* Problems may be difficult. (2) Given three distinct complex numbers α, β, γ. Show that they are collinear iff Im(αβ + βγ + γα) =0.

Transcription

1 Practice Exercise Set I * Problems may be difficult. ( Describe the sets whose points satisfy the following relations (a + = (b + < (c = (d arg + i = π 3 ( Given three distinct complex numbers α, β, γ. Show that they are collinear iff Im(αβ + βγ + γα =. (3 Prove that if = = 3,,, 3 distinct, then arg 3 3 = arg. (4 For what complex ( value will the following series converge n n (a Σ *(b Σ n= + n= + n (5 When can equality occur in the triangle inequality? That is under what conditions on, w will +w = + w? (6 What is the boundary of the set {:Re and Im are rational }? n (7 Establish the identity Σ α k β k = Σ n α k k= k= (This implies the auchy-schwar inequality n Σ k= n Σ k= β k Σ k<j n α k β k nσ k= α k β j α j β k for the case n =. α k nσ k= β k. (8 Suppose <a a... a n. Prove that the polynomial P ( =a n + a n a n has no root in the unit disk <. (Hint: onsider ( P (. *(9 Prove that if +i 9 +i =, then =. [ Hint: onsider 9.] **( Let P ( = n + c n c n with c,..., c n real. Suppose P (i <. Prove that there is a root x + iy of P ( in the set (x + y + 4y <.

2 Practice Exercise Set II (To receive a solution, you have to hand in some work. ( (exercise #5, p.7 Show that (a f( = Σ k k is continuous in <. k= (b g( = Σ is continuous in the right half plane Re >. k= k + ( Find lim x n, where n (a x n =+( n n n +. (b x n = cos nπ 4. (3 Find the radius of convergence of the following power series: (a Σ n!. n= (b Σ (n + n n. n= (4 Give an example of two power series Σ a n n and Σ b n n with radii of convergence R and R, n= n= respectively, such that the power series Σ (a n + b n n has a radius of convergence >R + R. n= (5 Explain why there is no power series f( = Σ c n n such that f( = for = n=, 3,,... and 4 f ( >. (6 Does there exist a power series f( = Σ c n n such that f( n= n = n and f( n = for n =,, 3, n3... *(7 If f( = Σ c n n satisfies f( n = n= n n, n =,, 3,..., compute the values of the derivatives + f (k (, k =,, 3,...

3 Practice Exercise Set III (To receive a solution, you have to hand in some work. ( Show that there are no analytic function f = u + iv with u(x, y =x + y. ( Suppose f is an entire function of the form f(x, y = u(x + iv(y. Show that f is a polynomial of degree at most one. *(3 What is the range of e if we take to lie in the infinite strip Im < π? What are the images for horiontal lines and vertical segments in Im < π under the e mapping? **(4 Discuss ( if it is possible to define log( continuously on \ [, ]. Also discuss the possibility for + log continuously defined on \ [, ]. *(5 Let G be a region and G = {: G} is the mirror image of G across the x-axis. If f: G is analytic, show that f : G defined by f ( =f( is analytic. (6 If f = u + iv is analytic on some domain, given u(x, y below, find the possibilities of v(x, y. (a u(x, y =x 3 3xy. (b u(x, y =e y cos x. (c u(x, y = log(x + y. y (d u(x, y = ( x + y. (7 Write in polar coordinates. Then f( = u(+ iv( = u(r, θ+ iv(r, θ. Establish the polar form of the auchy-riemann equations: u r = v r θ, v r = u r θ.

4 Practice Exercise Set IV (To receive a solution, you have to hand in some work. ( Find a conformal mapping from the open unit disk D = {: < } onto the following regions: (a the infinite strip < Im < i i *(b the upper semidisk <, Im > - ( Hint: Find a map from st quadrant onto the upper half semidisk. (c the slit disk D \ [, ( Hint: Use (b. *(d { } \ [, ] - ( Hint: Find a map from \ (, ] to ( {} \ [, ]. ( Let a<band T ( = ia ib. Define L = {:Im = b},l = {:Im = a},l 3 = {:Re =}. Determine which of the regions A, B,, D, E, F in Figure, are mapped by T onto the region U, V, W, X, Y, Z in Figure. A B L 3 ib D L E ia F L Y X Z Figure Figure ( Hint: Orient L 3 by (, ia, ib. T U V W

5 Practice Exercise Set V (To receive a solution, you have to hand in some work. f ( ( Suppose f( is analytic and f( < in a region Ω. Show that d = for every closed f( curve in Ω, assuming f is continuous. ( ompute = d, where the unit circle = is given the counterclockwise orientation. *(3 Define f( d = f( d. IfP ( is a polynomial and denotes the circle a = R (counterclockwise, show that P ( d = πir P (a. (4 Find d, where is a smooth curve from to not passing through the origin. *(5 Show that if f is a continuous real-valued function and f(, then π ( Hint: onsult the top half of p.45, and show that f sin t dt. (6 Evaluate the integral ( Hint: = + + d, where is the curve show below = f( d 4. i i

6 Practice Exercise Set VI (To receive a solution, you have to hand in some work. *( The following curve divides the plane into 4 regions. For each region, state the winding number of around points in that region. (Give answers by inspection, no computation needed. 3 4 Log( 49 + ( Find lim e (cos 5. *(3 If the range of an entire function lies in the right half plane Re w>, show that the function is a constant function. ( Hint: ompose a Möbius map. (4 Suppose a polynomial is bounded by in the unit disk. Show that all its coefficients are bounded by. (c.f. proof of Liouville s Theorem. (5 Find = sin d, = sin d (counterclockwise orientation. **(6 (Optional Let f( = u( + iv( (or f(x, y =(u(x, y, v(x, y be a one-to-one analytic function from the open unit disk D = y f v {: < } onto a domain G with finite area. x u u u (a Show that J f (x, y def x y = v v = f ( D G. x y (b For distinct nonnegative integers m, n, show m n da =(orthogonality relation, where da = r dr dθ = dx dy is the area differential. D (c Show that if f( = Σ c n n is the power series for f in D, then area of G = π Σ n c n. n= n=

7 Practice Exercise Set VII (Solutions will be distributed in class. ( Show that if f( is analytic in, there must be some positive integer n such that f( n n +. ( Suppose that f is analytic in the annulus:, that f for = and that f 4 for =. Prove that f( throughout the annulus. sin t (3 Show that f( = dt is entire function t *(a by applying Morera s Theorem (b by obtaining a power series expansion for f. (4 Show that if f( is continuous on the closed unit disk {: }, analytic on the open disk {: < } and real-valued on the unit circle {: =}, then f( is a constant function. ( Hint: The disk is conformally equivalent to the upper half plane. Reflection. *(5 Let D = {: < }. Iff: D D is analytic with at least two fixed points, prove that f(. ( Hint: May assume one of the fixed point is by composing with suitable Möbius transformations. *(6 Let f( be an entire function which is real on the real axis and imaginary on the imaginary axis, show that f( is an odd function, i.e. f( f(. ( Hint: onsider the coefficients of the power series of f( or make use the reflection property. **(7 If f is an entire function mapping the unit circle into the unit circle (i.e. f( = for =, then f( =e iθ n. ( Hint: f( has finitely many roots α,...,α n (repeated according to multiplicities in the unit disk. Recall α j α j = for =. Use modulus theorems to show f( n α j =eiθ Π j= α j.

8 Practice Exercise Set VIII (To receive a solution, you have to hand in some work. ( Suppose f is analytic on \{a,...,a n } and Γ is a simple closed curve surrounding a,..., a n as shown. For each a j, let j be a simple closed curve about a j inside Γ. Γ a a... an n Show that f( d = Σ n f( d, γ j= j where the orientation of Γ,,..., n are as shown. ( Identify the isolated singularities of the following functions and classify each as removable singularity, pole (and its order or essential singularity: (a 4 + (b cot (c e/ (d sin π. (3 Find the Laurent series of (4 Find =r 4 on (a < < 4, (b < <. sin d (counterclockwise orientation for r, π, π, 3π,... (5 Suppose f is analytic on \ { } and satisfies f( +. Prove f is constant. (6 If f has a pole at, show that e f( cannot have a pole at. (7 If f is analytic on R < <, wesay is a removable singularity, pole of order k, essential singularity of f( iff is a removable singularity, pole of order k, essential singularity of f(. (i Prove that an entire function with a pole at is a polynomial. (ii Prove that an analytic function on {} except for isolated poles must be a rational function.

9 Practice Exercise Set IX (To receive a solution, you have to hand in some work. ( Prove that the image of the plane under a nonconstant entire mapping f is dense in the plane. [ Hint: If f is not a polynomial, consider f(. ] ( Suppose that f is entire and that f( is real if and only if is real. Use the Argument Principle to show that f can have at most one root. [ Hint: Let Γ be the circle = R, R large, what is n(f Γ,? ] (3 Is there an analytic function f on {: } which sends the unit circle with counterclockwise orientation into the unit circle with clockwise orientation? *(4 If f is analytic on and inside a simple closed curve Γ, and f is one-to-one on Γ, then f is one-to-one inside Γ. [ Hint: If f Γ a simple closed curve? For w f Γ, let g( =f( w, what is n(g Γ,? ] (5 Show that if α and β are real, the equation n + α n + β = has n roots with positive real parts if n is odd, and n roots with positive real parts if n is even. (6 If a>e, show that the equation e = a n has n solutions inside the unit circle.

10 Practice Exercise Set X (To receive a solution, you have to hand in some work. ( Find dx, where n is a positive integer. [ Hint: Use the contour +xn R Reπi/n.] ( Find sin x x dx. [ Hint: Integrate ei i around a large semi-circle. ] (3 Find ln x x + dx. [ Hint: Use the contour -R -r r R -R+i (4 Find e x cos xdx. [ Hint: Use the contour -R π (You may need to know e x dx =..] R+i and f( =e.] R (5 Find cos x dx and sin x dx. [ Hint: Use the contour.] π/4 R (6 Find π/ dθ + sin θ. (7 Suppose f is analytic on r< <, then we define Res (f, = πi =R f( d (clockwise orientation. o (The clockwise orientation relative to is the counterclockwise orientation relative to. Equivalently, if f( = Σ a k k on r< <, then Res (f, = a. k= (a If f is meromorphic on with isolated poles at a,...,a n, show that Σ n Res (f,a j +Res (f, = j= (i.e. the sum of all residues on {} is. (b Show that Res (f, = Res (,. (c Find = sin( d. f(

11 Solution to Practice Exercise Set I ( (a The locus of all points whose distances from the two points a = and b = having a fixed quotient λ = is a circle with center on the line through a, b. On the real axis, = 3, { 3 satisfies the equation. So the circle is : = 4 }. 3 Alternatively, for = x + iy, (x + y = =4 + = 4[(x + + y ]. Simplifying we get (x y = 6 9. (b The locus of all points whose distances from the two points a = and b = having a fixed difference λ =( distance between a, b is a branch of a hyperbola having a, b as foci. If λ = distance between a, b (as is the case here, the branch degenerate to a ray (or an infinite slit. The set is the whole complex plane minus all real numbers greater than or equal to. Alternatively, for = x+iy, (x + + y (x + y = + < (x+ + y < [ + (x + y ] =4+4 (x + y +(x + y x < (x + y. If x<, then x < (x + y. If x, then (x < (x + y implies y. So the set is { x + iy: x<or(x and y }. (c = +. The locus of all points whose distances from two points a =andb = a b having a fixed product λ = is a lemniscate with foci at a and b. (The case λ> results in two curves, each about a focus (and as λ, the two curves shrink toward the foci; the case a b λ = results in a figure-eight curve with double point at a + b. The set is a lemniscate with foci at and having a double point at. 45 o Alternatively, for = r cis θ, (r cos θ +(r sin θ = =. Simplifying we get r =orr = cos θ. The range of θ possible are - / O α β α β α β / - 45 o - 3 π 4 θ π 4 or 3π 4 θ 5π 4. (d To interpret arg 3 ( = arg 3, write 3 = R cis α, = r cis β. Then arg [ ] 3 R = arg cis(α β = α β is 3 r the angle 3 (measured from the ray 3 counterclockwise to the ray 3. The set is the open arc of the circle containing all points such that i =6. -i ( γ ( α γ ( The line through β and γ is Im =. So α, β, γ collinear Im = β γ β γ ( (α γ(β γ Im β γ = Im(αβ γβ αγ + γ β γ = Im(αβ γβ αγ =. For any complex = x + iy, Im( = y =Im. So Im(αβ γβ αγ =Im(αβ + Im( γβ αγ = Im(αβ + βγ + γα = is the condition.

12 (3 O 3 (c.f. exercise (d This is just the complex way of expressing the geometry theorem that 3 =. (You should check also the case,. 3 are oriented clockwise. (4 (a Let w = +, then we know Σ w n converges iff w = n= ( < ( is closer to then Re >. (b ase : ( = r<. n + n n = r n n r (because n n + n. Apply ratio test to r n r n+ / r n rn Σ, we have lim n= rn n r n+ = r<. So rn Σ converges rn Σ n + n converges Σ n converges. +n ase : ( =. n + n n + = n (because +n + n n. So cannot +n converge to as n. Hence Σ n diverges. +n ase 3: ( >. For w =, w <, so by case, Σ n ( + = n Σ w n +( = Σ wn +w n w n converges. (5 Equality holds iff either =. w =or w is real. (6 For any complex w, every neighborhood B(w, r ofw contains a point with Re and Im rational and also a point not both Re and Im rational. So any complex w is in the boundary of the set. Therefore. the boundary of the set is all complex numbers. (7 L.H.S. = α β +α β =(α β +α β (α β +α β = α β +α β α β +α β α β + α β. R.H.S. = ( α + α ( β + β α β α β = α β + α β + α β + α β (α β α β (α β α β = α β + α β + α β + α β ( α β α β α β α β α β + α β. Alternative solution for n N, n by how hak-on. α k β j α j β k =(α k β j α j β k (α k β j α j β k = α k β j α k β k α j β j α j β j α k β k + α j β k. n n Σ Σ k= j= Since Σ n n Σ k= j= α k β j α j β k = n Σ n Σ k= j= α k β j α j β k = Σ ( α k β j α k β k α j β j α j β j α k β k + α j β k. k=j n = Σ i k<j n and Σ n n ( nσ ( nσ n Σ α k β k α j β j = α k β k α j β j = k= j= k= j= Σ α k β k. k= Therefore, Σ α k β j α j β k = Σ n α k n Σ β k k<j n k= k= both sides and rearranging terms we get the desired result. α k β j α j β k + Σ α k β j α j β k + Σ α k β j α j β k k<j n j<k n α k β j α j β k n Σ k= α k β k. ancelling the factor of on

13 (8 Suppose is a root of P ( and <, then = ( P ( = a n+ +(a a n +(a a n + +(a n a n + a n = a n [a n+ +(a a n + +(a n a n ] (* a n a n+ +(a a n + +(a n a n (** a n [a n+ +(a a n + +(a n a n ] (*** > a n [a +(a a + +(a n a n ] =, a contradiction (where (* α β α β, (** α + β α + β, and (*** <. (9 Let = x + iy, then 9 = 9 = i +i = + y + y + x x + y + y +. If <, then x + y < and + y + y + x > x + y + y +, forcing 9 >, a contradiction. If >, then x + y > and + y + y + x < x + y + y +, forcing 9 <, a contradiction. ( (We first observe that (x + y + 4y = x + y ++y x + y + y = i (x + iy i (x iy. Suppose the roots of P ( are r, r,..., r n. Because the coefficients are real, complex roots occur in conjugate pairs if any. Since > P (i = i r i r... i r n. For a real root r, i r = +r. So P ( must have complex roots. ( Now P (i = Π real roots ( i r Π complex roots in pairs i r i r. So there must be a pair of complex roots r = x+iy and r = x iy such that i r i r <. By the observation above, (x + y + 4y < as desired.

14 Solution to Practice Exercise Set II ( (a For fixed with <, there is a disk B(,R containing (R<. It suffices to show f is continuous on B(,R. (b For x =Re >, Σ k= k + For B(,R, < R <, k k kr k = M k. Now Σ M k = Σ k= k converges by the root test since lim krk = R<. So by Weierstrass M-test, k k= kr k Σ k k converges uniformly on B(,R to a continuous function f(. Since is k= arbitrary, f( is continuous on <. k + = (k + x + y k = M k. Since converges uniformly to a continuous function on Re >. ( (a If n is odd, x n = n n + asn.ifnis even, x n =+ So lim x n =3. n { (b lim x n = lim n,, { (3 if k = n! (a a k = (4 (b R = if k n!, lim k n = n + n lim n Σ M k = Σ k= k= k },,,,,,,... = lim {,,,... } =. lim n k ak = lim {,,,,,,... } =,R = =. = n n n n +. n n 3asn. n + Σ n= n has radius of convergence R =, Σ n= n has radius of convergence R =. However Σ ( n = Σ has radius of convergence >R + R. n= n= converges, (5 The center of the power series is at. Since lim n n =,f( Then f (. So f (. (6 The center of the power series is at. Observe that f(= for =, n =,, n 3,... Since lim n n =,f(. Similarly, f( n = n 3 forces f( However 3, so no such f(. (7 Set = n, then f( = = + + for =, n =,, 3,... Now lim =. So f( = n n n Σ c n n must be the power series of about the center. n= + Since Σ w n = n= w for w <, + = ( = Σ ( n = Σ ( n n, then f (k ( = { n= n= if k is odd k!c k = ( k/ k! if k is even.

15 Solution to Practice Exercise Set III ( v x = u y = y v = xy + (y v y = u not possible. x =x v =xy + (x ( Observe that u v u is a function of x and is a fucntion of y. Then x y x = v y u(x =ax + b, v(y =ay + c, f( =a +(b + ic. must be a constant. Then (3 If Im < π, then = x + iy ( π <y<π and e = e x (cos y + i sin y. So Re e = e cos y>. onversely, if Re w>, then w = r cis θ with π <θ<π and w = e, where =lnr + iθ is the infinite strip. Therefore, the range of e for Im < π is the right half plane Re w>. w = e w πi/ A horiontal line in Im < π has equation Im = c with π <c<π. Its image is the set e = e x+ic = e x (cos c + i sin c where <x<, i.e. the semicircle with the origin as center and radius e πi/ c. ( It s not possible to define log( continuously on \[, ] because arg( on the circle, say = 3, cannot be made continuous. (If arg(3 is defined, then going around = 3 once and returning to = 3 would force redefinition of arg(3. ( + Yes, it is possible to define log continuously on \ [, ] as ( ( + follows: define log =ln i Arg. The only possible place where this can be discontinuous is when + is negative or (where Arg w is discontinuous. However +. This segment is removed! (5 f = u + iv is analytic on G u x (x,y = v y (x,y and u y (x,y = v x (x,y for all (x,y in G and f x,f y continuous on G. For (a,b ing,f (a,b f(a, b =u(a, b iv(a, b =U(a,b +iv (a,b. [That is the real part of f is U(a,b =u(a, b and the imaginary part of f is V (a,b = v(a, b.] learly, fx,f y are continuous on G because f x,f y are continuous on G. Finally, we check auchy-riemann equations for f : U x (a,b = u x (a, b = v y (a, b = V y (a U ( u,b, y (a,b = y (a, b = v x (a, b = V x (a,b. Therefore, f is analytic on G. (6 (a v x = u y =6xy v y = u x =3x 3y v(x, y =3x y + (y v(x, y =3x y y 3 + (x } v(x, y =3x y y 3 +.

16 (b (c (d v x = u y = e y cos x v y = u x = e y sin x v(x, y } =e y sin x + (y v(x, y =e y sin x + (x v x = u y = y x + y v y = u x = x x + y v x = u y = y ( x (( x + y v y = u x = ( x (( x + y v(x, y = x ( x + y +. y v(x, y = arctan( x + (y v(x, y = arctan( y x + (x x v(x, y = ( x + y + (y x v(x, y = ( x + y + (x (7 x = r cos θ, y = r sin θ, u x = v y, u y = v x. u r = u x x r + u y v r θ = ( v x r x θ + v y = ( u y θ r y v r = v x x r + v y y r = v v cos θ + sin θ. x y u ( u r θ = x r x θ + u y = y θ r v(x, y =e y sin x +. v(x, y = arctan( y x +. y r = u u cos θ + sin θ. x y ( r sin θ+ u x (r cos θ = u u u cos θ + sin θ = x y r. ( v v ( r sin θ y x (r cos θ = v v v cos θ + sin θ = x y r.

17 Solution to Practice Exercise Set IV ( ( (a i + π Log (b T ( = ( (i π Log + ( i + + (c T ( =( ( / ( i i i + ( / ( T (= i( + i i( ( + + +i + i (d [ ( ( / ( T ( = i i i + ( + ] + i ( (( ( + T ( = += = ( +. + ( Observe that T ( =,T (ia =,T (ib =. Orient L 3 by (, ia, ib. The right side of L 3 is D E F. The right side of T (L 3 (with respect to (,, is U V W. The part of L 3 from to ia is mapped onto the segment from to. So we have the correspondence F V, E U, D W, Y, B X, A Z.

18 Solution to Practice Exercise Set V ( Observe that the range of f( is in the disk w <, which is contained in the slit plane \ (, (omplex plane minus the negative real axis and. Now Log w is analytic on the slit plane. So Log f( is defined and analytic. Then f ( f( d = (Log f( d = Log f((b Log f((a = because is closed (i.e. (a =(b. ( The unit circle (counterclockwise direction is given by (t = e it = cos t + i sin t ( t π, d = ie it dt = dt, so π π π d = (cos t + sin tdt= cos tdt= sin t dt =8. = (3 Observe that (f( +g( d = f( d + g( d and αf( d = α f( d. So it suffices to consider the special case P ( = n (the general case is obtained by using the linearity properties above. is given by (t =a + Re it, t π, d = ire it dt. (* where we use the fact P ( d = π (4 Since has the antiderivative (5 Suppose = = f( d = f( d = R = π π = π π n d = π (a + Re it n ire it dt (a n + na n Re it + + R n e int ire it dt = na n ir dt ( = πina n R = πir P (a. π (cos nt + i sin nt dt = ifn e int dt = π. dt =π if n = in \{}, f(e it ie it dt = Re iθ. Then ( f(e it ie i(t θ π dt =Re = π π π d = = =. = f(e it ( sin(t θ+icos(t θ dt f(e it sin(t θ dt f(e it sin(t θ dt sin(t θ dt = π sin t dt =4.

19 (6 i i 4 3 : (t =t, t : (t =e it, t from π down to 3 : (t =t, t 4 : (t =e it, t π - - d = dt =, 3 d = dt =, 4 d = = 4 3. d = d = π π e it e it ieit dt = i e it e it ieit dt =i e 3it dt = e3it 3 = π 3. e 3it dt = π 3 e3it = 4 3. π π

20 Solution to Practice Exercise Set VI ( a a a a n(, a= n(, a = n(, a = n(, a= ( Log(w +=w w + w3 3 w4 +...for w near. cos w = w 4! + w4 4! w6 +...for all w. 6! For near, Log(49 + cos 5 = ( Therefore, lim exp( Log(49 + cos 5 =e. = (3 f T Let f( be an entire function, whose range lies in the right half plane. Let T ( =, then T f( is + entire and has range in the unit disk, i.e. T f( for all. By Liouville s theorem, T f( constant. Therefore f T T f constant. (4 Let f( =a + a a n n be a polynomial such that f(. By corollary (, a k = k!f (k ( f( = d πi k+ π }{{} π =. = }{{} L M { sin (5 Let f( sin and g( = if, then by auchy s integral formula, if = sin f( sin d = d =πif( = and d = g( d =πig( = πi. = = (6 (a J f (x, y = u v x y u ( ( v u v y x = + = f ( because f ( = u x x x + i v x auchy-riemann equations. π π (b m n da =reiθ = r m e imθ r n e inθ r dr dθ = e i(m nθ dθ r m+n+ dr D = ei(m nθ π i(m n θ= m + n + =. (c area of G = J f (x, y dx dy (a = f ( da = Σ nc n n D D D n= da ( ( Σ Σ = mc m m nc n n da D m= n= ( Σ = m=n= n c n n + Σ mnc m c n m n da m n D (b = Σ n= n c n π = = r n r dr dθ = Σ n= n c n π n = π Σ n= n c n. and by the

21 Solution to Practice Exercise Set VII ( Suppose f( n = n + for all positive integer n. Since n has limit point, which is in { : }, so the identity theorem implies f( = =. However, this is not analytic at =, a + + contradiction. ( Define g( = f( for. Then g is continuous on (which is closed and bounded and analytic on < <. So the maximum modulus theorem implies that for, f( f(w = g( max g(w = max w =, w =, w =. (3 (a Let Γ be the boundary of a rectangle. For a fixed t, sin t is entire, so sin t =. Then Γ sin t sin t ( f( d = dt d = d dt = sin td dt =. Γ Γ t Γ t t Γ sin w sin w (Details: Since lim =, is bounded for w. Suppose sin w w w w w M for w. Then sin t t M for <t and sin t Γ t dt d <. These imply f( is continuous and interchange of integration is possible. (b sin t = ( n (t n+ Σ. For a fixed, <t,t lies in the closed disk B(,. So n= (n +! ( n n+ t n Σ converges uniformly to sin t as a function of t (power series converges uniformly in closed subdisks of domain of convergence. n= (n +! t Then f( = sin t t dt = Σ n= ( n n+ t n dt = Σ (n +! ( n n+ n= (n +! t n dt = ( n n+ Σ n= (n +!(n +. (4 Since { : } is closed and bounded, let M = max f(. Also let T ( be a Möbius transformation mapping UHP = { :Im } onto { : } and R onto { : = } (e.g. T ( = i. Then + i f T ( is continuous on UHP, analytic on UHF = { :Im> } and real-valued on R. By the Schwar reflection principle, f T ( can be extended to an entire function. So f T ( M for all, Liouville s theorem implies f T ( is constant. Therefore, f( is a constant. (5 Let w,w be two distinct fixed points of f in D. Let T ( = + w, then T is one-to-one map from +w D onto D. So there is such that T ( =w. The function T f T ( is an analytic function from D onto D and T f T ( = and T f T ( =. The equality case of Schwar lemma implies T f T ( e iθ. Using T f T ( =,wegete iθ =. So T f T ( =. Therefore, f T ( =T ( for all D. Therefore f(w w for all w D. (6 Since f( is entire, f( = Σ a n n for all, where a n = f (n (. Since f( is real on the real n= n! axis, Schwar reflection principle and the identity theorem imply f( =f( for all. So Σ a n n = n= Σ a n n = Σ a n n. By the uniqueness of power series, a n = a n for all n. n= n=

22 Define g( = if(i. Since f( is imaginary on the imaginary axis, g( is real on the real axis. Now g( = Σ i n+ a n n. By the above argument, we have i n+ a n = i n+ a n. If n is even, then n= i n+ a n = i n+ a n. This implies a n = a n = a n for even n. So a n = for all even n. Therefore f( = Σ a k+ k+ is odd. k= Alternatively, f( = f( for all and g(w = g(w for all w f( =f( for all and if(iw =if(iw = if(iw for all w f( =f( and f( = f( for all (set = iw f( =f( = f( for all f(u = f( u for all u (set u = (7 f( has finitely many roots α,...,α n (repeated according to multiplicities in the open unit disk and no roots on the unit circle since f( = for = (otherwise the roots have a limit point in the closed disk forcing f(. By the theorem following the orientation principle, α j α j = for f( =. The function g( = n α is analytic on D. (This is clear for α,...,α n because j Π j= α j n Π j= α j α j. If α k is a root of multiplicity m, then lim α k f( ( α k = f (m (α k (by l Hôpital s m m! rule and m of the α k s equal α k so that lim α k g( exists and is nonero, which is used to define g(α k. It follows that g is analytic at α k. By the maximum modulus theorem, for, g( max g(w = max w = w = n Π j= f(w w α j α j = max f(w =. w = Since g has no root in D = { : < }, the minimum modulus theorem implies, for, g( min g(w =. Therefore g( = for. Then g( = w = eiθ because g( =. Hence f( =e iθ Π n α j j= α j by the identity theorem. Since f is entire, α j = (otherwise f is not defined at. Therefore f( e iθ n. α j

23 Solution to Practice Exercise Set VIII ( Introducing n+ cross-cuts as shown below and applying auchy s theorem to the upper and lower simple closed curves, we get (after cancelling the integrals over the cross-cuts... a a... a n = f( d Σ n f( d. Γ j= j The result follows. ( (a Since 4 + =, there are a pole of order at, a pole of order at i and i, ( + i( i respectively. (b Since cot = cos cos and lim ( nπcos =, lim ( nπ =, there is a pole of order sin nπ sin nπ sin at nπ, where n is any integer. (c The isolated singularities are at and. There is a pole of order at. Since e/ doesn t have a limit as, so is an essential singularity. (d The isolated singularities are at all integers. Since lim sin π = π, lim sin π = π, lim ( nπ,n ± nπ sin π = π ±n, there are removable singularities at and, respectively, and pole of order at all other π integers. (3 (a 4 = 4( 4 <. (b ( + = 4 4 = ( 4 = ( = Σ 4( ( +( = ( k Σ, where k= 4 k+, where 4 k <. k= 4 k (4 Suppose (n +π <r< nπ. Since sin = 3! 3 + 5! 5... for (n +π < < nπ, sin d =πa =πi. (This is the same if π <r<. =r (5 Let f( = Σ a k k for \ { } = { :< < }.Fork =, k= Therefore, f a. a k π =r r + r f( d k+ r k = r + { as r if k> r k+ as r ifk< (6 If f has a pole of order k at, then f( = a k k + a k+ k +...= a k k ( + a k a = a kg( k }{{} k for =g( < <ε, where g is analytic on <ε,g( = and a k. Let = k ta k with t a real variable, then lim e f( = e + =+ and lim e f( = e =. So is an essential singularity of e f(. t + t

24 (7 (i f( entire f( = Σ a k k. f( k= has a pole of order N f( = Σ a k k=n.sof( = N k Σ a k k k= is a polynomial. (ii Since { } is closed and bounded, and the poles are isolated, there are only finitely many poles. Let the finite poles be p,...,p k and the respectiveorders be N,...,N k. Let P ( = ( p N...( p k Nk f(, then P is entire. Since is either a removable singularity or a pole of f(, the same is true for P.Ifis a removable singularity, then P is analytic at and P ( is a complex number. This implies P is bounded on, so by Liouville s theorem, P ( P ( P ( and f( = is a rational function. If is a pole, then by part (i, P ( ( p N...( p k Nk P ( is a polynomial, then f( = is a rational function. ( p N Nk...( p k

25 Solution to Practice Exercise Set IX ( Let g( =f(, then is an isolated singularity of g. If it is removable, then f has a removable singularity at and f will be bounded, forcing it to be a constant. If is a pole of g, then g( = Σ a k k near and f( = Σ N a j j. Since f is analytic at, a j = for j <, then f is k= N j= a polynomial. By the fundamental theorem of algebra, the image of the plane under f is the whole plane (hence the image is dense. If is an essential singularity of g, then asorati-weierstrass theorem implies the image of the plane under g (or f is dense in the plane. ( Suppose f has more than one roots. Let R be so large that there are more than one roots inside Γ, the circle = R. Since f(r,f( R are the only real valued on f Γ, the curve { f(re iθ : <θ<π } and { f(re iθ : π<θ<π } lie entirely on the upper or lower half plane. Then n(f Γ, = or. Therefore, f can have at most one root by the Argument Principle, a contraction. (Note f, so we may assume f(r and f( R by the identity theorem. (3 No, otherwise the Argument Principle implies f has poles inside the unit circle. (4 Since f is one-to-one on Γ,f Γ has no self-intersections, so it is a simple closed curve. For w f Γ, let g( =f( w, then n(g Γ, = n(f Γ,w = or depends whether w is inside or outside f Γ. If is inside Γ, then f( is not on f Γ by the open mapping theorem. Now is a root of g( =f( f(, hence n(g Γ, =. Then g has no other root. This implies f is one-to-one inside Γ. (5 ir onsider the contour on the left with R large. On γ,(t =Re it, π t π,f( = n + α n + β n,so γ arg f( =nπ. On γ,(t =it, wher t decreases from R to R. γ γ R -ir f o γ f o γ If n is odd, f(it = t n + β + iα t n,f(ir + i +,f( ir + i. (Note f(i β n =iα β n n,f( = β,f( i β n = iα β n n. (n π A sketch shows γ arg f( = π. So f has = n roots inside the π contour for R large, so f has n roots with positive real parts. If n is even, f(it =t n + β iα }{{} t n,f(ir + i +,f( ir + i. > A sketch shows γ arg f( =. So f has nπ = n roots inside the contour for R π large, so f has n roots with positive real parts. (6 Let f( =e a n and g( =a n, then for =, f(+g( = e = e Re e <a= a n = g( < f( + g(. So by Rouché s theorem f( has n roots inside the unit circle (because g has n roots inside the unit circle.

26 Solution to Practice Exercise Set X ( By residue theorem, Reπi/n ( R γ d ( (* γ + + =πi Res =πi γ γ + n =e πi/n + n n(e πi/n. n R On γ, = Re iθ, θ π n, d γ + n ML = πr R n asr +. n On γ, = xe πi/n, θ π n, d R γ + n = e πi/n R dx dx = eπi/n +xn +x n. Letting R +, (* ( e πi/n dx +x n =πi πi = n(e πi/n n n eπi/n. dx Therefore, +x n = πi n(e πi/n e πi/n = π n sin π. n ( R e i =+i + (i +, ei i! = 4i 3 +. (At, ei i has a removable singularity, where we assign the value to make it analytic. -R R onsider the contour shown. By auchy s theorem, ( R e i i e i i R + R R d (* =. R d x= e ix +ix = x dx. On R, e i e Im + R R. So e i R d πr asr +. Then R e i i ( d d = lim i =π. R + lim R + R Taking R +, (* implies sin x get dx = π x. R e ix + e ix x dx +π =. Since e ix + e ix = 4 sin x,we (3 Recall log =ln + i arg, arg π and so log i = i π i R. ( r R log (* log r By residue theorem, + + d =πires -R -r r R R r r R + =i + = iπ. log r + d ln r + π r }{{ }{{} πr asr +, log } + d ln R + π R πr asr +. L M r log R R + d = ln x + iπ r x + dx. Letting r ln x +, R +, (* x + dx+iπ dx x = + }{{} π/ i π. Therefore, ln x dx =. x + (4 -R+i γ R+i γ 3 γ -R R onsider f( =e. By auchy s theorem, ( R + e d (* =. R γ + γ γ 3

27 (5 (6 γ,γ 3 e d e R + }{{} asr +. M e γ d = R R R e (x+i dx = e x ix+ dx. R Letting R +, (* implies e x dx e e x (cos x i sin x dx =. Since π π, taking the real part, we get e x cos xdx= e. γ γ π/4 R So e i d γ ( R onsider f( =e i. By the auchy s theorem, e x dx = + e i d (* =. On γ γ γ, = Re iθ, θ π 4, e i = e R sin θ e 4R θ π by Jordan s inequality. π 4 e 4R θ π Rdθ = π 4R ( asr +. e R R On γ, = xe i π 4, x R, e i d = e x (e i π 4 dx. Letting R +, (* e ix dx γ e i π π 4 e x dx =. Since e x dx =, taking real and imaginary parts, we get cos x dx = π cos π π π 4 = 4, sin x dx = 4. π/ = i 4 dθ + sin θ = π 4 ( = d dθ + sin θ = 4 d + = f( d = n Σ j= Res(f,a j (clockwise orientation, where the last equality fol- (7 (a Res(f, = πi =R lows from residue theorem. (b Suppose f( = + a k = i d 4 ( = ( = π Res = = } {{ } = i ( ( + d Res = + + }{{} = = π k + + a f( = ( + a k k + + a + a + a So Res( (c Let f( = = a k k a f(, = a = Res(f,. sin(, then = orientation. Now f( = + a + a + on r< <, then + on< < r a a. 3 f( d = πi Res(f, = πi Res( f(, (counterclockwise ( = 3 ( 6 + = ( = 3 6.So sin( d = πi( = 6 =πi (counterclockwise orientation. 3 Alternatively, = sin( d = = d = ( d = πi πi = 6 3, = where all orientations are counterclockwise and the last integrand converges uniformly on =.