Solutions to Selected Exercises. Complex Analysis with Applications by N. Asmar and L. Grafakos

Solutions to Selected Exercises in Complex Analysis with Applications by N. Asmar and L. Grafakos

Section. Complex Numbers Solutions to Exercises.. We have i + i. So a and b. 5. We have So a 3 and b 4. 9. We have So a 5 8 and b 7. i + i because i i + i 4 + 4i + {}}{ i 3 + 4i. + i 3 7 i 3 3 + i 4 7 {}}{ i i 7 5 8 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 8. 4 + 3i + i i + i 8 + 4i + 6i 3 5 4 + 4 i + 3i + 3i 4 + 5 5 + 4 5 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.

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

Section. Complex Numbers 3 5. + 3iz iz i { + 3i i}z i + 3i + iz i 4iz i z i 4i z 4 So z 4. 9. 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

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 + 6 + 4i z + iz 3i i + iz i z + i z + 6 + 4i z + iz 3i + + iz + z + i z 3 + i i + i 3 + i i 3iz + 8 + i + i 3iz + i 8 i 3iz 6 z 6 3i z i x + 4x + 5 x 4 ± 4 4 5 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

Section. Complex Numbers 5 and we get pz z 4 + 4 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 + 3 4 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.

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 +. 5. 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 +. 9. 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 5 5 5.

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.

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

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.

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 4. 45. 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

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

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 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 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 4. 9. 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

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 3.53. 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 3 3 + 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 3 4 5 + i 5 i. Thus Re + i 3, and Im + i 3 5. π 3 + i sin 4 + i sin 3π 4

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

Section.3 Polar form 5 37. 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 3 7 7 cos5π/7 + i sin5π/7, z 4 7 7 cos7π/7 + i sin7π/7, z 5 7 7 cos9π/7 + i sin9π/7, z 6 7 7 cosπ/7 + i sinπ/7, and z 7 7 7 cos3π/7 + i sin3π/7.

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 6 6. 45. We have z + 3 3i w 3 3 cos π + i sin π Set w z +

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 6 3 3 3 + i 6 3 3 5 3 + i w 3 3 cos 5π 6 + i sin 5π 6 3 3 3 + i 6 3 3 5 3 + i w 3 3 3 cos 9π 6 + i sin 9π 6 3 3i Since z w, the solutions of the original problem are z w 6 3 3 5 3 + i z w 6 3 3 5 3 + + i z 3 w 3 3 3i. Section.3 Polar form 7 49. We have Now let z + z + i w 3 + 4i w 5 3 5 + 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

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 3 5 5 + 3 + 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.

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

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.

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 + 3 + i i z + 3 + i e 3iπ 4 z + 3 + i

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 +. 7. 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 +. 5. We are given S {z C : Rez >, Imz > } as shown by the shaded region below.

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.

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.

Section.4 Complex Functions 5 33. 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

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

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

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.

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.

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.

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.

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 <. 4 5. 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

Section.5 Sequences and Series of Complex Nunbers 33 9. 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

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. 7 + 3in a n+ + in a n a n+ 7 + 3in a n + in a n+ a n 7 + 3in + in n 58 58 + 4n 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

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. 3 5. 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

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 6. 6 4π + i sin 3 4π e i 4π 3. 3 3e 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.

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 + 7 4 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.

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

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

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

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.

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

Section.7 Trignometric and Hyperbolic Functions 43 sinhz + πi i siniz + πi i siniz π i siniz sinh z. 4. + 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.

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

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

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.

Section.8 Logarithms and Powers 47 3. 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

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 4. 9. 5 i e i Log 5 iln 5 +i Arg 5 e e i ln 5 33. 3i 4 3 4 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.

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

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.

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

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.

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 53 5. 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

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 3. 7. 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.

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.

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 4. 5. By the quotient rule and the formula for the derivative of a polynomial it follows z 3 z 3 + z 3 + + z 3 + 3z z 3 + 3z z 3 +. 9. 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 99. 7. 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.

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

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,

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

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.

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.

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