1. c From the part a we see that cos1 + i is represented by the point with the coordinates cos1 cosh1, sin1 sinh1, sin1+i is represented by the point with the coordinates sin1 sinh1, cos1 sinh1, and tan1 + i has coordinates 0, cos 1 cosh 1 + sin 1 sinh 1 cos 1 cosh 1 + sin 1 sinh. 1 9. We use the definition of sin z. We have for z x + iy sinz eiz e iz i eix+iy e ix+iy now we use polar representation i e y cosx+i sinx e y cos x+i sin x i e y cosx e y cos x + i e y sinx e y sin x i i i i cosx e y e y + sinx e y + e y Therefore we have sinz ux, y+ivx, y for ux, y sinx coshy and vx, y cosx sinhy. 13. For z x + iy we evaluate using the formulas REF15-16 for sin z and cos z as well as the formula REF17 for cos z tan z sin z sin zcos z cos z cos zcos z sin zcos z cos z [by formulas REF15-17] sin x cosh y + i cos x sinh ycos x cosh y + i sin x sinh y cos x + sinh y cos x sin xcosh y sinh y+i sinh y cosh ycos x + sin x cos x + sinh y [by properties REF8 and 7] cos x sin x + i sinh y cosh y cos x + sinh y Therefore we have tan z ux, y+ivx, y where ux, y cos x sin x and vx, y cos x+sinh y sinh y cosh y cos x+sinh y. 17. As in the Example 4, we will first find the image under f of a simple curve in the domain of definition, often a line segment or line. Then we will sweep the domain of definition with this curve and keep track of the area that is swept by the image. Fix α<y 0 <βand consider the horizontal
Complex Numbers and Functions line segment EF defined by: y y 0, π x π. Let u+iv denote the image of a point z x+iy 0 on EF. Using??, we get u + iv sinx + iy 0 sin x cosh y 0 + i cos x sinh y 0. Hence u sin x cosh y 0 and v cos x sinh y 0. Therefore 1 u cosh y 0 sin x and v sinh y 0 cos x. Note that v 0 because cos x 0 for π x π. Squaring both equations in 1 then adding them, we get u v + sin x + cos x 1. cosh y 0 sinh y 0 Hence as x varies in the interval π x π, the point u, v traces the upper semi-ellipse u cosh y 0 v + 1, v 0. sinh y 0 The u-intercepts of the ellipse are at u ± cosh y 0 and the v-intercept is at v sinh y 0.Asy 0 varies from α to β the upper semi-ellipses will vary between two upper semi-ellipse given by y 0 α and y 0 β whose u- and v-intercepts are at u ± cosh α, v sinh α and u ± cosh β, v sinh β. When y 0 changes from α to β continously the corresponding upper semi-ellipses will continously change from the ellipse corresponding to y 0 α to the ellipse corresponding to y 0 β and thus fill-in the shadowed area on Figure 4. 1. To prove??, we appeal to?? and???? and write sin z sinx + iy sin x cosiy + cos x siniy sin x cosh y + i cos x sinh y. To prove??, we use?? and the definition of the modulus of a complex number??, Section 1.. We also use the identity cosh y sinh y 1 for real hyperbolic functions. We get sin z sin x cosh y + cos x sinh y sin x 1 + sinh y + cos x sinh y sin x + sinh ycos x + sin x sin x + sinh y. Computing the square roots of both sides of the received identity we get?? proved. 5. To establish the identity we use the definition of sin z. sin z e iz e iz i eiz e iz i sin z. 9. sinz 1 + z sin z 1 cos z + cos z 1 sin z. Expanding the right-hand side of the identity above
Section 1.6 Trigonometric and Hyperbolic Functions 3 we have e iz1 e iz1 e iz + e iz + e + eiz1 iz1 e iz e iz 4i 4i [expanding the numerators we have] eiz1+z + e iz1 z e i z1+z e i z1+z 4i + eiz1+z + e iz z1 e iz1 z e iz1+z 4i [adding the fractions and canceling the middle terms we receive] eiz1+z e iz1+z 4i eiz1+z e iz1+z sinz 1 + z. i 33. Using already proved properties?? and?? of cos z we expand the right-hand side of the identity above cos z 1 cos z sin z 1 sin z cos z 1 cos z sin z 1 sin z cos z 1 cos z + sin z 1 sin z cos z 1 cos z + sin z 1 sin z sin z 1 sin z. 37. We use the definitions of cosh z and sinh z to get coshz + πi ez+πi + e z πi ez 1 + e z 1 sinhz + πi ez+πi e z πi ez 1 e z 1 ez e πi + e z e πi ez + e z ez e πi e z e πi ez e z cosh z, and sinh z. 41. First, we use the property?? or the problem 40 twice to show that the second, the third, and the fourth sides of the formula are equal to each other. We have cosh z + sinh z cosh z + cosh z 1 cosh z 1 1 + sinh z 1 1 + sinh z. Finally, we show that cosh z + sinh z is equal to the left-hand side. We have cosh z + sinh z ez + e z + ez e z 4 4 [adding the fractions] ez ++e z +e z +e z 4 ez + e z cosh z.
Complex Numbers and Functions 45. We use the definitions of the hyperbolic functions to prove this identity. We start from the right-hand side to show it is equal to the left-hand side. cosh z 1 cosh z + sinh z 1 sinh z + e ez1 z1 e z + e z + ez1 e z1 e z e z [expanding the numerators we get] ez1+z + e z1 z + e z1+z + e z1 z e + ez1+z z1 z e z1+z + e z1 z 4 4 [adding the fractions and simplifying we get] ez1+z +e z1 z + e ez1+z z1 z 4 coshz 1 + z. 49. We show that the left-hand side is equal to the right-hand side by using the result of the problem 46 and the facts that sinh z sinh z and cosh z cosh z which can proved by direct computation see the solution of a similar problem 5. By problem 46 the right-hand side is equal to sinh z 1 cosh z + cosh z 1 sinh z + sinh z 1 cosh z + cosh z 1 sinh z sinh z 1 cosh z + cosh z 1 sinh z + sinh z 1 coshz cosh z 1 sinh z sinh z 1 cosh z. 53. If tan z i had a solution then this would have meant that sin z i cos z. Now we use?? and?? to show that the assumption leads to a contradiction. Really, we would have for z x+iy sin x cosh y + i cos x sinh y i cosx cosh y + sin x sinh y, or equivalently sin xcosh y sinh y i cos xcosh y sinh y. We have a real number in the left-hand side and an imaginary number in the right-hand side. They can be equal only if they are both zeroes. Now we use the fact that the expressions in parenthesis never are equal to zero. Really, we compute cosh y sinh y ey + e y ey e y ey + e y e y + e y e y 0 because the exponential function is never equal to zero. At the same time it is impossible that sin x and cos x are both equal to 0 because of the trigonometric identity cos x + sin x 1. Therefore, we conclude that our assumption was wrong and tan z i does not have any solutions. Similar we show that the equation tan z i does not have solutions. We use again?? and?? to show that if tan z i for some z x + iy then sin x cosh y + i cos x sinh y i cosx cosh y sin x sinh y,
Section 1.6 Trigonometric and Hyperbolic Functions 5 or equivalently sin xcosh y + sinh y i cos xcosh y + sinh y which leads us to sin x + i cos xcosh y + sinh y 0. Now by direct computation we show that cosh y + sinh y e y 0 and sin x + i cos x 0 since we always have sin x + cos x 1. Thus, the assumption tan z i is wrong and the equation tan z i also does not have any solutions.
6 Chapter 1 Complex Numbers and Functions Solutions to Exercises 1.7 1. We will use the formula?? for evaluating log i. We have log i ln i + i arg i ln+i π +kπ where k is an integer. We remember that log z is a multiple-valued function. 5. Now we will use the formula?? for evaluating Log 3 + i 3. We have Log 3 + i 3 ln 3+i 3 + i Arg 3 + i 3 Since for the number z 3+i 3x + iy we have x>0 we compute Arg 3 + i 3 Arg z tan 1 y 3 x tan 1 3 π 6. Now we can substitute the found value to the formula 1 to find Log 3 + i 3 ln 3 +3+i π 6 ln 1 + i π 6. 9. We will use the formula?? to evaluate log π 1. We have log π 1 ln 1 + i arg π 1. So, we need to compute arg π 1. Since arg 1 kπ for all integer k we have π< arg π 1π<π+π. We substitute the found value to the formula 1 to get log π 1 0+i π i π. 13. We know that Therefore the solutions of the equations are for all integer k. e z 3 z log 3. z log 3 ln 3 + i kπ ln3+i kπ 17. The equation is equivalent to e z 5. We know that Therefore the solutions of the equations are for all integer k. e z 5 z log 5. z 1 log 5 1 ln 5 + i k +1π 1 1. We will use the formula?? for evaluating all three numbers Log 1 ln 1 + i Arg 1 ln 1 + i 00, k +1π ln 5 + i,
Section 1.7 Logarithms and Powers 7 Log i ln i + i Arg i ln1+i π i π, and Log i ln i + i Arg i ln1+i 3π i 3π. We see that for z 1 i and z i we have Log z 1 + Log z i π + i 3π i π. On the other hand Log z 1 z Log 1 0. So, in general we see Log z 1 z is not equal to Log z 1 + Log z. 5. We use the formula 9 for computing the principal value of z α. We have 5 1 i e 1 ilog 5. To evaluate this expression we need to compute the principal value of the logarithm of 5. We get Therefore, we conclude using the properties Log 5 ln 5 + i Arg 5 ln 5 + iπ. 5 1 i 1 iln 5+iπ e ln 5+π+i ln 5+π e e ln 5+π cos ln 5 + π+i sin ln 5 + π [since cosx + π cosx and sinx + π sinx] e ln 5 e π cos ln 5 i sin ln 5 5e π cos ln 5 + i sin ln 5 [since cos x cosx and sin x sinx] 5e π cosln 5 i sinln 5. 9. We use the definition of the complex power??to find i i e i log i iln i +i arg i e e i0+i π +kπ e π kπ for integer k. So, we see that i i has infinitely-many values and all of them are real numbers. 33. a Using the definition?? in Section 1.6 we evaluate 1+itan w 1+i sin w e cos w 1+i iw e iw i e iw +e iw 1+ eiw e iw e iw + e eiw + e iw iw e iw + e + eiw e iw iw e iw + e iw [adding two fractions and simplifying we get] e iw e iw + e iw.
8 Chapter 1 Complex Numbers and Functions b Acting similar to the solution of the part a we evaluate 1 i tan w 1 i sin w e cos w 1 i iw e iw i e iw +e iw 1 eiw e iw e iw + e iw eiw + e iw e iw + e iw eiw e iw e iw + e iw [subtracting two fractions and simplifying we get] e iw e iw + e iw. c In order to prove this formula for finding the yet unknown value w of tan 1 z we will use the definition tan 1 z w z tan w. To find the value of w we will explore the following expression So, we see that 1 iz 1+iz 1 i tan w 1+i tan w [now we use the formula from part a and b] and this equation is equivalent to e iw e iw +e iw e iw e iw +e iw e iw e iw e iw. 1 iz 1+iz e iw iw log 1 iz 1+iz, or if we divide both sides of the equation by i i we get an equivalent equation w i 1 iz log 1+iz, and since w tan 1 z we have proved the needed formula. 37. a We use the definition of log z. We have logz 1 { w : w Log z 1 +kπi for some integer k } {w : w ln z 1 + i Arg 1z } +kπi for some integer k {w : w ln z + i Arg 1z } +kπi for some integer k. Now we only need to notice that the principal value of the argument of z 1 is equal to the negative principal argument of z in the case if Arg z π. Really, if z re iθ then z 1 1 r e iθ for real r>0 and θ. So, if π <θ<πthen also π < θ <π. And, if Arg z π then similar we compute Arg z 1 π. In either case we have that either Arg z 1 Arg z or Arg z 1 Arg z +π. And if follows that log z 1 {w : w ln z i Arg z kπi for some integer k} { w : w lnz + i Arg z +kπi for some integer k}
Section 1.7 Logarithms and Powers 9 because if we add or subtract π from any number from the set above it is still in the same set. b We proceed similar to the solution of the part a. We have log z { 1 w : w Log z } 1 +kπi for some integer k z z { w : w ln z 1 z + i Arg z } 1 +kπi for some integer k z { w : w ln z 1 ln z + i Arg z } 1 +kπi for some integer k. z Now we notice that the difference between two numbers Arg z1 z and Arg z 1 Arg z is an integer multiple of π. Really, if z 1 r 1 e iθ1 and z r e iθ for r 1 > 0, r > 0, π < θ 1 π, and π <θ π then we also have z1 z r1 r e iθ1 θ. And, since the exponential function is πi-periodic we see that θ 1 θ differs from Arg z1 z by only an integer multiple of π. Thus, log z 1 z {w : w ln z 1 ln z + i Arg z 1 i Arg z +kπi for some integer k}. c To show that it is not true that logz log z we choose z 1. Then we evaluate log z log 1 {w : w kπi for some integer k} {w : w 4kπi for some integer k}. On the other hand logz log1 log 1 {w : w kπi for some integer k}. In particular, we see that πi is logz but not in log z since π is not an integer multiple of 4π. To show that logz log z we choose some complex number w from the set log z. By definition of the multiple-valued function fz log z it follows that e w z. If we square this identity we get e w z, or equivalently e w z. And this means that w is from the set logz.