Practice Exercise Set I * { Problems may be di�cult.(1) Describe the sets whose points satisfy the following relations(a) ����z � 1z + 1���� = 2 (b) jz + 1j � jz � 1j < 2 (c) jz2 � 1j = 1 (d) arg z � 1z + i = �3(2) Given three distinct complex numbers �, �, . Show that they are collinear i� Im(�� + � + �) = 0.(3) Prove that if jz1j = jz2j = jz3j, z1, z2, z3 distinct, then arg z3 � z2z3 � z1 = 12 arg z2z1 .(4) For what complex value z will the following series converge(a) 1�n=0� z1 + z�n *(b) 1�n=0 zn1 + z2n(5) When can equality occur in the triangle inequality? That is under what conditions on z, w will jz+wj =jzj+ jwj?(6) What is the boundary of the set fz: Re z and Im z are rationalg?(7) Establish the identity ���� n�k=1�k�k����2 = n�k=1 j�kj2 n�k=1 j�kj2 � �1�k<j�n ���k�j � �j�k��2 for the case n = 2.(This implies the Cauchy-Schwarz inequality ���� n�k=1�k�k����2 �r n�k=1 j�kj2r n�k=1 j�kj2.)(8) Suppose 0 < a0 � a1 � : : : � an. Prove that the polynomial P (z) = a0zn + a1zn�1 + : : :+ an has noroot in the unit disk jzj < 1. (Hint: Consider (1� z)P (z).)*(9) Prove that if 11z10 + 10iz9 + 10iz � 11 = 0, then jzj = 1.[ Hint: Consider z9. ]**(10) Let P (z) = zn + c1zn�1 + : : :+ cn with c1, : : :, cn real. Suppose jP (i)j < 1. Prove that there is a rootx+ iy of P (z) in the set p(x2 + y2 + 1)2 � 4y2 < 1.

Practice Exercise Set II (To receive a solution, you have to hand in some work.)(1) (exercise #15, p.17) Show that(a) f(z) = 1�k=0kzk is continuous in jzj < 1.(b) g(z) = 1�k=1 1k2 + z is continuous in the right half plane Re z > 0.(2) Find limn!1xn, where(a) xn = 1 + (�1)n 2nn+ 1.(b) xn = cos n�4 .(3) Find the radius of convergence of the following power series:(a) 1�n=0 zn!.(b) 1�n=0(n+ 2n)zn.(4) Give an example of two power series 1�n=0anzn and 1�n=0 bnzn with radii of convergence R1 and R2,respectively, such that the power series 1�n=0(an + bn)zn has a radius of convergence > R1 +R2.(5) Explain why there is no power series f(z) = 1�n=0 cnzn such that f(z) = 1 for z = 12 ; 13 ; 14 ; : : : andf 0(0) > 0.(6) Does there exist a power series f(z) = 1�n=0 cnzn such that f( 1n ) = 1n2 and f(� 1n ) = 1n3 for n = 1, 2, 3,: : :.*(7) If f(z) = 1�n=0 cnzn satis�es f( 1n ) = n2n2 + 1, n = 1, 2, 3, : : :, compute the values of the derivativesf (k)(0), k = 1, 2, 3, : : :.

Practice Exercise Set III (To receive a solution, you have to hand in some work.)(1) Show that there are no analytic function f = u+ iv with u(x; y) = x2 + y2.(2) Suppose f is an entire function of the form f(x; y) = u(x) + iv(y). Show that f is a polynomial ofdegree at most one.*(3) What is the range of ez if we take z to lie in the in�nite strip j Imzj < �2 ? What are the images forhorizontal lines and vertical segments in j Imzj < �2 under the ez mapping?**(4) Discuss if it is possible to de�ne log(z � 1) continuously on C n [�1; 1]. Also discuss the possibility forlog�z + 1z � 1� continuously de�ned on C n [�1; 1].*(5) Let G be a region and G� = fz: z 2 Gg is the mirror image of G across the x-axis. If f :G ! C isanalytic, show that f�:G� ! C de�ned by f�(z) = f(z) is analytic.(6) If f = u+ iv is analytic on some domain, given u(x; y) below, �nd the possibilities of v(x; y).(a) u(x; y) = x3 � 3xy2.(b) u(x; y) = e�y cosx.(c) u(x; y) = log(x2 + y2).(d) u(x; y) = y(1 � x)2 + y2 .(7) Write z in polar coordinates. Then f(z) = u(z) + iv(z) = u(r; �) + iv(r; �). Establish the polar form ofthe Cauchy-Riemann equations: @u@r = 1r @v@� ; @v@r = �1r @u@� :

Practice Exercise Set IV (To receive a solution, you have to hand in some work.)(1) Find a conformal mapping from the open unit disk D = fz: jzj < 1g onto the following regions:(a) the in�nite strip 0 < Imz < 1 i*(b) the upper semidisk jzj < 1; Imz > 01-1 0

i( Hint: Find a map from 1st quadrant onto the upper half semidisk. )(c) the slit disk D n [0; 1)0 1( Hint: Use (b). )*(d) C [ �1 n [�1; 1]

-1 1( Hint: Find a map from C n (�1; 0] to (C [ f1g n [�1; 1].)(2) Let a < b and T (z) = z � iaz � ib . De�ne L1 = fz: Imz = bg; L2 = fz: Imz = ag; L3 = fz: Re z = 0g.Determine which of the regions A, B, C, D, E, F in Figure 1, are mapped by T onto the region U , V ,W , X, Y , Z in Figure 2.0 1

Figure 1 Figure 2

DA

B

C

E

F

L

L

L 3

2

1

U W

X Z

V

Y

T

ia

ib( Hint: Orient L3 by (1; ia; ib). )

Practice Exercise Set V (To receive a solution, you have to hand in some work.)(1) Suppose f(z) is analytic and jf(z) � 1j < 1 in a region . Show that ZC f 0(z)f(z) dz = 0 for every closedcurve C in , assuming f 0 is continuous.(2) Compute Zjzj=1 jz � 1jjdzj, where the unit circle jzj = 1 is given the counterclockwise orientation.*(3) De�ne ZC f(z) dz = ZC f(z) dz. If P (z) is a polynomial and C denotes the circle jz � aj = R (counter-clockwise), show that ZC P (z) dz = �2�iR2P 0(a).(4) Find ZC 1z2 dz, where C is a smooth curve from 1 to �1 not passing through the origin.*(5) Show that if f is a continuous real-valued function and jf(z)j � 1, then ������� Zjzj=1 f(z) dz������� � 4.( Hint: Consult the top half of p.45, and show that ����Z f���� � Z 2�0 j sin tj dt. )(6) Evaluate the integral ZC zz dz, where C is the curve show below.( Hint: ZC = ZC1 + ZC2 + ZC3 + ZC4 )i

C4

C1

2C

C3

21-1-2 0

2i

Practice Exercise Set VI (To receive a solution, you have to hand in some work.)*(1) The following curve C divides the plane into 4 regions. For each region, state the winding number of Caround points in that region. (Give answers by inspection, no computation needed.)1 2 3

4

C(2) Find limz!0 ez Log(z49 + 1)(cos z25)� 1 .*(3) If the range of an entire function lies in the right half plane Rew > 0, show that the function is aconstant function.( Hint: Compose a M�obius map. )(4) Suppose a polynomial is bounded by 1 in the unit disk. Show that all its coe�cients are bounded by 1.(c.f. proof of Liouville's Theorem.)(5) Find Zjzj=1 sin zz dz; Zjzj=1 sin zz2 dz (counterclockwise orientation).**(6) (Optional) Let f(z) = u(z) + iv(z) (or f(x; y) = (u(x; y); v(x; y)))be a one-to-one analytic function from the open unit disk D =fz: jzj < 1g onto a domain G with �nite area.(a) Show that Jf (x; y) def= ������� @u@x @u@y@v@x @v@y ������� = jf 0(z)j2. y

x

D G

u

vf(b) For distinct nonnegative integers m, n, show ZD zmzn dA = 0 (orthogonality relation), where dA =r dr d� = dx dy is the area di�erential.(c) Show that if f(z) = 1�n=0 cnzn is the power series for f in D, then area of G = � 1�n=1njcnj2.

Practice Exercise Set VII (Solutions will be distributed in class.)(1) Show that if f(z) is analytic in jzj � 1, there must be some positive integer n such that f( 1n ) 6= 1n+ 1.(2) Suppose that f is analytic in the annulus: 1 � jzj � 2, that jf j � 1 for jzj = 1 and that jf j � 4 forjzj = 2. Prove that jf(z)j � jzj2 throughout the annulus.(3) Show that f(z) = Z 10 sin 2tt dt is entire function*(a) by applying Morera's Theorem(b) by obtaining a power series expansion for f .(4) Show that if f(z) is continuous on the closed unit disk fz: jzj � 1g, analytic on the open disk fz: jzj < 1gand real-valued on the unit circle fz: jzj = 1g, then f(z) is a constant function.( Hint: The disk is conformally equivalent to the upper half plane. Re ection. )*(5) Let D = fz: jzj < 1g. If f :D ! D is analytic with at least two �xed points, prove that f(z) � z.( Hint: May assume one of the �xed point is 0 by composing with suitable M�obius transformations. )*(6) Let f(z) be an entire function which is real on the real axis and imaginary on the imaginary axis, showthat f(z) is an odd function, i.e. f(z) � �f(�z).( Hint: Consider the coe�cients of the power series of f(z) or make use the re ection property. )**(7) If f is an entire function mapping the unit circle into the unit circle (i.e. jf(z)j = 1 for jzj = 1), thenf(z) = ei�zn.( Hint: f(z) has �nitely many roots �1; : : : ; �n (repeated according to multiplicities) in the unit disk.Recall ���� z � �j1� �jz ���� = 1 for jzj = 1. Use modulus theorems to show f(z) = ei� n�j=1 z � �j1� �jz . )

Practice Exercise Set VIII (To receive a solution, you have to hand in some work.)(1) Suppose f is analytic on C n fa1; : : : ; ang and � is a simple closed curve \surrounding" a1, : : :, an asshown. For each aj, let Cj be a simple closed curve about aj inside �.Γ

a1

a2 anC C

C

...

12

n

Show that Z f(z) dz = n�j=1 ZCj f(z) dz,where the orientation of �, C1, : : :, Cn are as shown.(2) Identify the isolated singularities of the following functions and classify each as removable singularity,pole (and its order) or essential singularity:(a) 1z4 + z2 (b) cot z (c) e1=z2z � 1 (d) z2 � 1sin�z .(3) Find the Laurent series of 1z2 � 4 on (a) 0 < jz � 2j < 4, (b) 2 < jzj <1.(4) Find Zjzj=r sin 1z dz (counterclockwise orientation) for r 6= 0, 1� , 12� , 13� , : : :.(5) Suppose f is analytic on C n �0 and satis�es jf(z)j �pjzj+ 1pjzj . Prove f is constant.(6) If f has a pole at 0, show that ef(z) cannot have a pole at 0.(7) If f is analytic on R < jzj <1, we say1 is a removable singularity, pole of order k, essential singularityof f(z) i� 0 is a removable singularity, pole of order k, essential singularity of f(1z ).(i) Prove that an entire function with a pole at 1 is a polynomial.(ii) Prove that an analytic function on C [ f1g except for isolated poles must be a rational function.

Practice Exercise Set IX (To receive a solution, you have to hand in some work.)(1) 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(1z ). ](2) Suppose that f is entire and that f(z) is real if and only if z is real. Use the Argument Principle toshow that f can have at most one root.[ Hint: Let � be the circle jzj = R, R large, what is n(f � �; 0)? ](3) Is there an analytic function f on fz: jzj � 1g which sends the unit circle with counterclockwise orien-tation into the unit circle with clockwise orientation?*(4) If f is analytic on and inside a simple closed curve �, and f is one-to-one on �, then f is one-to-oneinside �.[ Hint: If f � � a simple closed curve? For w 62 f � �, let g(z) = f(z) �w, what is n(g � �; 0)? ](5) Show that if � and � 6= 0 are real, the equation z2n + �2z2n�1 + �2 = 0 has n � 1 roots with positivereal parts if n is odd, and n roots with positive real parts if n is even.(6) If a > e, show that the equation ez = azn has n solutions inside the unit circle.

Practice Exercise Set X (To receive a solution, you have to hand in some work.)(1) Find Z 10 dx1 + xn , where n � 2 is a positive integer. [ Hint: Use the contour Re

0 R

2π i/n . ](2) Find Z 10 sin2 xx2 dx. [ Hint: Integrate e2iz � 1� 2izz2 around a large semi-circle. ](3) Find Z 10 lnxx2 + 1 dx. [ Hint: Use the contour-r r R-R

. ](4) Find Z 10 e�x2 cos 2x dx. [ Hint: Use the contourR

R+i

-R

-R+i and f(z) = e�z2 . ](You may need to know Z 10 e�x2 dx = p�2 .)(5) Find Z 10 cos x2 dx and Z 10 sinx2 dx. [ Hint: Use the contour0 R

π/4. ](6) Find Z �=20 d�1 + sin2 � .(7) Suppose f is analytic on r < jzj <1, then we de�ne Res (f;1) = 12�i Zjzj=R f(z) dz (clockwise orienta-tion).

oo

0

(The clockwise orientation relative to 0 is the counterclockwise orientation relative to1.) Equivalently, if f(z) = 1�k=�1 akzk on r < jzj <1, then Res (f;1) = �a�1.(a) If f is meromorphic onC with isolated poles at a1; : : : ; an, show that n�j=1Res (f; aj)+Res (f;1) = 0(i.e. the sum of all residues on C [ f1g is 0.)(b) Show that Res (f;1) = Res (� 1z2 f(1z ); 0).(c) Find Zjzj=1 1sin(1z ) dz.

Solution to Practice Exercise Set I(1) (a) The locus of all points z whose distances from the two points a = 1 and b = �1 having a �xedquotient � = 2 is a circle with center on the line through a, b. On the real axis, z = �3, �13 satis�esthe equation. So the circle is �z: ����z + 53 ���� = 43�.Alternatively, for z = x+ iy, (x� 1)2 + y2 = jz � 1j2 = 4jz + 1j2 = 4[(x+ 1)2 + y2].Simplifying we get (x + 53)2 + y2 = 169 .(b) The locus of all points z whose distances from the two points a = �1 and b = 1 having a �xeddi�erence � = 2 (� distance between a, b) is a branch of a hyperbola having a, b as foci. If � =distance between a, b (as is the case here), the branch degenerate to a ray (or an in�nite slit). Theset is the whole complex plane minus all real numbers greater than or equal to 1.Alternatively, for z = x+iy, p(x+ 1)2 + y2�p(x� 1)2 + y2 = jz+1j�jz�1j < 2 () (x+1)2+y2 < [2 +p(x� 1)2 + y2]2 = 4 + 4p(x� 1)2 + y2 + (x� 1)2 + y2 () x� 1 <p(x� 1)2 + y2.If x < 1, then x � 1 < 0 � p(x� 1)2 + y2. If x � 1, then 0 � (x � 1)2 < (x � 1)2 + y2 impliesy 6= 0. So the set is �x+ iy:x < 1 or (x � 1 and y 6= 0).(c) jz2�1j = jz�1jjz+1j. The locus of all points z whose distances from two points a = 1 and b = �1having a �xed product � = 1 is a lemniscate with foci at a and b. (The case � >r ja� bj2 resultsin two curves, each about a focus (and as � ! 0, the two curves shrink toward the foci); the case� = r ja� bj2 results in a �gure-eight curve with double point at a+ b2 .) The set is a lemniscatewith foci at 1 and �1 having a double point at 0.2

1/21/2-2

- 45

o45

o

Alternatively, for z = r cis �, (r2 cos 2��1)2+(r2 sin 2�)2 = jz2�1j2 = 1.Simplifying we get r = 0 or r2 = 2 cos 2�. The range of � possible are��4 � � � �4 or 3�4 � � � 5�4 .α−β

α−β

α β

z

z

z - z31

12z

3z2

- z3

O

(d) To interpret arg z3 � z2z3 � z1 �= arg z2 � z3z1 � z3�, write z2 � z3 = R cis�,z1 � z3 = r cis �. Then arg z2 � z3z1 � z3 = arg �Rr cis(�� �)� = � � � isthe angle 6 z1z3z2 (measured from the ray ��!z3z1 counterclockwise tothe ray ��!z3z2).-i

1

The set is the open arc of the circle containing all points z suchthat 6 � iz1 = 60�.(2) The line through � and is Im� z � � � � = 0. So �, �, collinear () Im��� � � � = 0 ()Im�(�� )(� � )j� � j2 � = Im(�� � � � � + j j2)j� � j2 = 0 () Im(�� � � � � ) = 0.For any complex z = x+ iy, Im(�z) = �y = Imz. So Im(�� � � � � ) = Im(��) + Im(� � � � ) =Im(�� + � + �) = 0 is the condition.

(3)O

z

zz

1

2

3

(c.f. exercise 1(d)) This is just the complex way of expressing the geometry theoremthat 6 z1z3z2 = 12 6 z10z2. (You should check also the case z1, z2. z3 are orientedclockwise.)(4) (a) Let w = z1 + z , then we know 1�n=0wn converges i� jwj = ���� z � 0z � (�1) ���� < 1 (() z is closer to 0 then�1 () Re z > �12).(b) Case 1: (jzj = r < 1). ���� zn1 + z2n ���� � jzjn1� jzj2n = rn1� r2n (because 1 � jz2nj � j1 + z2nj). Applyratio test to 1�n=1 rn1� r2n , we have limn!1 rn+11� r2n+1 � rn1� r2n = r < 1. So � rn1� r2n converges) � ���� zn1 + z2n ���� converges ) � zn1 + z2n converges.Case 2: (jzj = 1). ���� zn1 + z2n ���� � jzjn1 + jzj2n = 12 (because j1 + z2nj � 1 + jz2nj). So zn1 + z2n cannotconverge to 0 as n!1. Hence � zn1 + z2n diverges.Case 3: (jzj > 1). For w = 1z , jwj < 1, so by case 1, � zn1 + z2n = � ( 1w )n1 + ( 1w )2n = � wn1 +w2nconverges.(5) Equality holds i� either z = 0. w = 0 or zw is real.(6) For any complex w, every neighborhood B(w; r) of w contains a point z0 with Re z0 and Im z0 rationaland also a point z1 not both Re z1 and Im z1 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. = j�1�1+�2�2j2 = (�1�1+�2�2)(�1�1+�2�2) = j�1j2j�1j2+�1�2�2�1+�2�1�1�2+ j�2j2j�2j2.R.H.S. = (j�1j2+ j�2j2)(j�1j2+ j�2j2)�j�1�2��2�1j2 = j�1j2j�1j2+ j�1j2j�2j2+ j�2j2j�1j2+ j�2j2j�2j2�(�1�2��2�1)(�1�2��2�1) = j�1j2j�1j2+ j�1j2j�2j2+ j�2j2j�1j2+ j�2j2j�2j2� (j�1j2j�2j2��1�2�2�1��2�1�1�2 + j�2j2j�2j2).Alternative solution for n 2N, n � 2 by Chow Chak-On.j�k�j � �j�kj2 = (�k�j � �j�k)(�k�j � �j�k) = j�kj2j�jj2 � �k�k�j�j � �j�j�k�k + j�jj2j�kj2.n�k=1 n�j=1j�k�j � �j�kj2 = n�k=1 n�j=1(j�kj2j�j j2 � �k�k�j�j � �j�j�k�k + j�jj2j�kj2).Since n�k=1 n�j=1j�k�j � �j�kj2 = �1�k=j�n j�k�j � �j�kj2 + �1�k<j�n j�k�j � �j�kj2 + �1�j<k�n j�k�j � �j�kj2= 2 �i�k<j�n j�k�j � �j�kj2and n�k=1 n�j=1�k�k�j�j = � n�k=1�k�k�� n�j=1�j�j� = ���� n�k=1�k�k����2.Therefore, 2 �1�k<j�n j�k�j ��j�kj2 = 2 n�k=1 j�kj2 n�k=1 j�kj2�2 ���� n�k=1�k�k����2. Cancelling the factor of 2 onboth sides and rearranging terms we get the desired result.

(8) Suppose z is a root of P (z) and jzj < 1, then0 = j(1� z)P (z)j = j � a0zn+1 + (a0 � a1)zn + (a1 � a2)zn�1 + � � �+ (an�1 � an)z + anj= jan � [a0zn+1 + (a1 � a0)zn + � � �+ (an � an�1)z]j(*)� an � ja0zn+1 + (a1 � a0)zn + � � �+ (an � an�1)zj(**)� an � [a0jzjn+1 + (a1 � a0)jzjn + � � �+ (an � an�1)jzj](***)> an � [a0 + (a1 � a0) + � � �+ (an � an�1)] = 0;a contradiction (where (*) j�� �j � j�j � j�j, (**) j�+ �j � j�j+ j�j, and (***) jzj < 1).(9) Let z = x+ iy, then jzj9 = jz9j = ����11� 10iz11z + 10i ���� =s121 + 220y + 100y2 + 100x2121x2 + 121y2 + 220y + 100.If jzj < 1, then x2 + y2 < 1 and 121 + 220y + 100y2 + 100x2 > 121x2 + 121y2 + 220y + 100, forcingjzj9 > 1, a contradiction.If jzj > 1, then x2 + y2 > 1 and 121 + 220y + 100y2 + 100x2 < 121x2 + 121y2 + 220y + 100, forcingjzj9 < 1, a contradiction.(10) (We �rst observe thatp(x2 + y2 + 1)� 4y2 =px2 + y2 + 1 + 2ypx2 + y2 + 1� 2y = ji� (x+ iy)jji�(x � iy)j.) Suppose the roots of P (z) are r1, r2, : : :, rn. Because the coe�cients are real, complexroots occur in conjugate pairs if any. Since 1 > jP (i)j = ji � r1jji � r2j : : : ji � rnj. For a real root r,ji� rj = p1 + r2 � 1. So P (z) must have complex roots.Now jP (i)j = � �real roots ji � rj�� �complex roots in pairs ji� rjji� rj�. So there must be a pair of complex rootsr = x+iy and r = x�iy such that ji�rjji�rj < 1. By the observation above,p(x2 + y2 + 1)2 � 4y2 < 1as desired.

Solution to Practice Exercise Set II(1) (a) For �xed z0 with jz0j <, there is a disk B(0; R) containing z0 (R < 1). It su�ces to show f iscontinuous on B(0; R).0

z

For z 2 B(0; R), jzj < R < 1, ��kzk�� � kRk = Mk. Now 1�k=0Mk = 1�k=0kRkconverges by the root test since limk!1 kpkRk = R < 1. So by Weierstrass M-test,1�k=0kzk converges uniformly on B(0; R) to a continuous function f(z). Since z0 isarbitrary, f(z) is continuous on jzj < 1.(b) For x = Re z > 0, ���� 1k2 + z ���� = 1p(k2 + x)2 + y2 � 1k2 = Mk. Since 1�k=1Mk = 1�k=1 1k2 converges,1�k=1 1k2 + z converges uniformly to a continuous function on Re z > 0.(2) (a) If n is odd, xn = 1� 2nn+ 1 !�1 as n!1. If n is even, xn = 1 + 2nn+ 1 ! 3 as n!1.So limn!1xn = 3.(b) limn!1xn = lim(p22 ; 0;�p22 ;�1;�p22 ; 0; p22 ; 1; 0; : : :) = lim�1; 1; 1; : : : = 1.(3) (a) ak = � 1 if k = n!0 if k 6= n! , limk!1 kpjakj = lim�1; 1; 0; 0;0; 1; : : : = 1, R = 11 = 1.(b) R = 1limn!1 npn + 2n = 1limn!1 np2n nr n2n + 1 = 12.(4) 1�n=0 zn has radius of convergence R1 = 1, 1�n=0�zn has radius of convergence R2 = 1.However 1�n=0(1� 1)zn = 1�n=0 0 has radius of convergence 1 > R1 +R2.(5) The center of the power series is at 0. Since limn!1 1n = 0, f(z) � 1 + 0z + 0z2+ : : :. Then f 0(z) � 0. Sof 0(0) 6> 0.(6) The center of the power series is at 0. Observe that f(z) = 0 + 0z + z2 + 0z3 + : : : for z = 1n , n = 1, 2,3, : : :. Since limn!1 1n = 0, f(z) � z2. Similarly, f(� 1n ) = 1n3 forces f(z) � 0+ 0z+ 0z2 � z3+ 0z4+ : : :.However z2 6� �z3, so no such f(z).(7) Set z = 1n , then f(z) = 1z21z2 + 1 = 11 + z2 for z = 1n , n = 1, 2, 3, : : :. Now limn!1 1n = 0. So f(z) =1�n=0 cnzn must be the power series of 11 + z2 about the center 0.Since 1�n=0wn = 11� w for jwj < 1, 11 + z2 = 11� (�z2) = 1�n=0(�z2)n = 1�n=0(�1)nz2n, then f (k)(0) =k!ck = � 0 if k is odd(�1)k=2k! if k is even .

Solution to Practice Exercise Set III(1) @v@x = �@u@y = �2y ) v = �2xy + C1(y)@v@y = @u@x = 2x) v = 2xy + C2(x) 9>=>;) not possible.(2) Observe that @u@x is a function of x and @v@y is a fucntion of y. Then @u@x = @v@y must be a constant. Thenu(x) = ax+ b, v(y) = ay + c, f(z) = az + (b+ ic).(3) If j Imzj < �2 , then z = x + iy (��2 < y < �2 ) and ez = ex(cos y + i sin y). So Re ez = ez cos y > 0.Conversely, if Rew > 0, then w = r cis � with ��2 < � < �2 and w = ez, where z = ln r + i� is thein�nite strip. Therefore, the range of ez for j Imzj < �2 is the right half plane Rew > 0.w = e zz w

πi/2

−πi/2

A horizontal line in j Imzj < �2 has equation Im z = c with��2 < c < �2 . Its image is the set ez = ex+ic = ex(cos c +i sin c) where �1 < x <1, i.e. the semicircle with the originas center and radius ec.(4)31-1-3

It's not possible to de�ne log(z�1) continuously onCn[�1; 1] because arg(z�1)on the circle, say jzj = 3, cannot be made continuous. (If arg(3� 1) is de�ned,then going around jzj = 3 once and returning to z = 3 would force rede�nitionof arg(3� 1).)Yes, it is possible to de�ne log�z + 1z � 1� continuously on C n [�1; 1] asfollows: de�ne log�z + 1z � 1� = ln ����z + 1z � 1 ����+ iArg�z + 1z � 1�. The only possibleplace where this can be discontinuous is when z + 1z � 1 is negative or 0 (whereArgw is discontinuous). However z + 1z � 1 � 0 () �1 � z � 1. This segmentis removed!(5) f = u+ iv is analytic on G () @u@x (x0; y0) = @v@y (x0; y0) and @u@y (x0; y0) = �@v@x (x0; y0) for all (x0; y0)in G and fx; fy continuous on G.For (a0; b0) in G�; f�(a0; b0)f(a0;�b0) = u(a0;�b0) � iv(a0;�b0) = U (a0; b0) + iV (a0; b0). [That isthe real part of f� is U (a0; b0) = u(a0;�b0) and the imaginary part of f� is V (a0; b0) = �v(a0;�b0).]Clearly, f�x ; f�y are continuous on G� because fx; fy are continuous on G.Finally, we check Cauchy-Riemann equations for f�: @U@x (a0; b0) = @u@x (a0;�b0) = @v@y (a0;�b0) =@V@y (a0; b0), @U@y (a0; b0) = ��@u@y (a0;�b0)� = @v@x (a0;�b0) = @V@x (a0; b0). Therefore, f� is analyticon G�.(6) (a) @v@x = �@u@y = 6xy@v@y = @u@x = 3x2 � 3y29>=>;) v(x; y) = 3x2y +C1(y)v(x; y) = 3x2y � y3 +C2(x)�) v(x; y) = 3x2y � y3 +C:

(b) @v@x = �@u@y = e�y cos x@v@y = @u@x = �e�y sinx 9>=>;) v(x; y) = e�y sinx+C1(y)v(x; y) = e�y sinx+C2(x)�) v(x; y) = e�y sinx+C:(c) @v@x = �@u@y = �2yx2 + y2@v@y = @u@x = 2xx2 + y2 9>=>;) v(x; y) = 2 arctan(yx ) + C1(y)v(x; y) = 2 arctan(yx ) + C2(x)9=;) v(x; y) = 2 arctan(yx ) + C:(d) @v@x = �@u@y = y2 � (1� x)2((1� x)2 + y2)2@v@y = @u@x = 2(1� x)((1� x)2 + y2)2 9>>=>>;) v(x; y) = x� 1(1� x)2 + y2 + C1(y)v(x; y) = x� 1(1� x)2 + y2 + C2(x)) v(x; y) = x� 1(1� x)2 + y2 + C:(7) x = r cos �, y = r sin �, @u@x = @v@y , @u@y = �@v@x . @u@r = @u@x @x@r + @u@y @y@r = @u@x cos � + @u@y sin �.1r @v@� = 1r�@v@x @x@� + @v@y @y@� � = 1r��@u@y (�r sin �) + @u@x (r cos �)� = @u@x cos � + @u@y sin � = @u@r .@v@r = @v@x @x@r + @v@y @y@r = @v@x cos � + @v@y sin �.�1r @u@� = �1r�@u@x @x@� + @u@y @y@�� = �1r�@v@y (�r sin �) � @v@x (r cos �)� = @v@x cos � + @v@y sin � = @v@r .

Solution to Practice Exercise Set IV(1) (a) z 7! i�1� z1 + z� z 7! 1� Log zT (z) = 1� Log�i�1� z1 + z��(b) z 7!ri�1� z1 + z� z 7! z � 1z + 1T (z) = �ri�1� z1 + z�� i� � ��ri�1� z1 + z�+ i�(c) T (z)=�pi( 1�z1+z )�i����pi( 1�z1+z )+i� z 7! z2T (z) = "�ri�1� z1 + z�� i� � ��ri�1� z1 + z�+ i�#2(d) z 7! �1� z1 + z�2 z 7! z � 1 z 7! 1z z 7! 2z + 10 -1 -1 0 -1 1T (z) = 2��1� z1 + z�2 � 1��1 + 1 = ��z2 + 12z � = �12�z + 1z�.(2) Observe that T (1) = 1, T (ia) = 0, T (ib) = 1. Orient L3 by (1; ia; ib). The right side of L3 isD [E [F . The right side of T (L3) (with respect to (1; 0;1)) is U [V [W . The part of L3 from1 toia is mapped onto the segment from 1 to 0. So we have the correspondence F $ V , E $ U , D $W ,C $ Y , B $ X, A$ Z.

Solution to Practice Exercise Set V(1)1

Observe that the range of f(z) is in the disk jw � 1j < 1, which is contained inthe slit plane C n (�1; 0) (Complex plane minus the negative real axis and 0).Now Logw is analytic on the slit plane. So Log f(z) is de�ned and analytic. ThenZC f 0(z)f(z) dz = ZC(Log f(z))0 dz = Log f(z(b))�Log f(z(a)) = 0 because C is closed(i.e. z(a) = z(b)).(2) The unit circle (counterclockwise direction) is given by z(t) = eit = cos t + i sin t (0 � t � 2�),jdzj = jieitj dt = dt, soZjzj=1 jz � 1jjdzj= Z 2�0 q(cos t� 1)2 + sin2 t dt = Z 2�0 p2� 2 cos t dt = 2 Z 2�0 ����sin t2���� dt = 8:(3) Observe that ZC(f(z) + g(z)) dz = ZC f(z) dz + ZC g(z) dz and ZC �f(z) dz = �ZC f(z) dz. So it suf-�ces to consider the special case P (z) = zn (the general case is obtained by using the linearity propertiesabove). C is given by z(t) = a+ Reit, 0 � t � 2�, dz = iReit dt.ZC P (z) dz = ZC zn dz = Z 2�0 (a +Re�it)niReit dt= Z 2�0 (an + nan�1Re�it + � � �+Rne�int)iReit dt= Z 2�0 nan�1iR2 dt (�)= �2�inan�1R2= �2�iR2P 0(a):(*) where we use the fact Z 2�0 eint dt = 8>><>>:Z 2�0 (cos nt+ i sin nt) dt = 0 if n 6= 0Z 2�0 1 dt = 2� if n = 0.(4) Since 1z2 has the antiderivative �1z in C n f0g, ZC 1z2 dz = �1z ���z=�1z=1 = 2.(5) Suppose Zjzj=1 f(z) dz = Z 2�0 f(eit)ieit dt = Rei�. Then��� Zjzj=1 f(z) dz��� = R = Z 2�0 f(eit)iei(t��) dt = Re�Z 2�0 f(eit)(� sin(t� �) + i cos(t� �)) dt�= � Z 2�0 f(eit) sin(t� �) dt� Z 2�0 ��f(eit) sin(t � �)�� dt� Z 2�0 j sin(t� �)j dt = Z 2�0 j sin tj dt = 4.

(6)i

C4

C1

2C

C3

21-1-2 0

2iC1: z(t) = t, �2 � t � �1C2: z(t) = eit, t from � down to 0C3: z(t) = t, 1 � t � 2C4: z(t) = 2eit, 0 � t � �ZC1 zz dz = Z �1�2 1 dt = 1, ZC2 zz dz = Z 0� eite�it ieit dt = i Z 0� e3it dt = e3it3 ���0�= 23.ZC3 zz dz = Z 21 1 dt = 1, ZC4 zz dz = Z �0 2eit2e�it2ieit dt = 2i Z �0 e3it dt = 23e3it����0= �43 .ZC zz dz = 1 + 23 + 1� 43 = 43.

Solution to Practice Exercise Set VI(1)a a a

an(C; a) = �1 n(C; a) = 1 n(C; a) = �1 n(C; a) = 0(2) Log(w + 1) = w � w22 + w33 � w44 + : : : for w near 0. cosw � 1 = �w22! + w44! � w66! + : : : for all w.For z near 0, z Log(z49 + 1)cos z25 � 1 = z(z49 � z982 + : : :)�z502 + z10024 � : : : = 1� z492 + : : :�12 + z5024 + : : :.Therefore, limz!0 exp(z Log(z49 + 1)cos z25 � 1 ) = e�2.(3)f

1

T

Let f(z) be an entire function, whose range lies in theright half plane. Let T (z) = z � 1z + 1, then T � f(z) isentire and has range in the unit disk, i.e. jT �f(z)j � 1for all z. By Liouville's theorem, T �f(z) � constant.Therefore f � T�1 � T � f � constant.(4) Let f(z) = a0 + a1z + : : :+ anzn be a polynomial such that jzj � 1 ) jf(z)j � 1. By corollary (1),jakj = ��k!f (k)(0)�� = ������� 12�i Zjzj=1 f(z)zk+1 dz������� � 12� 11|{z}M 2�|{z}L = 1.(5) Let f(z) sin z and g(z) = ( sin zz if z 6= 01 if z = 0, then by Cauchy's integral formula,Zjzj=1 sin zz dz = Zjzj=1 f(z)z � 0 dz = 2�if(0) = 0 and Zjzj=1 sin zz2 dz = Zjzj=1 g(z)z dz = 2�ig(0) = 2�i.(6) (a) Jf (x; y) = @u@x @v@y � @u@y @v@x = �@u@x�2 +�@v@x�2 = ��f 0(z)��2 because f 0(z) = @u@x + i @v@x and by theCauchy-Riemann equations.(b) ZZD zmzn dA z=rei�= Z 2�0 Z 10 rmeim�rne�in�r dr d� = Z 2�0 ei(m�n)� d�Z 10 rm+n+1 dr= ei(m�n)�i(m � n) ����2��=0 1m + n+ 2 = 0.(c) area of G = ZZD��Jf (x; y)�� dx dy (a)= ZZD��f 0(z)��2 dA = ZZD ���� 1�n=1ncnzn�1����2 dA= ZZD� 1�m=1mcmzm�1�� 1�n=1ncnzn�1� dA= ZZD� 1�m=n=1 n2jcnj2jzj2n�2 + �m6=nmncmcnzm�1zn�1� dA(b)= 1�n=1n2jcnj2 Z 2�0 Z 10 r2n�2r dr d� = 1�n=1n2jcnj22� 12n = � 1�n=1njcnj2.

Solution to Practice Exercise Set VII(1) Suppose f( 1n ) = 1n+ 1 for all positive integer n. Since 1n has limit point 0, which is in �z: jzj � 1,so the identity theorem implies f(z) = 11z + 1 = z1 + z . However, this is not analytic at z = �1, acontradiction.(2) De�ne g(z) = f(z)z2 for 1 � jzj � 2. Then g is continuous on 1 � jzj � 2 (which is closed and bounded)and analytic on 1 < jzj < 2. So the maximum modulus theorem implies that for 1 � jzj � 2,jf(z)jjzj2 = jg(z)j � maxjwj=1;2 jg(w)j = maxjwj=1;2 jf(w)jjwj2 = 1:(3) (a) Let � be the boundary of a rectangle. For a �xed t; sin zt is entire, so Z� sin zt = 0. ThenZ� f(z) dz = Z� Z 10 sin ztt dt dz = Z 10 Z� sin ztt dz dt = Z 10 1t�Z� sin zt dz�dt = 0.(Details: Since limw!0 sinww = 1; sinww is bounded for jwj � jzj. Suppose ���� sinww ���� � M for jwj � jzj.Then ���� sin ztt ���� � M jzj for 0 < t � 1 and Z� Z 10 ���� sin ztt ���� dt dz <1. These imply f(z) is continuousand interchange of integration is possible.)(b) sin zt = 1�n=0 (�1)n(zt)2n+1(2n+ 1)! . For a �xed z0; 0 < t � 1; tz0 lies in the closed disk B(0; jz0j). So1�n=0 (�1)nz2n+10 t2n(2n+ 1)! converges uniformly to sin z0tt as a function of t (power series converges uni-formly in closed subdisks of domain of convergence). Thenf(z0) = Z 10 sin z0tt dt = Z 10 1�n=0 (�1)nz2n+10 t2n(2n+ 1)! dt = 1�n=0 (�1)nz2n+10(2n+ 1)! Z 10 t2n dt = 1�n=0 (�1)nz2n+10(2n+ 1)!(2n+ 1) :(4) Since �z: jzj � 1 is closed and bounded, let M = maxjzj�1 jf(z)j. Also let T (z) be a M�obius transformationmapping UHP = �z: Imz � 0 onto �z: jzj � 1 and R onto �z: jzj = 1 (e.g. T (z) = z � iz + i ). Thenf �T (z) is continuous on UHP, analytic on UHF = �z: Imz > 0 and real-valued on R. By the Schwarzre ection principle, f � T (z) can be extended to an entire function.So jf �T (z)j � M for all z, Liouville's theorem implies f �T (z) is constant. Therefore, f(z) is a constant.(5) Let w1; w2 be two distinct �xed points of f in D. Let T (z) = z +w11 +w1z , then T is one-to-one map fromD onto D. So there is z0 such that T (z0) = w2. The function T�1 �f �T (z) is an analytic function fromD onto D and T�1 � f �T (0) = 0 and T�1 � f �T (z0) = z0. The equality case of Schwarz lemma impliesT�1 � f � T (z) � ei�z. Using T�1 � f � T (z0) = z0, we get ei� = 1. So T�1 � f � T (z) = z. Therefore,f � T (z) = T (z) for all z 2 D. Therefore f(w) � w for all w 2 D.(6) Since f(z) is entire, f(z) = 1�n=0anzn for all z, where an = f (n)(0)n! . Since f(z) is real on the realaxis, Schwarz re ection principle and the identity theorem imply f(z) = f(z) for all z. So 1�n=0anzn =1�n=0anzn = 1�n=0 anzn. By the uniqueness of power series, an = an for all n.

De�ne g(z) = if(iz). Since f(z) is imaginary on the imaginary axis, g(z) is real on the real axis.Now g(z) = 1�n=0 in+1anzn. By the above argument, we have in+1an = in+1an. If n is even, thenin+1an = �in+1an. This implies an = an = �an for even n. So an = 0 for all even n. Thereforef(z) = 1�k=0a2k+1z2k+1 is odd.Alternatively, f(z) = f(z) for all z and g(w) = g(w) for all w) f(z) = f(z) for all z and if(iw) = if(iw) = �if(iw) for all w) f(z) = f(z) and f(z) = �f(�z) for all z (set z = iw)) f(z) = f(z) = �f(�z) for all z) f(u) = �f(�u) for all u (set u = z)(7) f(z) has �nitely many roots �1; : : : ; �n (repeated according to multiplicities) in the open unit disk andno roots on the unit circle since jf(z)j = 1 for jzj = 1 (otherwise the roots have a limit point in theclosed disk forcing f(z) � 0). By the theorem following the orientation principle, ���� z � �j1� �jz ���� = 1 forjzj = 1. The function g(z) = f(z)n�j=1 z � �j1� �jz is analytic on D. (This is clear for z 6= �1; : : : ; �n becausen�j=1 z � �j1� �jz 6= 0. If �k is a root of multiplicitym, then limz!�k f(z)(z � �k)m = f (m)(�k)m! 6= 0 (by l'Hopital'srule) and m of the �k's equal �k so that limz!�k g(z) exists and is nonzero, which is used to de�ne g(�k).It follows that g is analytic at �k.) By the maximummodulus theorem, for jzj � 1,jg(z)j � maxjwj=1 jg(w)j = maxjwj=1 jf(w)j���� n�j=1 w � �j1� �jz ���� = maxjwj=1 jf(w)j = 1:Since g has no root in D = �z: jzj < 1, the minimum modulus theorem implies, for jzj � 1; jg(z)j �minjwj=1 jg(w)j = 1. Therefore jg(z)j = 1 for jzj � 1. Then g(z) = ei� because jg(z)j = 1. Hencef(z) = ei� n�j=1 z � �j1� �jz by the identity theorem. Since f is entire, �j = 0 (otherwise f is not de�ned at1�j ). Therefore f(z) � ei�zn.

Solution to Practice Exercise Set VIII(1)......

a1 a

2an

Introducing n+1 cross-cuts as shown below and applying Cauchy's theoremto the upper and lower simple closed curves, we get (after cancelling theintegrals over the cross-cuts)0 = Z� f(z) dz � n�j=1ZCj f(z) dz:The result follows.(2) (a) Since 1z4 + z2 = 1z2(z + i)(z � i) , there are a pole of order 2 at 0, a pole of order 1 at i and �i,respectively.(b) Since cot z = cos zsin z and limz!n�(z � n�)cos zsin z = 1; limz!n�(z � n�)2 cos zsin z = 0, there is a pole of order 1at n�, where n is any integer.(c) The isolated singularities are at 0 and 1. There is a pole of order 1 at 1. Since e1=zz � 1 doesn't havea limit as z ! 0, so 0 is an essential singularity.(d) The isolated singularities are at all integers. Since limz!1 z2 � 1sin�z = � 2� , limz!�1 z2 � 1sin�z = 2� ,limz!n�;n6=�1(z � n�)z2 � 1sin�z = �n2�2 � 1� 6= 0, there are removable singularities at 1 and �1, respec-tively, and pole of order 1 at all other integers.(3) (a) 1z2 � 4 = 14(z � 2) 1(1 + z � 24 ) = 14(z � 2)(1� z � 24 + (z � 24 )2 � : : :) = 1�k=�1 (z � 2)k4k+2 , where����z � 24 ���� < 1.(b) 1z2 � 4 = 1z2(1� 4z2 ) = 1z2 (1 + 4z2 + 16z4 + : : :) = 1�k=1 4k�1z2k , where ���� 4z2 ���� < 1.(4) Suppose 1(n+ 1)� < r < 1n� . Since sin 1z = 1z � 13!z3 + 15!z5 � : : : for 1(n+ 1)� < jzj < 1n� ,Zjzj=r sin 1z dz = 2�a�1 = 2�i. (This is the same if 1� < r <1.)(5) Let f(z) = 1�k=�1 akzk for C n �0 = �z: 0 < jzj <1. For k = 0,jakj � 12� Zjzj=r jf(z)jjzjk+1 jdzj � pr + 1prrk = r + 1rk+ 12 ! 0� as r!1 if k > 0as r! 0 if k < 0Therefore, f � a0.(6) If f has a pole of order k at 0, then f(z) = a�kzk + a�k+1zk�1 + : : : = a�kzk (1 + a�k+1a�k + : : :| {z }=g(z) ) = a�kg(z)zk for0 < jzj < ", where g is analytic on jzj < "; g(0) = 1 and a�k 6= 0. Let z = kpta�k with t a real variable,then limt!0+ ef(z) = e+1 = +1 and limt!0� ef(z) = e�1 = 0. So 0 is an essential singularity of ef(z).

(7) (i) f(z) entire ) f(z) = 1�k=0akzk. f(1z ) has a pole of order N ) f(1z ) = 0�k=N akzk . So f(z) = N�k=0akzkis a polynomial.(ii) Since C [ �1 is closed and bounded, and the poles are isolated, there are only �nitely manypoles. Let the finite poles be p1; : : : ; pk and the respectiveorders be N1; : : : ; Nk. Let P (z) =(z � p1)N1 : : : (z � pk)Nkf(z), then P is entire. Since 1 is either a removable singularity or a poleof f(z), the same is true for P . If1 is a removable singularity, then P is analytic at 1 and P (1)is a complex number. This implies P is bounded on C, so by Liouville's theorem, P (z) � P (1)and f(z) = P (1)(z � p1)N1 : : : (z � pk)Nk is a rational function. If 1 is a pole, then by part (i), P (z)is a polynomial, then f(z) = P (z)(z � p1)N1 : : : (z � pk)Nk is a rational function.

Solution to Practice Exercise Set IX(1) Let g(z) = f(1z ), then 0 is an isolated singularity of g. If it is removable, then f has a removablesingularity at 1 and f will be bounded, forcing it to be a constant. If 0 is a pole of g, then g(z) =1�k=�N akzk near 0 and f(z) = N�j=�1 a�jzj . Since f is analytic at 0, a�j = 0 for j < 0, then f isa polynomial. By the fundamental theorem of algebra, the image of the plane under f is the wholeplane (hence the image is dense). If 0 is an essential singularity of g, then Casorati-Weierstrass theoremimplies the image of the plane under g (or f) is dense in the plane.(2) Suppose f has more than one roots. Let R be so large that there are more than one roots inside �, thecircle jzj = R. Since f(R); f(�R) are the only real valued on f � �, the curve �f(Rei�): 0 < � < �and �f(Rei�):� < � < 2� lie entirely on the upper or lower half plane. Then n(f � �; 0) = 0 or 1.Therefore, f can have at most one root by the Argument Principle, a contraction.(Note f 6� 0, so we may assume f(R) 6= 0 and f(�R) 6= 0 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 �; f � � has no self-intersections, so it is a simple closed curve. For w 62 f � �,let g(z) = f(z)�w, then n(g ��; 0) = n(f ��; w) = 1 or 0 depends whether w is inside or outside f ��.If z0 is inside �, then f(z0) is not on f � � by the open mapping theorem. Now z0 is a root ofg(z) = f(z)�f(z0), hence n(g ��; 0) = 1. Then g has no other root. This implies f is one-to-one inside�.(5)γ

2

γ1

R

-iR

iR

f γ2

f o γ2

o

Consider the contour on the left with R large. On 1; z(t) = Reit;��2 � t � �2 ; f(z) =z2n + �2z2n�1 + �2 � z2n, so � 1 arg f(z) = 2n�. On 2; z(t) = it, wher t decreasesfrom R to �R.If n is odd, f(it) = �t2n + �2 + i�2t2n�1; f(iR) � �1 + i0+; f(�iR) � �1 + i0�.(Note f(ij�j 1n ) = i�2j�j 2n�1n ; f(0) = �2; f(�ij�j 1n ) = �i�2j�j 2n�1n .)A sketch shows � 2 arg f(z) = �2�. So f has (2n� 2)�2� = n � 1 roots inside thecontour for R large, so f has n� 1 roots with positive real parts.If n is even, f(it) = t2n + �2| {z }>0 �i�2t2n�1; f(iR) � +1� i0+; f(�iR) � +1� i0�.A sketch shows � 2 arg f(z) = 0. So f has 2n�2� = n roots inside the contour for Rlarge, so f has n roots with positive real parts.(6) Let f(z) = ez � azn and g(z) = azn, then for jzj = 1; jf(z) + g(z)j = jezj = eRe z � e1 < a = jaznj =jg(z)j < jf(z)j + jg(z)j. So by Rouch�e's theorem f(z) has n roots inside the unit circle (because g hasn roots inside the unit circle).

Solution to Practice Exercise Set X(1)Re

0 R

2π i/n

γ1

− γ2

By residue theorem,�Z R0 + Z 1 + Z 2� dz1 + zn (*)= 2�i Resz=e�i=n� 11 + zn� = 2�i 1n(e�i=n)n�1 :On 1, z = Rei�, 0 � � � 2�n , ����Z 1 dz1 + zn ���� �ML = 1Rn � 1 2�Rn ! 0 as R! +1.On 2, z = xe2�i=n, 0 � � � 2�n , Z� 2 dz1 + zn = �Z R0 e2�i=n1 + xn dx = �e2�i=nZ R0 dx1 + xn .Letting R! +1, (*) ) (1 � e2�i=n)Z 10 dx1 + xn = 2�i 1n(e�i=n)n�1 = 2�i�n e�i=n.Therefore, Z 10 dx1 + xn = 2�in(e�i=n � e��i=n) = �n sin �n .(2)R-R

CR e2iz = 1+ 2iz + (2iz)2! + � � �, e2iz � 1� 2izz2 = �2� 4i3 z+ � � �. (At 0, e2iz � 1� 2izz2has a removable singularity, where we assign the value �2 to make it analytic.)Consider the contour shown.By Cauchy's theorem,�Z R�R+ ZCR�e2iz � 1� 2izz2 dz (*)= 0: Z 0�R e2iz � 1� 2izz2 dz x=�z= Z R0 e�2ix � 1 + 2ixx2 dx:On CR, ����e2iz � 1z2 ���� � e�2 Im z + 1R2 � 2R2 . So ����ZCR e2iz � 1z2 dz���� � 2R2�R ! 0 as R ! +1. ThenlimR!+1 ZCR e2iz � 1� 2izz2 dz = limR!+1��2i ZCR dzz � = 2�.Taking R! +1, (*) implies Z 10 e2ix + e�2ix � 2x2 dx+ 2� = 0. Since e2ix + e�2ix � 2 = �4 sin2 x, weget Z 10 sin2 xx2 dx = �2 .(3)-r r R-R

CR

Cr

i

Recall log z = ln jzj+ i arg z, 0 � arg z � � and so log i = i�2 .By residue theorem, �Z �r�R � ZCr + Z Rr + ZCR� log zz2 + 1 dz (*)= 2�iResz=i log zz2 + 1 = i�22 .����ZCr log zz2 + 1 dz���� � j ln rj+ �1� r2| {z }M �r|{z}L ! 0 as r ! 0+, ���� log zz2 + 1 dz���� � j lnRj+ �R2 � 1 �R ! 0 as R ! +1.Z �r�R log zz2 + 1 dz = Z Rr lnx+ i�x2 + 1 dx. Letting r ! 0+, R! +1, (*)) 2Z 10 lnxx2 + 1 dx+i� Z 10 dxx2 + 1| {z }�=2 =i�2 . Therefore, Z 10 lnxx2 + 1 dx = 0.(4)R

R+i

-R

-R+i

−γ3

γ1

γ2

Consider f(z) = e�z2 . By Cauchy's theorem,�Z R�R+ Z 1 + Z 2 � Z 3�e�z2 dz (*)= 0:

������ Z 1; 3 e�z2 dz������ � e�R2+1| {z }M ! 0 as R ! +1. Z 2 e�z2 dz = Z �RR e�(x+i)2 dx = � Z R�R e�x2�2ix+1 dx.Letting R! +1, (*) implies Z 1�1 e�x2 dx� e Z 1�1 e�x2 (cos 2x� i sin 2x) dx = 0. Since Z 1�1 e�x2 dx =p�, taking the real part, we get Z 10 e�x2 cos 2x dx = p�2e .(5)0 R

π/4

γ1

−γ2

Consider f(z) = eiz2 . By the Cauchy's theorem, �Z R0 + Z 1 � Z 2�eiz2 dz (*)= 0. On 1, z = Rei�, 0 � � � �4 , ��eiz2 �� = e�R2 sin 2� � e�4R2�� by Jordan's inequality.So ����Z 1 eiz2 dz���� � Z �40 e�4R2�� Rd� = �4R(1� e�R2 )! 0 as R! +1.On 2, z = xei�4 , 0 � x � R, Z 2 eiz2 dz = Z R0 e�x2(ei �4 dx). Letting R ! +1, (*) ) Z 10 eix2 dx �ei�4 Z 10 e�x2 dx = 0. Since Z 10 e�x2 dx = p�2 , taking real and imaginary parts, we get Z 10 cos x2 dx =p�2 cos �4 = p2�4 , Z 10 sinx2 dx = p2�4 .(6) Z �=20 d�1 + sin2 � = 14 Z 2�0 d�1 + sin2 � = 14 Zjzj=1 �i1� 14(z � 1z )2 dzz = Zjzj=1 iz(z2 � 2z � 1)(z2 + 2z � 1) dz= i4� Zjzj=1 dzz2 � 2z � 1 � Zjzj=1 dzz2 + 2z � 1� = ��2 � Resz=1�p2 1z2 � 2z � 1| {z }= 1�2p2 � Resz=�1+p2 1z2 + 2z � 1| {z }= 12p2 � = �2p2(7) (a) Res(f;1) = 12�i Zjzj=R f(z) dz = � n�j=1Res(f; aj) (clockwise orientation), where the last equality fol-lows from residue theorem.(b) Suppose f(z) = � � �+ a�kzk + � � �+ a�1z + a0 + a1z + � � � on r < jzj <1, then� 1z2 f(1z ) = � 1z2 (� � �+ a�kzk + � � �+ a�1z + a0 + a1z + � � �) on 0 < jzj < 1r= � � � � a�kzk�2 � � � � � a�1z � a0z2 � a1z3 � � � � .So Res(� 1z2 f(1z ); 0) = �a�1 = Res(f;1).(c) Let f(z) = 1sin(1z ) , then Zjzj=1 f(z) dz = �2�iRes(f;1) = �2�iRes(� 1z2 f(1z ); 0) (counterclockwiseorientation). Now � 1z2 f(1z ) = � 1z2(z � z36 + � � �) = � 1z3� 11� z26 + � � �� = � 1z3 (1 + z26 + � � �) =� 1z3 � 16z � � � �. So Zjzj=1 1sin(1z ) dz = �2�i(�16 ) = �i3 (counterclockwise orientation).Alternatively, Zjzj=1 1sin(1z ) dz = Zjzj=1 11z � 16z3 + � � � dz = Zjzj=1 (z + 16z + � � �) dz = 162�i = �i3 ,where all orientations are counterclockwise and the last integrand converges uniformly on jzj = 1.