Solutions to Selected Exercises. Complex Analysis with Applications by N. Asmar and L. Grafakos
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1 Solutions to Selected Exercises in Complex Analysis with Applications by N. Asmar and L. Grafakos
2 Section. Complex Numbers Solutions to Exercises.. We have i + i. So a and b. 5. We have So a 3 and b We have So a 5 8 and b 7. i + i because i i + i 4 + 4i + {}}{ i 3 + 4i. + i 3 7 i i 4 7 {}}{ i i i 7 3. Multiplying and dividing by the conjugate of the denominator, i.e. by i + i we get 4 + 3i i So a 3 and b i + i i + i 8 + 4i + 6i i + 3i + 3i i 3 + 8i 7. Multiplying and dividing by the conjugate of the denominator, i.e. by x iy x + iy we get x + iy x iy x + iy x + iy x iy x + iy x + iy x + y x + xyi + y i x + y x y + xyi x + y x y x + y + xy x + y i.
3 Chapter A Preview of Applications and Techniques So a x y x + y and b xy x + y.. Let z j x j + i y j, where x j, y j are real numbers and j,, 3. a We have b We have z + z x + x + i y + y x + x + i y + y z + z. z z x x y y + i x y + x y x x y y + i y x + y x z z. c The associative property x + x + x 3 x + x + x 3 is valid for real numbers and so it holds for the real and imaginary parts of z, z, z 3. Consequently we have z + z + z 3 z + z + z 3. d Note that z z z 3 x x y y + i x y + x y x 3 + i y 3 x x y y x 3 x y + x y y 3 + i x x y y y 3 + x y + x y x 3 and also that z z z 3 x + i y x x 3 y y 3 + i x y 3 + x 3 y x x x 3 y y 3 y x y 3 + x 3 y + i x x y 3 + x 3 y + y x x 3 y y 3. We simply check now that the real and imaginary parts of the complex numbers z z z 3 and z z z 3 coincide. Thus z z z 3 z z z 3. e Notice that while z z + z 3 x + i y x + y + i x 3 + y 3 x x + y y x 3 + y 3 + i x x 3 + y 3 + y x + y z z + z z 3 x + i y x + i y + x + i y x 3 + i y 3 x x y y + i x y + x y + x x 3 y y 3 + i x y 3 + x 3 y. The real and imaginary parts of these numbers are equal so the distributive property holds. z z + z 3 z z + z z 3
4 Section. Complex Numbers iz iz i { + 3i i}z i + 3i + iz i 4iz i z i 4i z 4 So z iz + i 4 iz + i 4 Conjugating both sides iz + i 4 Using problem 34 iz 4 i z 4 i i z 4 i i 4i + i z z 4i i i So z 4i. 33. We are given iz + z 3 + i z + iz + i From the first equation we obtain z 3 + i iz 3 + i z + iz
5 4 Chapter A Preview of Applications and Techniques We then substitute the expression for z into the second equation to get Thus So z i and z. 37. We have z + i3 + i z + iz + i z i z + iz 3i i + iz i z + i z i z + iz 3i + + iz + z + i z 3 + i i + i 3 + i i 3iz i + i 3iz + i 8 i 3iz 6 z 6 3i z i x + 4x + 5 x 4 ± ± 4 4 ± i ± i By the quadratic formula 4. We have a n z n + a n z n + + a z + a a n z n + a n z n + + a z + a a n z n + a n z n + + a z + a [Since a, a,, a n, a n are all real] a n z n + a n z n + + a z + a [by using the property that z n z n ] a n z n + a n z n + + a z + a. 45. We already know that z + i is a root of pz z 4 + 4, then then by problem 48, z i is also a root of pz. Thus hz z iz + i z z + divides pz
6 Section. Complex Numbers 5 and we get pz z z z + z + z + z iz + iz + iz + + i Hence, pz z iz +iz + iz ++i and its roots are +i, i, +i, i. 49. Let z x + iy such that z 3 + 4i. We have x + iy 3 + 4i x y + xyi 3 + 4i By comparing the real and imaginary parts we get the following equations x y 3 xy 4 From the first equation we have y x + 3 We now consider the second equation and substitute the expression for y xy 4 xy x y 4 Squaring both sides x x Substituting the expression for y x 4 + 3x 4 x 4 + 3x 4 u + 3u 4 Put u x u + 4u u Negative root is discarded since u x is non-negative x x ± Now from the relation xy, we compute the value of y y x ± We thus conclude that + i and i are the two square roots of 3 + 4i.
7 6 Chapter Complex Numbers and Functions Solutions to Exercises.. Note that z i, z i +i and z i +i. Thus z has coordinates,, z has coordinates, and z has coordinates,. Also z Note that z i + i. Hence z + i i and z + i i. Thus z has coordinates,, z has coordinates, and z has coordinates,. Also z We use the property that ab a b twice below: {}}{{}}{{}}{ + i i + 3i + i i + 3i + i i + 3i. 3. Using the properties z z z z and z z we have i i i i i i
8 Section. The Complex Plane 7 7. The equation z i 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.. Let z x + iy and we have z 4 x + iy 4 x + iy 4 x + y 6 Squaring both sides The inequality z 4 represents a closed disc with radius 4 units and centre at z as shown below. 5. Let z x + iy and we have < z i < < x + iy i < < x + iy < < x + y < The inequality < z i < represents a puntured open disc with radius units, centre at z + i and puctured at z + i as shown below.
9 8 Chapter Complex Numbers and Functions 9. a Let z x + iy and we have Rez a Rex + iy a x a Thus the equation Rez a represents the vertical line x a. b Let z x + iy and we have Imz b Imx + iy b y b Thus the equation Imz b represents the horizontal line y b. c Let z x + iy and let z x + iy, z x + iy be distinct points. We have z z + tz z t is a real variable x + iy x + iy + tx + iy x iy x + iy {x + tx x } + i{y + ty y } By comparing the real and imaginary parts we get From the first equation we have x x + tx x y y + ty y x x tx x x x x x t
10 Section. The Complex Plane 9 From the second equation we have y y ty y y y y y t Now by eliminating t from both the equations, we get x x y y x x y y x x x x y y y y which represents the equation of a line passing through distinct points x, y and x, y. 33. We know that z.z z z z. z Dividing both sides by z Thus we get that z z z. 37. a We get the estimate cos θ + i sin θ cos θ + i sin θ by the triangle inequality. Now we notice that i sinθ i sin θ sin θ sin θ. So, cos θ + i sin θ cos θ + sin θ. We know from algebra that sin θ, and cos θ Therefore we have cos θ + sin θ +, 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 is true for any number θ. Therefore we have cos θ + i sin θ cos θ + sin θ.
11 Chapter Complex Numbers and Functions 4. We want to know something about z 4 given some information about z. Notice that an 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 4 z 3. Therefore by the inequality z z z z we have z 4 z 3 z 3 Now since we know z we get that z 3 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: z a By triangle inequality we have n v j w j j n v j w j. And now we use the properties z z z z and z z and get j 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 n n v j w j j v j w j j n n n v j w j v j w j. j j b Using the hint we start from the obvious inequality n j j v j w j. Expanding the right hand side of this inequality we have n n v j w j v j v j w j + w j j n j j v j + w j n v j w j. j
12 Section. The Complex Plane Therefore n v j w j j n j v j + w j n n v j + w j j j n n v j w j. + j j c Using the hint we can consider v v, v,..., v n and w w, w,..., w n and look at them as at vectors. If we define v n j v j and w n j w j. Then it turns out we need to prove the following inequality: n v j w j v w. j In order to do so we define new vectors U u u and W w w. So, the coordinates of these vectors are correspondingly U j u j u and W j w j w. Then we have n U j j n u j u j u n j u j u u, and similar we get n j W j. Therefore we can apply part b to the vectors V and W. We have n n n V j W j V j W j. The left side is equal to j j n j v j v j j w j w v w The right side is equal to n v j n w j v w j n v j w j. j n v v j n w w j j j n n v j v w w j j u v. v w Hence we can rewrite the inequality we received in the form: n v j w j. v w So, if we multiply both sides by v w we get the needed inequality. j j
13 Chapter Complex Numbers and Functions Solutions to Exercises.3. 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 In Cartesian coordinates z is represented by 3 cos 7π 7π, 3 sin as shown below. 5. Let z 3 3 i. Then we have r Also, we can find the argument by evaluating cos θ x r 3 3 and sin θ y r 3 3. From the Table in the Section.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 4 + i sin 3π Again we can denote z i. Then we have r. We can evaluate cos θ x r / and sin θ y r / /. And, from the Table in the Section.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
14 Section.3 Polar form 3 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 cos π + i sin π. 3. We have z x + i y 3 + i. Since x > we compute Arg z tan y x tan Hence we express arg z.53 + πk for all integer k. 7. First, we need to express the number z 3 + i in the polar form. We compute r 3 +. And since 3 <, and > we see that Arg z tan 3 tan 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: 5π 5π i 3 cos + i sin 6 6 5π 5π 3 cos 3 + i sin 6 6 5π 5π 8 cos + i sin 6 6 3π 3π 8 cos + i sin. 6 6 Where in the last identity we used the fact that Arg 3 + i 3 5π 6 π 3π 6 should be in the interval π, π].. First, we find the modulus and the argument of the number z + i. We have r +. And, since for z x + iy + i we get x > we compute Arg z tan y x tan tan π 4. Hence z can be expressed in the polar form + i π π cos + i sin. 4 4 After this we can use De Moivre s identity to get π cos 4 + i 3 3π cos i 5 i. Thus Re + i 3, and Im + i 3 5. π 3 + i sin 4 + i sin 3π 4
15 4 Chapter Complex Numbers and Functions 5. a Set z z. Then Argz z Arg, whereas Argz + Argz π + π π. Thus Argz z Argz + Argz. z b Set z and z. Then Arg Arg, whereas Argz Argz z z π π. Thus Arg Argz Argz. z c Set z. Then Arg z Arg, whereas Argz Arg π. Thus Arg z Argz. d Set z. Then Arg z Arg, whereas Argz+π Arg +π π +π π. Thus Arg z Argz + π. 9. We know that for any complex numbers z r cos θ + i sin θ, and z r cos θ + i sin θ we have z z r [cosθ + i sinθ ]r [cosθ + i sinθ ] r r [cosθ + θ + i sinθ + θ ]. We can use this property several times to get the identity above step by step. z z z n z z z 3 z n r cosθ + i sinθ r cosθ + i sinθ z 3 z n r r cosθ + θ + i sinθ + θ z 3 z 4 z n r r cosθ + θ + i sinθ + θ r 3 cosθ 3 + i sinθ 3 z 4 z n r r r 3 cosθ + θ + θ 3 + i sinθ + θ + θ 3 z 4 z 5 z n r r r n cosθ + θ + + θ n + i sinθ + θ + + θ n z n r r r n cosθ + θ + + θ n + i sinθ + θ + + θ n r n cosθ n + i sinθ n r r r n r n cosθ + θ + + θ n + θ n + i sinθ + θ + + θ n + θ n. 33. We have z i z cos π + i sin π So by the formula for the n-th roots with n, we find the roots to be z cos π 4 + i sin π 4 z cos 5π 4 + i sin 5π 4
16 Section.3 Polar form In order to solve this equation we need to find the 7-th root of 7. Thus we express 7 in the polar form. 7 7cosπ + i sinπ. Hence by the formula for the n-th root with n 7 we have that the solutions are z 7 7 cosπ/7 + i sinπ/7, z 7 7 cos3π/7 + i sin3π/7, z cos5π/7 + i sin5π/7, z cos7π/7 + i sin7π/7, z cos9π/7 + i sin9π/7, z cosπ/7 + i sinπ/7, and z cos3π/7 + i sin3π/7.
17 6 Chapter Complex Numbers and Functions 4. We have z 4 i z 4 cos 5π 4 + i sin 5π 4. So by the formula for the n-th roots with n 4, we find the roots to be z 8 cos 5π 6 z 8 cos 3π 6 z 3 8 cos π 6 z 4 8 5π + i sin 6 + i sin 3π 6 + i sin π 6 cos 9π 9π + i sin We have z + 3 3i w 3 3 cos π + i sin π Set w z +
18 So by the formula for n-th roots for n 3, we find the roots to be w 3 3 cos π 6 + i sin π i i w 3 3 cos 5π 6 + i sin 5π i i w cos 9π 6 + i sin 9π 6 3 3i Since z w, the solutions of the original problem are z w i z w i z 3 w 3 3 3i. Section.3 Polar form We have Now let z + z + i w 3 + 4i w i4 5 z ± 4 i ± 3 + 4i Using the quadratic formula w 5cos θ + sin θ where π < θ < π and cos θ 3 5, sin θ 4 5
19 8 Chapter Complex Numbers and Functions So the principal square root is w 5 cos θ + i sin θ Now we use the half-angle identity i.e. cos θ + cos θ ±, sin θ cos θ ± compute and w i 5 + i. Thus the original solutions are z + w z w i i. 53. We have z 4 + iz + i u + iu + i Set u z u + i ± + i 4i Using the quadratic formula u + i ± i + i ± + i u The principal square root of i is + i, see problem 5 u, i.
20 Section.3 Polar form 9 Since z u, we get z i z i z 3 i cos 3π + i sin 3π cos 3π 4 + i sin 3π 4 + i z 4 i z 3 i. 57. By De Moivre s identity for n 3 we have cos 3θ + i sin 3θ cos θ + i sin θ 3 cos 3 θ + 3i cos θ sin θ + 3i cos θ sin θ + i 3 sin 3 θ cos 3 θ 3 cos θ sin θ + i3 cos θ sin θ sin 3 θ Now by comparing the real and imaginary parts we get cos 3θ cos 3 θ 3 cos θ sin θ sin 3θ 3 cos θ sin θ sin 3 θ. 6. We have z n and by the formula for n-th roots we get z n cos + i sin kπ kπ ω k cos + i sin n n where k,, n. 65. We prove the binomial formula for complex numbers by induction on n. The statement for n can be easily verified. By induction hypothesis, we assume that the statement is true for n and we prove it for n +. a + b n+ a + b n.a + b n n a n m.b m a + b m m
21 Chapter Complex Numbers and Functions By expanding the product we see that the coefficient of a n+ i b i in a + b n+ is equal to the sum of the coefficient of a n i b i in a + b n i.e. n i and the coefficient of a n+ i b i in a + b n i.e. n i. Thus we get n n n a + b n+ a n+ + + a n+ m b m + b n+ m m m n n + n n n + a n+ + a n+ m b m + b n+ use + m m m m m n+ n + n + n + a n+ m b m By convention. m n + m We have thus proved the statement for n + which shows that the binomial theorem holds true for complex numbers.
22 Section.4 Complex Functions Solutions to Exercises.4. We are given fz iz + + i. Now f + i i + i + + i + i f + i i + i + + i f i i i + + i 3 f i i i + + i 3 + i. 5. Let S be the square with vertices + i, + i, i, i which is shown below. Then f[s] is also a square with vertices, 3, 3 + i, + i which is shown below. 9. Observe that fz iz i i z i e 3iπ 4 z i
23 Chapter Complex Numbers and Functions Thus fz is obtained by rotating z clockwise by 3π 4, then stretching by a factor of and then translating unit up and 3 units to the right. 3. We are given fz z z + i. Now set z x + iy and we get fx+iy x+iy x+iy+i x +ixy y x iy+i x y x+ixy y+ Therefore ux, y x y x and vx, y xy y We are given fz 3 Argz. Now set z x + iy and we get fx + iy 3 Argx + iy Therefore 3 tan y x if x > ux, y 3 tan y x + 3π if x <, y 3 tan y x 3π if x <, y < and vx, y.. Let fz az + b and we have f 3 + i a + b 3 + i a 3 + i b Also we have f3i + 6i 3ia + b + 6i 3i3 + i b + b + 6i Substituting the expression for a 3 + 9i + 3ib + 6i 3ib 3i b So a 3 + i + i. Thus fz + iz We are given S {z C : Rez >, Imz > } as shown by the shaded region below.
24 Section.4 Complex Functions 3 An arbitrary point of S is of the form z x + iy, where x > and y >. Then under the mapping fz z + i, we have fz x + iy + i x + i y We hence note that Refz ranges in the interval < Refz < and Imfz ranges in the interval < Refz <. So as shown by the shaded region below. f[s] {z C : Rez <, Imz < } 9. S is a square with vertices + i, + i, i, i which is shown below.
25 4 Chapter Complex Numbers and Functions We see that the transformation f is given by fz 3z + e iπ 3z + i Also note that f + i 3 6i f + i 3 f i 3 f i 3 6i f[s] is a square with vertices given by 3, 6, 3,, 3,, 3, 6 which is shown below.
26 Section.4 Complex Functions We are given S {z C : < z 3, π 3 Argz π 3 region below. } as shown by the shaded If we write z rcos θ + i sin θ, then fz z r cos θ + i sin θ. Hence the polar
27 6 Chapter Complex Numbers and Functions coordinates of w fz ρcos φ + i sin φ are 3 < ρ <, and π 3 Arg w π 3. As r increases from to 3, ρ decreases from to 3 ; and as θ goes from π 3 up to π 3, φ decreases from π 3 to π 3. Thus as shown by the shaded region below. f[s] {z C : 3 < z <, π 3 Argz π 3 } 37. We are given S {x + iy : x } and is represented by the shaded region below. Now consider a vertical strip L x {x + iy : x x } of S. Let z be an arbitrary point of L x and assume x >. Then under the mapping fz z, we have fz x + iy x y + ix y
28 Section.4 Complex Functions 7 We set u Refz x y and v Imfz x y. By eliminating y, we get an algebraic relation between u and v, i.e. Thus we get u x v 4x f[l x ] {u + iv : u x v } which is a leftward-facing parabola with vertex at x, and v-intercepts at, ±x. Also it is not hard to see that f[l ], ]. Since S L x, we get f[s] x, ] f[l x ] x < {u + iv : u 4 v }. 6 4x x {u + iv : u x v 4x } In other words f[s] is the region enclosed within the parabola u 4 v 6 by the shaded region below. as represented 4. We are given S {x + iy : y }. Now consider a horizontal strip L y {x + iy : y y } of S. Let z be an arbitrary point of L y and assume y >. Then under the mapping fz z, we have fz x + iy x y + ixy
29 8 Chapter Complex Numbers and Functions We set u Refz x y and v Imfz xy. By eliminating x, we get an algebraic relation between u and v, i.e. Thus we get u v 4y y f[l y ] {u + iv : u v which is a rightward-facing parabola with vertex at y, and v-intercepts at, ±y. Also it is not hard to see that f[l ] [,. Since S L y, we get f[s] y f[l y ] [, <y {u + iv : u v 6 4 } 4y y } y {u + iv : u v In other words f[s] is the region enclosed within the parabola u v 4 as represented 6 by the shaded region below. 4y y } 45. We are given S {z C \ {} : π 4 Argz 3π 4 } {}. We consider the strip L r {z : z r, π 4 Argz 3π 4 } of S. Let z be an arbitrary point of L r, then under the mapping fz iz we get fz i r cos θ + i sin θ r cos θ + π + i sin θ + π.
30 Section.4 Complex Functions 9 As θ increases from π 4 to 3π 4, θ + π increases from π to π. Thus f[l r ] {z C : z r, π Argz π} and f. Since S {} L r, we get f[s] {f} <r < <r < f[l r ] {z C \ {} : π Argz π} {} In other words f[s] is the lower half complex plane including the real axis as shown by the shaded region below. 49. 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 z z + b, g z cos θ + i sin θz, and g 3 z rz we can find a representation for fz: fz g cos θ + i sin θrz g g rz g g g 3 z. And we notice that g is a translation, g is a rotation, and g 3 is a dilation. 53. If a real number z is positive then we can express it in the polar form z r rcos + i sin for r >. And it follows that Argz. If a real number z is negative then we can express it in the polar form z r rcosπ + i sinπ for r >. And it follows that Argz π. So, it follows that the image of the set S is two numbers and π.
31 3 Chapter Complex Numbers and Functions 57. We need to solve the equation z z which is equivalent to z. The solutions of the last equation are the square roots of which are + and. Hence the fixed points are ±. 6. a We know that w is in f[l]. This means that there is z m + i n with integers m and n such that z w. By the way we can compute w z m + i n m + i mn + i n m n + i mn Now take m n and n m. Then for z m + i n we compute z m + i n m n + i m n n m + i mn Hence w is also from f[l]. m n + i mn w. If we take z z m i n then we find Hence w is from f[l]. z z z w. Finally, we can compute Re w + i Im w Re w 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 + i n w z m + i n m + i mn + i n 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.
32 Solutions to Exercises.5. Note that Section.5 Sequences and Series of Complex Nunbers 3 lim a i sin n π n lim n n n lim n n. Therefore the sequence {a n } n converges to. 5. Note that lim a n lim n n cos n in n lim n cos n + in n + n lim n n Therefore the sequence {a n } n converges to. 9. a Let L lim a n. Then given ɛ > we have a n L < ɛ for n > N ɛ. Then n a n+ L < ɛ for n > N ɛ. Therefore lim a n L lim a n+. n n b We are given a i and a n+ 3 + a n lim a n+ lim n n lim a n+ n 3 + a n 3 + lim a n n L 3 + L L + L 3 L + 3L L, 3 by part a We shall now show that L 3 is absurd. Claim : Rea n Proof. We shall prove the statement by induction on n. Since Rea, the statement is true for n. Hence assume that the statement is true for some n. Then we have 3 Re a n+ Re + a n 3 Re + a n 3 + Re a n which completes the proof. Thus the claim rules out the possibility of L 3. Hence lim n a n.
33 3 Chapter Complex Numbers and Functions 3. We have n3 3 i + i n 3 i + i 3 3 i + i 3 + i n n +i 3 i + i. + i 3 i 3 i i + i 3 + i convergent since geometric series formula + i < 7. Let S N N n n + in + i N n N n + in + i denote the N-th partial sum. Then we have n n + i n + i n + in + i N n + i n + i n + i N + i Note that lim S N N + i lim N N + i + i The series n n + in + i is convergent since the partial sums S N converge and n n + in + i lim S N N + i.. The series + 3i n is convergent because it is a geometric series and the modulus n 4 of the common ratio is + 3i 4 < We apply root test to the series + in n. Note that n n ρ lim + in n n n lim + in n n n lim n n + i lim 4 + n n > Since ρ > we conclude that the series + in n is divergent. n n
34 Section.5 Sequences and Series of Complex Nunbers Note that e n ie n e n e n n ie n n e n n ie n n Using the root test on n en n gives lim n Similarly, for n en n, we have lim n so both of these series converge. Thus, converges. 33. Note that n n n e n n lim n e n <. n e n n lim n e n <, e n ie n e n e n n i n n z n is a geometric series and it converges iff n z < z < Now if z satisfies the inequality z < we get n z n e n n n z geometric series formula z. 37. Note that n is a geometric series and it converges iff z n z < z > z 5 > dividing both sides by
35 34 Chapter Complex Numbers and Functions Now if z satisfies the inequality z 5 > we get z n z z n n n z z z. geometric series formula 4. We are given that the n-th partial sum s n i n. Thus lim n s n Therefore the series converges to. 45. We have lim n i n lim n n in a n+ + in a n a n in a n + in a n+ a n 7 + 3in + in n n 4 n 4 + n 4 lim a n+ 58 a n lim n n 4 + n 4 n < Hence by the ratio test we conclude that the series a n converges. n
36 Solutions to Exercises.6 Section.6 The Complex Exponential 35. a We have e iπ cos π + i sin π. b We have e iπ cosπ + i sinπ. c We have 3e +iπ 3e e iπ 3e cosπ + i sinπ 3e 3 e. d We have e ln3+i π 3 eln3 e i π 3 3 cos π + i sin π i a We have b We have c We have d We have cos θ i sin θ cos θ + i sin θ e iθ. π π sin θ + i cos θ cos θ + i sin θ e i π θ. cos θ + i sin θ e iθ e iθ. cos θ + i sin θ cos3θ + i sin3θ eiθ e 3iθ eiθ 3iθ e iθ. 9. a We have e z e +i e e i e cos + i sin e cos + ie sin. b We have 3ie z 3ie i 3ie e i 3iecos + i sin 3e sin + i3e cos. c We have d We have e z e z e z.e z e z +z e +i+ i e. ez z e +i i e i cos + i sin. 3. a We have 3 3i 3 i 3 cos 5π + i sin 4 5π 3 e i 5π 4. 4
37 Complex Numbers and Functions b We have c We have d We have 3i 3 + i cos i 3 5π + i sin 6 cos 5π e i 5π π + i sin 3 4π e i 4π e i 3 cos i sin 3cosπ + + i sinπ + 3e iπ+. 7. Consider a vertical line segment of S, { L x x + iy : y π }. An arbitrary point z x + iy of L x is mapped to fz e z e x +iy e x e iy. Now as y varies from to π, the image fz traces a quarter circle having radius ex. Thus we get { fl x e x e iθ : θ π }. Therefore fs f 3 x 3 3 x 3 L x fl x { e x e iθ : θ π } 3 x 3 { re iθ : e 3 r e 3, θ π }. Consider a vertical line segment of S, L x {x + iy : y π} An arbitrary point z x + iy of L x is mapped to fz e z e x +iy e x e iy Now as y varies from to π, the image fz traces a semicircle having radius e x. Thus we get { } f[l x ] e x e iθ : θ π.
38 Section.6 The Complex Exponential 37 Therefore [ f[s] f <x < <x < <x < L x ] f[l x ] { } e x e iθ : θ π {z C : Imz } \ {}. 5. a We have e z i e z i e z 7π cos 4 e z e ln e i 7π 4 + i sin 7π 4 e z ln i 7π 4 z ln i 7π ikπ k Z 4 z ln + i k π k Z. b We have e z i e z cos π + i sin π e z e i π e z i π z i π ikπ k Z z i k + π k Z 4 9. Let z x + iy and we have e z e x x < Rez < e x+iy e x, see Exercise 33.
39 38 Chapter Complex Numbers and Functions Solutions to Exercises.7. a cos i eii + e ii e + e cosh cos cosh i sin sinh b sin i eii e ii i e e i e e i i sinh sin cosh + i cos sinh cos π π π cos cosh i sin sinh sin π π π sin cosh + i cos sinh c cosπ + i eiπ+i + e iπ+i eiπ e + e iπ e cos π + i sin π e + cos π i sin π e e + e e e cos π i sin π cosπ cosh i sinπ sinh
40 Section.7 Trignometric and Hyperbolic Functions 39 d sinπ + i eiπ+i e iπ+i i eiπ e e iπ e i cos π + i sin π e cos π i sin π e i e + e e e sin π + i cos π sinπ cosh + i cosπ sinh π cos + πi ei π +πi + e i π +πi e π i e π + e π i e π cos π + i sin π e π + cos π i sin π eπ e π + e π cos π e π i e π sin π π π cos coshπ i sin sinhπ π sin + πi ei π +πi e i π +πi i e π i e π e π i e π i cos π + i sin π e π cos π i sin π eπ i e π + e π sin π e π + i e π cos π π π sin coshπ + i cos sinhπ 5. cos + i cos cosh i sin sinh sin + i sin cosh + i cos sinh sin + i tan + i cos + i sin cosh + i cos sinh cos cosh i sin sinh cos + i cos + sinh sin + i sin + sinh
41 4 Chapter Complex Numbers and Functions 9. sinz sinx + iy sinx + iy sinx coshy + i cosx sinhy 3. tan z sin z cos z sinx + iy cosx + iy sinx coshy + i cosx sinhy cosx coshy i sinx sinhy sinx coshy + i cosx sinhy cosx coshy i sinx sinhy cosx coshy + i sinx sinhy cosx coshy + i sinx sinhy sinx cosx cosh y sinx cosx sinh y + isinhy coshy sin x + sinhy coshy cos x cos x + sinh y sinx cosx + i sinhy coshy cos x + sinh y sinx cosx sinhy coshy cos x + sinh + i y cos x + sinh y 7. Consider the horizontal strip L y { x + iy : π x π } The mapping fz sin z maps an arbitrary point of L y to fx + iy sinx + iy sin x cosh y + i cos x sinh y Set u sin x cosh y and v cos x sinh y. Observe that u cosh y v + sinh y Now as x varies in the interval π x π, the point u, v traces an upper semi-ellipse with u-intercepts ± cosh y, and v-intercept, sinh y. Thus { u v f[l y ] u + iv : +, v } cosh y sinh y
42 Section.7 Trignometric and Hyperbolic Functions 4 Thus [ f[s] f α y β α y β L y ] f[l y ] { u v u + iv : +, v } cosh y sinh y α y β { u v u v u + iv : +, + }. cosh α sinh α cosh β sinh β. sin z sinx + iy eix+iy e ix+iy i e y+ix e y ix i e y e ix e y e ix i e y cos x + i sin x e y cos x sin x i e y + e y e y e y sin x + i cos x sin x cosh y + i cos x sinh y sin z sinx + iy sin x cosh y + i cos x sinh y sin x cosh y + cos x sinh y sin x + sinh y + sin x sinh y sin x + sinh y.
43 4 Chapter Complex Numbers and Functions 5. cos θ eiθ + e iθ e cos 3 iθ + e iθ θ 3 cos 3 θ e3iθ + 3e iθ + 3e iθ + e 3iθ 8 cos 3 cos3θ + i sin3θ + cosθ + i sinθ + cosθ i sinθ + cos3θ i sin3θ θ 8 cos 3 cos 3θ + 3 cos θ θ 4 9. sin z cos z + cos z sin z e iz e iz e iz + e iz i + e iz + e iz e iz e iz eiz +z + e iz z e iz z e iz +z + e iz +z e iz z + e iz z e iz +z 4i eiz +z e iz +z i sinz + z i 33. e iz e iz e iz e iz sin z sin z i i e iz +z e iz z e iz z + e iz +z eiz z + e iz z eiz +z + e iz +z cosz z cosz + z. 37. coshz + πi cosiz + πi cosiz π cosiz cosh z
44 Section.7 Trignometric and Hyperbolic Functions 43 sinhz + πi i siniz + πi i siniz π i siniz sinh z sinh z + i siniz sin iz cos iz cosh z cos iz cosiz coshz cosiz cos iz sin iz cosiz + i siniz cosh z + sinh z. 45. cosh z cosh z + sinh z sinh z cosiz cosiz + i siniz i siniz cosiz cosiz siniz siniz cosiz iz cosiz z coshz z. 49. sinh z cosh z i siniz cosiz i siniz cosiz isiniz + iz + siniz iz i siniz + z i siniz z sinhz + z + sinhz z. 53. a Let S + z + + z n.
45 44 Chapter Complex Numbers and Functions Then zs z + z + + z n+. By subtracting the second equation from the first, we get S zs + z + + z n z + z + + z n+ zs z n+ S zn+ z z. b + e iθ + + e inθ ein+θ e iθ By Problem 53a. e in+θ e iθ e iθ e iθ ein+θ e iθ e iθ e iθ e in+θ e iθ i e iθ e iθ i i ein+θ e iθ sin θ c From part b, we get + e iθ + + e inθ i ein+θ e iθ sin θ iθ ie e in+ θ sin θ i cos θ i sin θ cos n + sin θ + sin n + θ cos θ sin θ + i θ i sin n + θ cos n + θ sin θ
46 Section.7 Trignometric and Hyperbolic Functions 45 By taking the real and imaginary parts of the above identity, we get Re + e iθ + + e inθ + sin n + θ sin θ + cos θ + + cos nθ + sin n + θ sin θ sin n + + cos θ + + cos nθ θ sin θ Im + e iθ + + e inθ cos θ cos n + θ sin θ sin θ + + sin nθ cos θ cos n + θ sin θ
47 46 Chapter Complex Numbers and Functions Solutions to Exercises.8. a logi ln i + i arg i π ln + i + kπ k Z. b log 3 3i ln 3 3i + i arg 3 3i ln 3 5π + i 4 + kπ k Z. c log 5e i π 7 ln 5e i π 7 + i arg 5e i π 7 π ln 5 + i 7 + kπ k Z. d log 3 ln 3 + i arg 3 ln 3 + i π + kπ k Z. 5. If we know log z, to find Log z, it suffices to choose the value of log z with the imaginary part lying in the interval π, π]. π Log i ln + i Log 3 3i ln 3 Log 5e i π π 7 ln 5 + i 7 Log 3 ln 3 + iπ. i 3π 4 9. Note that log ln + i arg ikπ k Z If we know log z, to find log 6 z, it suffices to choose the value of log z with the imaginary part lying in the interval 6, 6 + π]. Thus log 6 iπ.
48 Section.8 Logarithms and Powers We have e z 3 e z e Log 3 e z Log 3 z Log 3 ikπ k Z z Log 3 + ikπ k Z ln 3 + ikπ k Z 7. We have e z + 5 e z 5 e z Log 5 e z Log 5 e z Log 5 ikπ k Z z Log 5 + ikπ k Z π ln 5 + i + kπ k Z.. Note that π Log i Log i i Log ln + i Arg iπ π Log i ln i + i Arg i i π Log i ln i + i Arg i i. π iπ + i Log + Log i. 5. Log z + Log z 3π ln z + i Arg z + ln z + i Arg z 3π ln z + i Arg z 3π Arg z Arg z By comparing the real and imaginary parts, we get two relations ln z 3π Arg z
49 48 Chapter Complex Numbers and Functions From the first relation we get ln z 3π z e 3π z e 3π 4 From the second relation we get Arg z Arg z Thus z z. cos Arg z + i sin Arg z e 3π i e i Log 5 iln 5 +i Arg 5 e e i ln i i 4 8 Thus 3i 4 8 has a unique value. 37. Set z. Then Log z Log Log ln + i Arg iπ Again Thus we see that Log z Log ln + i Arg iπ iπ Log Log 4. By problem 38, the image of the punctured plane C \ {} under the mapping fz log 3π z is S 3π {z x + iy : 3π < y 5π}. 45. The domain of the map is S {z r e iθ : r, π < θ π}. The mapping fz Log z maps an arbitrary point z r e iθ of S to fz Log r e iθ ln r e iθ + i Arg r e iθ ln r + iθ.
50 Section.8 Logarithms and Powers 49 Set u ln r and v θ. As r varies between r, u varies between ln u and as θ varies between π < θ π, v also varies between π < v π. Thus f[s] {u + iv : ln u, π < v π}. 49. a We have b From part b, we have tan w sin w cos w e iw e iw tan w i i tan w eiw e iw e iw + e iw + i tan w + eiw e iw e iw + e iw e iw e iw + e iw e iw + e iw c From parts a and b we get i tan w eiw e iw e iw + e iw i tan w eiw e iw e iw + e iw e iw e iw + e iw + i tan w e iw e iw tan w + e iw e iw + e iw e iw + i tan w tan w eiw + iz iz eiw Set z tan w + iz log log e iw iz + iz log iw iz w i iz log + iz
51 5 Chapter Complex Numbers and Functions 53. a z p q e p q log z e p ln z +i arg z q e p q ln z +i Arg z+kπ k Z e p q Log z+ikπ k Z p Log z ikpπ q e e q k Z b Set E n e inpπ q. Now E n+q e in+qpπ q e inpπ q +ipπ e inpπ q e ipπ e inpπ q E n Since E n E n+q, there can be at most q values for z p q. c Lets suppose that E j E l for some j < l q. Thus e ijpπ q e ilpπ q e il jpπ q il jpπ ikπ for some integer k q pl j k q pl j d It is impossible for to be an integer since gcdp, q and < l j < q. q Therefore E n are distinct for n q and hence z p q has q distinct values.
52 Solutions to Exercises. Section. Regions of the Complex plane 5. We notice that the set {z : z } is the closed disk of radius one centered at the origin. The interior points are {z : z < }, or the open unit disk centered at the origin. And, the boundary is the set {z : z }, or the unit circle centered at the origin. 5. We see that the set {z : Re z > } is a right half-plane. Since the inequality Re z > is strict, this set is open. Really, if some z is in from the set {z : Re z > }. Then for any number < r < Re z and any complex number z such that z z < r we have Re z > Re z r >. Therefore z is also from the same set {z : Re z > }. This justifies that the set {z : Re z > } is open. Also, it is clear from its picture that any two points from the set {z : Re z > } can be connected even by a straight line segment, the simplest polygonal line. So, the set {z : Re z > } is also connected. Since it is simultaneously open and connected this set is a region. 9. We see that the set A {z : z, Arg z < π 4 } {} is an infinite sector with the vertex z included but the rays {z : z, Arg z ± π 4 } are not included. Since any open disc with the center at the origin z is not contained in the set A it follows this set is not open. On the other hand since the rays {z : z, Arg z ± π 4 } are not in this set A it follows the set A is also not closed. It is clear from the picture that any two points from the set A can be connected by a straight line segment. So, the set A is connected. Since it is not open the set A is not a region. 3. One of simple examples is the following. Let us take the sets A {} and B {}. Each of these sets is a point. Any single point is a connected set but it is obviously that A B {} {} consists of only two points and they cannot be connected by a polygonal line inside A B. 7. To prove this statement we need to show that if C \ S is closed then any point z of the set S is an interior point of S. Let us assume this is not true and there it is a point z of S which is not an interior point. This means that every neighborhood of z contains at least one point not from S. Thus, every neighborhood of z contains at least one point from the complement of S, the set C \ S. And z is not from C \ S. By definition of boundary of a set it follows that z is from the boundary of the set C \ S. But we know that the set C \ S is closed and thus it contains all its boundary. And, thus z must be in C \ S which is impossible since z is from the set S. This gives a contradiction to our assumption. Hence the assumption was wrong and the set S is open.. To show that the set A B is a region we need to show that is open and connected. Let z is a point of A B. Thus, z is either from A or from B. If z is from A then since A is open it follows some neighborhood of z is contained in A and thus also in A B. Similar if z is from B then some neighborhood of z is contained in B and thus in A B. It implies that z is an interior point of A B. Hence the set A B is open. To show that A B is connected we need to show that any two points z and z of A B can be connected by a polygonal line. Let the point z 3 is some point from the non-empty intersection of A and B. Then z 3 and z are from the same set A or B. Therefore they can be connected by a polygonal line l. Also z 3 and z are from the same set A or B. And they can be connected by a polygonal line l. Now it is obvious that z and z are connected by
53 5 Chapter Analytic Functions the polygonal line l l passing through z 3. Hence A B is connected. Since A B is simultaneously open and connected it is a region.
54 Solutions to Exercises.. Using properties of limits and the fact that lim z i z i we have lim z i 3z + z 3 lim z + lim z z i z i 3 lim z + lim z z i z i 3i + i 3 + i 4 + i. Section. Limits and Continuity First we add to fractions under the sign of limit. We use the identity z + z iz + i. We have z i z + z + i z iz + i z iz + i z + i z iz + i. We again use the properties from the Theorem..7 and the fact that lim z i z i to get lim z i z i z + z + i lim z i z iz + i lim z i z + i lim z i z i lim z i z + i i lim z i z ii [since lim fz lim z c z c fz ] i. i 9. We evaluate using properties of limits and the fact that Arg z is always real to get lim Arg z 3 z lim Arg z z 3 lim Arg z. z 3 We know that in general Arg z is discontinuous on the ray, ]. But it turns out that the function fz Arg z is continuous on the open ray,. Really if z is from the open ray, and z approaches to z from the second quadrant then Arg z approaches to π. Therefore Arg z also approaches π. Now if z approaches to z from the third quadrant then Arg z approaches to π. But Arg z approaches π again. So, in either case Arg z approaches π. Therefore we conclude lim z z Arg z π. If we take z 3 we have lim Arg z z z π. lim Arg z z 3
55 54 Chapter Analytic Functions 3. Since z approaches thus z we can divide by z both the numerator and the denominator. We have lim z z + 3i z + lim z + z 3i + z + lim z z 3i + lim z z 3i i 3i i We have lim z z lim z. z Now we notice that if fz approaches if z approaches then gz fz approaches. Also if lim z gz then lim z gz. Using these properties we compute lim. z z Since multiplying by a non-zero constant does not change approaching to we have lim z z.. We know that a real exponential function fx e x has the properties and lim fx, x + lim fx +. x So, we see that even for real z the function fz has different limits in the positive and negative directions. Hence, it also cannot have a unique limit when z approaches infinity in the complex plane C. 5. Approaching z from the positive x-axis and the negative x-axis, respectively, thus Since lim y x + z z lim y x 9. We need to show that lim y x + lim y x z z lim x + z z lim x z, therefore the limit lim z z z z x x, lim fz L lim f z z x x. does not exist. L; z Suppose the left statement is true. This means that for any ε > there is an R > such that z > R it follows fz L < ε. Now let ε > be any. Let us denote δ R where R is chosen as above. Then it follows that for any z < δ we have z > R. Thus it follows f z L < ε. Hence lim z f z L.
56 Section. Limits and Continuity 55 Now we suppose that the right statement is true. Again we pick any ε >. We can choose δ > such that if z < δ then f z L < ε. Let us denote R δ. If z > R then z < R δ. So, fz L f L < ε. z Hence lim z f z L. 33. If z 3i then the function hz z i z++3i is continuous in z. The discontinuity at z 3i is not removable because z i 3i i 4i and z + + 3i. And we have lim z 3i z i z + + 3i 4i lim z 3i z + + 3i 4i. 37. We use the definition of sin z using the exponential function. We have sin z eiz e iz. i From the Example..9 we know that e z is continuous everywhere in C. Since e iz fgz for gz iz and fz e iz and both functions are continuous it follows from Theorem..3 that e iz is also continuous. Similar we show that the function e iz is continuous. Then by Theorem..3 is continuous in C. sin z eiz e iz i i eiz i e iz 4. a First, assume that f is continuous and A is open. Let z be from f [A]. This means that fz is in A. Since A is open there is ε > such that the ε-neighborhood is a subset of A, or B ε fz A. Since f is continuous at z there is a δ > such that if z z < δ then fz fz < ε. Thus fz is in A. And therefore z which is in f [fz] is in f [A]. Therefore the whole δ-neighborhood of z is in f [A]. And it follows that f [A] is open. Second, assume that f [A] is open whenever A is open. Let z be any complex number and ε > be any too. Then by assumption the ε-neighborhood B ε fz of fz is open too. Therefore f [B ε fz ] is open. In particular, there is δ > such that B δ z f [B ε fz ]. It follows that for any z such that z z < δ we have fz fz < ε. Hence f is continuous. b First, assume that f [A] is closed whenever A is closed. To show that f is continuous by parta it is enough to show that f [B] is open whenever B is open. So, let open B be given. Since the complement of an open set is closed problem 7 Section. we have A C \ B is closed. Therefore, by assumption f [A] is closed too. Now we use the identity f [B] C \ f [A] which is true for any set B and A C \ B. Really, if z is in f [B]. Then fz is in B C \ A. Therefore fz is in the complement of f [A]. On the other hand if z is in C \ f [A] then fz cannot be in A C \ B. And, thus fz is in B. So, f z is in f [B]. Since f [A] is closed it follows from and Problem 7 Section. that the complement set f [B] is open. Thus f is continuous. Second, assume that f is continuous and B is closed. Then A C \ B is open. Therefore by part a it follows that f [A] is open. By it follows that f [B] C \ f [A]. f [A] is open. Then by Problem 7 Section. it follows that its complement, f [B] is closed.
57 56 Chapter Analytic Functions Solutions to Exercises.3. If we take gz 3z + z and hz z then we have fz 3z + z ghz. By the chain rule and the formula that z n nz n for a positive integer n we have f z g hzh z 6z + 6z + 6z By the quotient rule and the formula for the derivative of a polynomial it follows z 3 z 3 + z z 3 + 3z z 3 + 3z z We have fz z /3 z /3. So, if we take gz z /3 and hz z then fz hgz. So, by the chain rule and the formulas for the derivatives of polynomial and n-th root we get f z h gzg z gz 3 z 3/3 z /3 3 z /3 3 z /3. 3. If we take fz z and z then the limit is the definition of the derivative of f at z since fz. By the formula for the derivative of z n for a positive integer n we have z lim z z f z z If z is from the region {z : z < } then the function f coincides with the function gz z on some neighborhood of z. Since gz is differentiable everywhere then the limit used in the definition of differentiability of f coincides with the corresponding limit for g which we know is equal to g z. So, inside the region {z : z < } we have f z. Now if z is from the region {z : z > } then the function f coincides with the function hz z on some neighborhood of z. So, using the same argument as above we get that inside this region {z : z > } we have f z z z. The question of differentiability remains for z from the circle {z : z }. We have fz z. So, if z then z z and since f coincides with hz z inside the region {z : z > }. And hz is continuous everywhere. So, if z approaches z inside the region {z : z > } then fz hz approaches hz z which we know is not equal to fz. So, f is not continuous at a such z. Now if z then both g and h coincide at this point. Now let C be any path which approaches z inside the region {z : z < } for example C {z : z r, < r < } and C be any path which approaches z inside the region {z : z > } for example C {z : z r, r > }. Since g and h are differentiable everywhere we have fz fz gz gz lim lim z z z z z z z on z z C z on C g z and fz fz hz hz lim lim z z z z z z z on z z C z on C h z z.
58 Section.3 Analytic Functions 57 Since we have different limits when we approach z from different directions, it follows that the derivative of fz does not exist at z. Finally, we see that the derivative of f exist everywhere except the circle {z : z } and { f if z <, z z if z >,. We take gz z p/q and fz z q in Theorem 4 and get hz fgz z p/q q z p there. Since fz for any z and z p/q for z and not on the negative real axis we have f gz. We also know that gz e p q Log z+ikπ for some integer k. So, g is a composition of two continuous functions on its domain. Therefore g is also continuous. Now we can use Theorem 4 to get d dz zp/q h z f gz pzp qgz q pz p qz pzp z p/qq qz p z p/q [if we divide the numerator and the denominator by z p and multiply them by z p/q z] p qz zp/q. 5. a Since i + 7 i 6 +, we have b We have lim z i z + 7 z 6 + z 3 + 3iz + i 3z + + i lim z i z i 7 z + 6 z 6z 5. zi 3z + 3iz + i 3 3i. zi
59 58 Chapter Analytic Functions Solutions to Exercises.4. If x, y, then we can compute the derivatives directly since x + y. We have Similar u x xy xx + y xyx + y x x + y yx + y x y x + y. u y xy yx + y xyx + y y x + y xx + y xy x + y. Now if x, y, then we can notice that for any x we have Therefore ux, x x. + ux, u, u x, lim lim. x x x x Therefore the partial derivative of u with respect to x exists at, and is equal to. Similarly, we notice that for any y we have Therefore u, y y + y. u, y u, u y, lim lim. y y y y Hence u y exists at, and is equal to. To show that ux, y is not continuous at, we pick a point x, x for any number x. We compute ux, x x x + x. Therefore if x approaches zero then x, x approaches, but ux, x / does not approach u,. Therefore u is discontinuous at,. 5. Since f and g are differentiable at x x and y y correspondingly we have and fx fx + f x x x + ε x x x gy gy + g y y y + ε y y y where ε x as x x and ε y as y y. Therefore we have ux, y fxgy fx + f x x x + ε x x x gy + g y y y +ε y y y fx gy + gy f x x x + fy g x + mx, y,
60 Section.4 Differentiation of Functions of Several Variables 59 where mx, y f x g y x x y y +ε x x x gy + g y y y + ε y y y +ε y y y fx + f x x x. To show that that u is differentiable at z x, y it is enough to check that lim z z mz z z. x x First, we show that lim y y z z z z. To see it we will use the squeeze theorem and the inequality ab a +b. To prove the inequality we start from the obvious inequality a b which is equivalent to a ab + b or if we add ab to both sides we get a + b ab and if we divide by we get a +b ab which is the needed inequality. Now for a x x and b y y we get x x y y x x + y y z z We also can put a x x and b y y to receive similar x x y y x x + y y by Pythagorian theorem. z z. If we multiply the whole inequality by then we need to reverse the inequality and we get x x y y z z. If we combine the inequalities and we conclude that z z If we divide these inequalities by z z we get z z x x y y z z. x x y y z z z z. We have lim z z z z and lim z z z z. So, by squeeze theorem We notice that Now we find that x x y y lim. z z z z lim z z gy + g y y y + ε y y y lim y y gy + g y y y + ε y y y gy. ε x x x gy + g y y y + ε y y y lim z z z z ε x x x lim lim gy + g y y y + ε y y y gy z z z z z z
61 6 Chapter Analytic Functions since x x z z is bounded by and ε x if z z. Similarly, we find that ε y y y fx + f x y y lim z z z z ε y y y lim lim fx + f x x x fx. z z z z z z Finally adding the limits - we conclude that lim z z mz z z. 9. Project Problem: Is it true that if u y x, y for all x, y is a region Ω, then ux, y φx; that is, u depends only on x? The answer is no in general, as the following counterexamples show. For x, y in the region Ω shown in Figure 8, consider the function ux, y { if x >, sgny if x, where the signum function is defined by sgny,,, according as y <, y, or y >. Show that u y x, y for all x, y in Ω but that u is not a function of x alone. Note that in the previous example u x does not exist for x. We now construct a function over the same region Ω for which the partial exist, u y, and u is not a function of x alone. Show that these properties hold for { if x >, ux, y e /x sgny if x. Come up with a general condition on Ω that guarantees that whether u y on Ω then u depends only on x. [Hint: Use the mean value theorem as applied to vertical line segments in Ω.] For any point in the region x, y Ω, the partial derivative u y x, y at this point can be denoted as ux, y + h ux, y u y x, y lim. h h By the definition of the function ux, y, If x >, then u y x, y ; If x < and y < or y >, since the region does not contain the boundary, thus u y x, y ; Otherwise, consider the line segments along the y-axis. Since the region Ω does not contain the origin, for y >, x, we have Similarly, for y <, x, we have u, y + h u, y u y, y lim lim. h h h h u, y + h u, y u y x, y lim lim. h h h h Hence, it implies that the partial derivative satisfies u y x, y for all x, y Ω, but ux, y depends on both x and y.
62 Consider the u x in the previous example. Since If y > similar applying to y <, then Section.4 Differentiation of Functions of Several Variables 6 u + h, y u, y u x, y lim. h h uh, y u, y lim lim lim, h + h h + h h h which implies that u x does not exist for x. Now, consider the new function from this example. For the partial derivative u y x, y, If x >, then u y x, y u x x, y ; If x <, then we can also obtain that u y x, y, whenever y > or y < since e /x >, x R\{}. And for u x, since e /x is differentiable at R\{}, thus u x also exists. But u is not a function of x alone. Therefore, the properties hold for this new ux, y. Assume that the region Ω is convex or path-connected. Fix any x x, such that we can obtain a vertical line segment in Ω. Then for any x, y, x, y Ω, since the region Ω is convex, then the vertical line segment joins x, y and x, y is contained in Ω. By mean value theorem, we have ux, y ux, y u y x, ξy y, for some ξ between y and y. Since u y on Ω, then ux, y ux, y. Since x and then x, y, x, y Ω are all arbitrarily chosen, thus u depends only on x.
63 6 Chapter Analytic Functions Solutions to Exercises.5. For z x + iy we get z ux, y + ivx, y for ux, y x and vx, y y. Differentiating u with respect to x and y, we find u x, u y. Differentiating v with respect to x and y, we find v x, v y. Comparing these derivatives we see clearly that u x v u y and y v x. Hence the Cauchy-Riemann equations are satisfied at all points. The partial derivatives are clearly continuous everywhere, so by Theorem.4.4, ux, y and vx, y are differentiable everywhere.. Appealing to Theorem.5., we conclude that z is analytic at all points, or, entire. We compute the derivative as f x + iy u x x, y + iv x x, y, giving 5. Let fz e z, z x + iy. Then f z + i. e z e x iy e x cos y + i sin y e x cosy i e x siny ux, y + ivx, y for ux, y e x cosy and vx, y e x siny. Differentiating, we have u x ex cosy, v x ex siny, u y ex siny, v y ex cosy. If the Cauchy-Riemann equations are to be satisfied, we must have e x cosy and e x siny. However, this implies that siny cosy, and sin and cos are never simultaneously zero. Thus, the Cauchy-Riemann equations cannot be satisfied at any point, and fz is nowhere analytic. 9. For z x + iy we get ze z x + iye x+iy x + iye x cos y + i sin y e x x cos y y sin y + iy cos y + x sin y ux, y + ivx, y for ux, y e x x cos y y sin y and vx, y e x y cos y + x sin y. product rule with respect to x, we find u x x ex x + e x x x x ex x cos y e x y sin y e x x + e x cos y e x y sin y e x x cos y + cos y y sin y. cos y x ex y sin y Differentiating u using the
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