.

2 + 2 2 + 2

Fundamentals of Complex Analysis with Applications to Engineering

and Science 3rd Edition Saff SOLUTIONS MANUAL

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CHAPTER 2: Analytic Functions

EXERCISES 2.1: Functions of a Complex Variable

1. a. to= (3z2 - 31,2 + 5% + 1) + i(6zt, +Sy+ 1)

b. tD = z2 : y2 + i ( z2 ! y2) 1 z . -y + 1

c. to=-= + ,----- z - i z (y - 1 ) z (y - 1 )

d 2%2 - 2y

2 + 3 . 4zy

• tD = J(z-1)2 + y2 + a J(z-1)2 + y2

e. w = e3z: cos 3y + ii- sin 3y

f. w = (ec + e-c) cosy+ i(� - e-c) siny

= 2 cosh z cosy + i2 sinh z sin y

2. a. C

b. C \ {O}

c. c \ {i,-i}

d. C \ {1}

e. C

f. c

3. a. Rew> 5

b. Imw > 0

c. lwl > 1

d. The intersection of fwl < 2 and -,r < Arg w < r/2

4. a. Taking IJ from O to 2r, the points % = rei' traveise the circle

lzl = r exactly once in the oounterclockwise direction. For the

same values of 9 the points to= 1 , = !.e-il traverse the circle

re• r

lw I = .!:. exactly once in the clockwise direction, hence the mapping . r 1s onto.

b. For z = re �

on the ray Argz = B , w = 1

/J- = 1 e-

� .s on

' 0 0 - - • 0 i

re'"o r theray Argw = -90• Taking values O < r < oo shows that this mapping goes onto the ray Argw = -Bo.

Z- 2 Z lz 2 Z

-----·---------·-----------·- ...

4 (c) lz-� = I 2�0� � z = I+ ei9

. F(z) = 1/z. = 1/(l+ eia)

. - .0 + e ) I { 2(1 +cos0)} = Yi -i(Yi)sin0/(l +cos0) which ts a vertical line at x = Yi.

5. a. domain: C

range: C \ {O}

1 1 b. f(-z)=Cz=-=•

ez f(z)

c. circle lw I = e

d. ray Arg w = 1r / 4

e. infinite sector O � Arg w � 1r / 4

6. a. 1(!)=!(!+-)=!(z+.!.)=J(z)

b. For z = ei8 on the unit circle lzl = 1, J(z) = i (i8 + e!s) = cos 0.

For all values of (}, this ranges over the real interval [-1, 1]. ·

c. For z = rei8 on the circle lzl = r, J(z) = i (re;e + r:;s) -

i (r +;) cos O + ii (r - ;) sine. Setting u and v equal to the

real and imaginary parts of this expression, respectively, one gets a pair of parametric equations that are equivalent to the ellipse

u2 v2

[!(r + �)]2 + [Hr_ �)]2 = 1, which has foci at ±1.

a.::j Q7. b.

-2 -l

J. - l.

------------------- ------

c.

8. a. b.

.s

c.

9.

10.

a.

a. translate by i, rotate 1r /4

c. translate by i, reduce by 1/2

-l -o.s o.s

'). - L,(

b.

b. reduce by 1/2, rotate 1r /4

d. reduce by 1/2, rotate 1r /4,

translate by i

-1 -o.1s-o.s-o.2s 0.25 o.s o. 5

2

,

)

l l

11. a. translate by -3, rotate -r/2

c. translate by -3, magnify by 2

b.magnify by 2,

rotate -1r/2

d. magnify by 2,· rotate -r /2,

translate by -3

-1

12. Let a = pe•, F(z) = pz, G(z) = e•z, a.nd H(z) = z + b. Then

H(G(F(z))) = az + b.

13. (a) w = u + iv=z2

= (1 + iy)2 = 1 -y2+ i2y

u = 1-y2, v = 2y �y = v/2 � u = 1- v /4 a parabola in thew-plane.

(b) w = u + iv=z2 = (x + iy)

2 = (x + i!x)2 =x2 - l/x

2 + 2i

u = x2 - l/x

2 v = 2 a straight line in thew-plane.

(c) w = u + iv=z2 = (1 + ei0)2 = (1 + 2ei9 + ei29 = (e-i9 + 2 + ei6)ei9

= (2 + 2cos0)ei9 = 2( 1 + cos0)ei

9 a cardioid in the w-plane.

14. (a) x, = 2x/(lzi2 + 1), x2 = 2y/(lzl2

+ 1), x, = (lzl2

- 1) /(lzi2 + 1)

w = ei<Pz = xcoscp-ysincp + i(xsincp+ycoscp), lwl = lzl

X1 = (xcoscp-ysincp)/(lz 2 + 1), Xi= (xsincp+ycoscp)/(lz

2 + 1), X3 = x,

X1 = (xjcoso-x-simp), Xi= (x.sinqi+x-costp), X3 = X3 which corresponds

to a rotation of an angle cp about the x3 axis.

(b) w = -1/z. lwl = 1/lzl. w = -1/(x+iy) = -x/lzl + iy/lzl

X1 = -Xj , Xi= X2, X3 = -x, so that (X1, Xi, ,!3) is obtained from

(xr, x2, X3) by a 180° rotation about the x2 axis.

15. w = (1 +z)/(1 - z) = (1 +x + iy)/(1 - x - iy) = (1-lzl2 + i2y)/(1- 2x + lzi2)

lwl2

= ( 1 + 2x + lzl2)/( 1- 2x + lzl2).

Cxt, .!2, ,!3) = (-x3, x2, X1) so that Cx1, Xi, XJ).is obtained by a 90°

counterclockwise rotation about the x2 axis.

2-5

4

1.6. w = (1 - iz)/(1 + iz) = (1- ix + y)/(1 + ix - y) == (1-lzl2 + i2x)/(1- 2y + lzi2)

lwi2 = (1 + 2y + lzl2)/(l- 2y + lzl

2).

(!1, � ' � ) = (-x3, -x 1, x2) so that (!1, � ' � ) is obtained as a 90°

counterclockwise rotation about the x2 axis followed by a 90° counterclockwise rotation about the x-axis.

17. Any circle or line in the z-plane corresponds to a line or circle on the

stenographic projection onto he Riemann sphere. The function w=l/z

rotates the Riemann sphere 180° about the x 1 axis. Lines and circles on

the rotated sphere project to lines and circles in the w-plane. As a result

lines and circles in the z-plane map to lines and circles in the w-plane.

EXERCISES 2.2: Limits and Continuity

1. The first five terms are, respectively, i, - 1 , -i, sequence converges to O in a spiral-like fashion.

0.

o.

- .25 -0.2 -0.lS -0.l -0.0S

-o.

1 , and

16

i . The

32

2. 2i, -4, -Bi, 16, 32i; divergent because terms grow in modulus without bound.

3

20

10

10 15

;;:i - b

z:

3. If limn-+oo Zn= z0, then for any e>O, there is an integer N such that

lzn - z01<E for all re-N. For the same integer N we have

lx, - Xol<=lzn - zol<E and lyn - Yol<=lzn - zol<E for all ro-N, Therefore,

limn-+oo x- = Xo and limn-+oo v« = Yo,

If limn-+oo Xn = Xo and limn-+oo Yn = Yo, then for any £1 >0 and E2 >0 there are

integers N 1 and N2 such

lx, - xol<E1 for all n > N 1 and lyn - y01<E2 for all n > N2. Given any E > 0;

let Et= £12 and £2 = £12. Then

lzn - zol<=E lx, - xol + lyn - yol < E1 + E2 = E for all n > maximum/Ni, N2).

Thus limn-+oo Zn = Zo.

4. If Zn= x, + iyn � Zo = Xo + iyo, then x; � Xo and Yn � Yo (see Problem 3).

bi = Xn - iyn � Xo - iyo = io.

If bi= Xn - iy, � io = Xo - iyo, then Xn � Xo and Yn � Yo (see Problem 3) ..

Zn = x, + iyn � Xo + iyo = Zo. Thus Zn � Zo if and only if bi � io.

� .!!.m, l� I = 0 ==> There exists an integer N such that

11� 1- OI = lz.f < E whenever n > N. ==> ··Jin - OI < e whenever

n > N. ==> lim.n-oo Zn = O, and conversely.

l. � -+ 0 as n -+ oo by problem 3, since the real-valued.. sequence

1%31-+ 0 as n-+ oo. On the other hand, if l.zol > 1, then 1� 1-+ oo as n -+ oo so diverges.

1, a. converges to O

b. does not converge

c. converges to 1r

d. converges to 2 + i e. converges to O

f. does not converge

8. Given e > 0, choose 6 = e/6. Then whenever

0 < fz - (1 + i)f < 6,

f6z -4- (2 + 6i)f = 6lz - (1 + i)I < 6(e/6) = e

'1, Given e > 0, choose 6 = 1 : e' Whenever O < lz - (-i)I < 6 notice

that lzl > 1 - 6 and

1;-i1 = 1(-;) (i+z)j = 1!11z- (-i)I < (A) 6 = e

�-7

.

= =

z-zo g z im g z g zo

2i

10. Given that f and g aie mntinuoua a.t Zo,

..l_i.m...., /(z) ±g(z) = .l..i.m.... /(z) :l: .l..i.m... g(z) = /(Zo) ±g(.ro)

==> /(z) ± g(z) is continuous at Zo·

lim /(z)g(z) lim /(z) lim g(z) /(Zo)g(Zo) ..1-uo s--11 ..1-t.SC,

====> /(z)g(z) is continuous at Zo·

. f(z) }i� f(z) f(zo) .

lim -) = r ( ) = -(-, provided g(z0) i= 0 z-zo

f(z) . . ==> g( z) is continuous at z0•

Hi a. -Si

b . - 7 .

c. 6i

d. -1/2

e. 2zo

f. 4v'2

I�. Clearly Argz is discontinuous at z = 0. Let a> 0 be any real number

and consider the sequence

Zn= -a - i/n n = 1, 2, ... , which converges to - a.

For each n, -1r < Arg Zn < -1r /2, but Arg (-a) = 1r.

13, lim f ( z) exists for all z i= -1; f is continuous for all z i= 0, -1; f has z-zo a removable discontinuity at z = 0.

1 � . let z0 be any complex number. Given e > 0 choose S = c. Then

whenever lz - zol < S,

lg(z) - g(zo)I = lz - zol = lz - zol = lz - zol < e.

15.. Given e > 0 choose S so that lf(z) - f(zo)I < e whenever lz - zc,I < S.

Then, whenever lz - zo I < &

a. lf(z) - f(zo)I = lf(z) - f(zo)I = lf(z) - f(zo)I < c

b. IRef(z) - Ref(zo)I == IRe(f(z)- f(zo))I � lf(z)- f(zo)I < e

c. llmf(z) - lmf(zo)I = llm(f(z) - f(zo))I � lf(z) - f(zo)I < e

d. 11/(z)l - lf(zo)II $ lf(z)- f(.zo)I < c

.-1 .--1 a-1 % + 1/ e-1

0 0

L/

16 Gmn £ > 0, chooee Oo > 0 such that 1/(g(z))- /(g(Zo))I <£whenever

.... �--1,(z) -- g(.ic,)1--< 6c,. Now choose- 6 > 0 such that (g(z) - g(.ro)I < 6o whenever fz - Zol < 6. Then 1/(g(z)) - /(g(zo))I < e whenever

lz- Zol < 6; hence /(g(z)) is continuous at zo. ·

ti. No: Observe that although

1 - -+ 0 and

i - -+ 0 as n --+ oo,

n n .

J (!) - 1 + 2i and J (!) - 2i; thus .=:AJ(z) does not exist.

-- --1-8:·· - -·Iflimz-+zoftz} =wo-; then-given E>O there exists B>O such that· -

lf(z)-wol<E for all lz-z01<o. Notice that

lf(z)-wQI = lf(z)-wQI = lf(z)-wol<Efor all lz-zol<O .. So that limz-+zof<zl=wo.

limx-+xo,y-+o µ(x,y) = limz-+zo ((f(z)+fu.l)/2) = (wo+wo)/2 = µo.

limx-+xo,y-+o U(x,y) = limz-+zo ({f(z)-fu.l)/2i) = (Wo-Wo)/2i = Uo.

Thus, limx-+xo,y-+o µ(x,y) = µo and limx-+xo,y-+o u(x,y) = Uo.

Conversely, if limx-+xo,y-+o µ(x,y) - = µo and limx-+xo,y-+o u(x,y) = ·uo, then

(by Theorem 1.) µo + iuo =limx-+xo,y-+o µ(x,y) + ilimx-+xo,y-+o u(x,y) =

limz-+zo ((f(z)+fu.l)/2) + limz-+zo ((f(z)-fu.l)/2) = limz-+zo f(z) = Wo.

Also µo - iuo = limx-+xo,y-+o µ(x,y) - ilimx-+xo,y-+o u(x,y) = limz-+zo

((f(z)+fu.l)/2) - limz-+zo ((f(z)-f(z})/2) = limz-+zo f(z) = Wo.

Thus, limz-+zof(z) =wo,

.... - ··-· .. ··-1---· ···-·- -· · ··- --·--· ··- z 1 I 'l- -- - i, since lim

2 3 .:_ -- and· lim :&fl= -1. -

10, For any Zo in the complex plane,

..l.i.m.....· < = ·l-im.. es 'COS r + i -li-m.. t!' sin r = e"' 008 ,..v. + ii"' sin �IIA = -� . ..... W-.0

'l I, a. 1

b. 0

c. -1r/2 + i d. 1

d.l, By contradiction: Suppose zJi.� f(z) # w0• Then there is an. e > O

for which there exists a sequence { zn} such that lzn - z0

I < .!.. but

lf(zn) - wol > c. For this sequence, lim Zn= z but lim f(z ) � w n-oo n-+oo n , '

contrary to hypothesis. ··

. ········--····--·-··-· -------

23. If Zn� =. then for any M>O there exist an integer N such lznl >M for al

n > N. Consider the chordal distance X(Zn,oo) = 2l\!(lzi + 1) < 2/-\!(lz/)

= 2/lznl < 2/M -c e for all n > N. Thus Zn� oo as n � oo is equivalent to

X(Zn,oo) � 0 as n � oo,

24. If limH20f(z) =oo, then for any M>O there exis O > 0 such that lf(z)I >M for

all lz-z01 < 0. Consider X(f(z),oo) = 2/-\!(lf(z)l2 + 1) < 2/-\!(lf(z)l2) = 2/lf(z)I <

2/M -c e for all lz-z01 < 0. Thus limz-nof(z) =oo, is equivalent to

limH""X(f(z),oo) = 0.

25. (a) oo

does not exist.

(b) 3 (c) oo (d) 00 (f) the limit

... -·

EXERCISES 2.3: Analyticity. -

I. Let �z = z - zo so that �z--+ 0 � z--+ z0

• Then

. f(zo + �z) - f(z0)

Irm =L� Az-o �Z

given e > 0, there is a 8 > 0 such that

f(zo + �z) - f(zo) �z - L < e wheneverjzxs - OI < 8 �

f(z) - f(zo) -----...;� - L < swheneverjz - zol < 8 �

Z- Zo

lim f(z) - f(zo) = L. z-zo z - Zo

2 If '( ) f ( z) - f ( zo) · ,• A z =

z-� - f (zo), then A(z) --+ 0 as z--+ zo and

f(zo) + f'(zo)(z - zo) + A(z)(z - z0) = f(z).

3. }i..� f ( z) = }i.� [f( zo) + !' (zo) ( z - z0) + A ( z) ( z - z0) J

= f(zo) + 0 + 0 = f(zo).

J - J o

4.

) )

+

Ii Re(z + !l.z) - Re(z) _ fun Re(!l.z) _ { 1, if ll.z = !l.x

a. tu� o !l.z - az-o !l.z - 0, if tu = ill.y

b. lim Im(z + &) - Im(z) = lim Im(!l.z) = { 0,. � tl.z = �x

a.-o !l.z tu-o tl.z -i, if !l.z = illy

c. Case 1, z = 0.

lim IO+ !l.zl - IOI = lim ./(ll.z 2

+ (!l.y 2

= { ±!, if Llz = !l.xtu-o tl.z Az-o !l.x + i!l.y -i, if !l.z = ±illy

Case 2, z # O.

lim lz + !l.zl - lzl A.s-o !l.z

= Jim V-x_+_/l.x_) + !l._y_2 - Jx2 + y2

Az-o tl.x+i!l.y

= lim (z + ll.z)2 + (y + ll.y)2 - (x2 + rr) Az-0 (!:ix+ itl.y)(./(x + &;)2 + (y + !l.y)2 + v-2-+-y--)-

= lim 2xll.x + (ll.x)2 + 2y!l.y + (ll.y)

2

Az-0 (fl.x + ifl.y)( ./(x + fl.z:)2 + (y + fl.y)2 + ,V-2-+-y--.).

_ { ,/ /:

3

y2 , if !:..z = !::.x, z :f: 0

- . 'Y , if !l.z = i!l.y, z # 0 i,vx""+y""

5. Rule 5: (/ ± g)'(.zo) = fun (/ ± g)(Zo + � ) - (! ± g)(Zo) A.s-0 z

_ fun [/(Zo + ll.z) - f(Zo) ± g(zo + tl.z) - g(Zo)] Az-O liz liz

- /'(Zo) ± g'(zo)

Rule 7: (/g)'(Zo) = lim fg(Zo + !l.z)- fg(Zo) As...o !l.z

J_-\\

i zo ixz A

�z

- •

0

[ A -

= lim { f(zo + �z)g(zo + �z) - f(zo + �z)g(zo)

.6.z-o �z

f(zo + �z)g(zo) - f(zo)g(zo)}

+ �z

= 1.m

{!( + A ) [g(zo + �z) - g(zo)].6.z-o uz

+ 9 ( zo ) [f(zo + �z) - f(zo)]}

= f(zo)g'(zo) + g(zo)f'(zo)

6. Let n > 0 be an integer.

Then d z-n =

d ( 1 ) - - =

-nzn- I

(using Rule 8) = -nz-n-1

dz dz z" z2n

7. a. 18z2 + 16z + i

b. -12z(z2 - 3i)-7

-iz4 + (2 + 27i)z2 + 21rz + 18

c. ( iz3 + 2z + 1r )2

-(z + 2)2(5z2 + (16 + i)z - 3 + Si)

d. -------(z-

2-+-i-z-+-1

-)5-------

e. 24i(z3 - 1)3(z2 + iz)99(53z4 + 28iz3

- 50z - 25i)

8. Let z = z0 + �z. Then

1.Im

lf(z) - f(zo)I _-I1.m

f(zo + �z) - f(zo) I--

If'( z

)Iz-zo lz - zol .6.z-o �z

0•

lim arg[f(z) - f(zo)] - arg(z - z ) - lim arg [f(z) - f(zo)] z-+zo z-zo z - zo

. f(zo + �z) - f(zo)]arg 1Im

.6.z-+0 uz

'1.- l 1.

arg[f'(zo)]

Zo - Zo =

+ + =

9. a.. 2 - 3i

b. ±i

-1 ± i"'15 c.

2

1

d. 2'1

1:_ IZo + ll.zl2 - l.zol2

10.

al.l-l-ll

o az

_ fun (Zo + az)(zo + az}- ZoZo

4%-+0 ll.z

_ Jim (zo+�ZZo+L\z)={Zo+Zo ifaz=dz az-o �z if l::iz ill.y

H Zo = 0, then the difference quotient is

lim (0 0 ll.z) 0. � ....o

11. a.. nowhere analytic

b. nowhere analytic

c. analytic except a.t z = 5

d. everywhere analytic

e. nowhere analytic

f. analytic except at z = 0

g. nowhere analytic

h. nowhere analytic

12. The case when n = 1 is trivial. Assume that the result holds for all

positive integers less than or equal to n and define

Q(z) = P(z)(z-z.a+1)· Since Q'(z) = P'(z)(z-:tn+i)+P(z), it follows

that

Q'(z) P'(z) 1 1 1 1

Q(z) = P(z) + z- Z.+1 - z - Z1 + z - Z2 + ... + z - Zn+t

2 - I '3

13. a, b, d, f, and g are always true

14 lim f ( z) = lim [f ( z) - f ( zo)] / ( z - zo) = !' ( zo)

· z-zo g(z) z-zo [g(z) - g(z0)]/(z - z0) g'(zo)

15 . �5

16. Any point on the line through z1 and z2 has the form

z = -� + iv'3 (� - t), t real (see Section 1.3, Exercise 18). However,

f(z2) - f(zi) = 0 but f'(w) = 3w2 # 0 on the line in question.

17. F'(zo) = f(zo)(gh)'(zo) + f'(zo)gh(zo)

= f(zo)[g(zo)h'(zo) + g'(zo)h(zo)] + f'(zo)g(zo)h(zo)

= f'(zo)g(zo)h(zo) + f(zo)g'(zo)h(zo) + f (zo)g(zo)h'(zo)

EXERCISES 2.4: The Cauchy-Riemann Equations

OU av 1. a. ax = 1 # oy = -1

OU av b. ax = 1 # ay = 0

OU av c. oy = 2 # - ax = 1

au 2 2 av au av au av

2. -ax = 3x + 3y - 3 = -ay , but -ay = 6xy = -ax. Therefore -ay = -•axonly when x = 0 or y = 0. This means h is differentiable on the axes

but h is nowhere analytic since lines are not open sets in the complex

plane.

3 . aux = 6 x + 2 = oayv' auy = -6y = - oaxv.

Si ce t

hese part

.a

l der

.vat

.vesm i i i

exist and are continuous for all z and y, g is analytic. g can be written

as g( z) = 3z2 + 2z - 1.

J. - I�

4_ IJu = lim u(ax,O) - u(O,O) = lim _!_ = O IJx � o ll.x � t:J,.z

lJu = lim u(0,6.y)- u(O,O) = Jim _!_ = O. l}y At,-0 fl.y Ay-0 fl.y

Similarly : = O and : = O.

However, when 6.z-+ 0 through real values (!:,,.z = 6.x)

lim /(0 + ax) - /(0) = 0, � -+O fl.z

while along the real line y = x (ll.z = ll.z + i� )

lirn /(0 + ll.z) - /(0) _

lim (Az)'''(Az)5/3+t(Az)5/3(Az)'/3

2(� A-f _

A.a-+0 11z Az-+O ll.x(l + i) 1

- 2·

Therefore f is not differentiable a.t z = 0.

5. : = 2F-r[zcos(2zy)-ysin(2zy)J =: : = -2F-rr,cos(2z,) + zsin(2zy)J = -:

f is entire because these first partials exist and are continuous for all

x and y.

2es2-,2 (x + iy)[cos(2zy) + i sin(2zy )]

- 2e<r-,2)ei2sfl(x + iy)

- 2ze.e2

(This derivative could have been obtained directly, since /(z) = ez2 .)

6. z = re18 ==> x = rcosfJ and y = rsinfJ and

f(z) = u(z(r,IJ),y(r,11)) + iv(z(r,6),y(r,6))

8u = ouoz + IJuoy = ou cosB + 8u sinlJ

fJr IJz or 8y 8r IJx lJy "'2-\ S"'

= = - =

i

+

=

Similar applications of the chain rule yield

OU OU . OU oB = fJx(-rsmB) + fJyrcosB

ov fJv ov . - = -cosB+-smB or ox oy

ov av . av-ao = -(-rsmO) +-rcosB

ax oy

Replace the partial derivatives on the right sides of the equations for

:� and :� by their Cauchy- Riemann counterparts to obtain:

-OU ov

cos() - ov .

s B = -1 -ov

or -

- m

r {)()oy ox

-fJv

= - Bu

cosB+ Bu .

s B = --1 •ou

or -

- m

r 80oy ox

7. Let h(z) = f(z) - g(z). Then his analytic in D and h'(z) = 0 so his a constant function.

h(z) = c = f(z) - g(z) � f(z) = g(z) + c

. ou fJu fJv

8. u(x,y)=cmD�-=Oand-=0=--. Hence

f'(z) = :: + i:: ox fJy ox

= 0 so f is constant in D.

9. By contradiction. If f is analytic in a domain D then v( x, y) = 0 (a

constant)=> f is constant (by condition 8) => u is constant. (However, there is no open set in which u(x, y) = jz2

- z] is constant).

. av av OU 10. lmf(z) 0 m D ===} - = 0 � - 0

. ax ay ax

===} !'( z ) = Box

u + . ofJxv =

O � f i.s constant i

.n

D .

1 � - 11. Ref(z) =

2[J(z) f(z)] is real valued and analytic if both f and f

are analytic. Hence Ref(z) is constant by Exercise 10. It follows that

f(z) is constant by Exercise 8.

lJy

1 81! 1 1 olf 1 2 2 av

8y'

- = m 0% = -m -

= -o:e- - --

(zo,,o)

fJy - - 0% - ax - l)y - ox • ox - .

+ = + =

12. 1/(z)I constant in D ==> l/(z)l2 = u2 + v2 is constant in D. Hu = 0

or v = 0 in D, then f is constant by Exercises 8 and 10. Otherwise,

a1112

ox - a1112

_

8y

ou av 2u-ox+2v-ox=0

lJu 8v ov 8u 2u-oy 2v- -2u-8x 2v-ox 0

==> 2 2

-v- - -u- = 0 = (u +v )-- 2 IJx 2 lJy 8x

8v 8u ov ou==> ax =O===> ax= oy = 8y =0

� f'(z) =:: +i:: = 0

==> f is constant in D.

13. f/(z)I is analytic and real-valued, so the result follows from Exercises

10 and 12.

14. If the line is vertical then Re/(z) is constant a.nd this reduces to Prob-

lem 8. H the line is not vertical, then v(x,y) = mu(x,y) + b, and

8v /Ju 8v -ox=m-ax=m-

It follows that

OV IJy

au -8y = -m-

lJv

2 0V lJy.

8v _ 0 _ 8u _ IJv _ ou a.nd /'(z) _ ou + /Jv _ 0

Hence f(z) is constant.

auav 8ulJv 15. J(zo,Yo)

811 8y 0%

= [::czo,Yof + [:(zo,Yo>J2 = l/'(Zo)l2

(':ul-si,n,g Equation (1))

2 2 +

)

=

16. aJ 8f8x 8f8y

a. ae = ax ae + aJae

= (au+ iav � + (au+ iav) � ax ax 2 8y 8y 2i

= H::+ :) +H:> :;) aJ aJ ax BJ ay

a,,, = ax a,,, + By a,,, = (au+ iav) �+(au + iav) (-1)

ax ax 2 ay ay 2i

= � (au _ av) + i (au + av) 2 ax ay 2 ay ax

aJ 1 (au fJv) 1 (au av) b. a,,, = 0 <=> 0 = fJx - 8y and O = ay ax

au fJv fJu fJv

<=> -ax = -

ay and {)y - - fJx

EXERCISES 2.5: Harmonic Functions

a2u a2u

1. a. u(x, y) = x2 - y

2 + 2x + 1, ax2 = 2 = - ay2 ==} �u = 0

a2v fJ2v v(x, y) = 2xy + 2y, fJx2 = 0 = - {)y2 ==} �v = 0

x a2u 2x(x2 - 3y2) a2u

b.u(x,y)= x2

+y 2' a2

x = (2

x +y 2)3 =-a2

y ==}�u=O

y a2v -2y(3x2 - y

2 a2vv(x,y)=-

x 2

+y2' a

x2= (

x 2

+y2)3 =-a2

y ==}�v=O

fJ2u c. u(x, y) = ex cosy, ax2

a2u ex cosy= - fJy2

==} �u = 0

fJ2v fJ2v v(x,y) = exsiny, fJx2 = exsiny = -ay2 ==> �v = 0

2. h(x,y) = ax2 + bxy - ay2

-= =-�

= = - oz =

3. a. u = Re(-iz), v = -x + a, where a is a constant

b. u = Re(-ie"), v =-�cosy+ a

c. u = R. e (-iz2 - 1·z - z) , v = - 1 (z2 - 112) - ( x + y ) + a

2 2 d. It is straightforward to verify that �u = 0.

a8ux= cosxcoshy =

lJv 8y

=> v(x,y) = j cosxcoshydy = cosxsinhy + tf,(x)

: = sin:i:sinhy = -: = sinzsinhy + tp'(z) :::> tp(z) = a

Thus, v(x,y) = cosxsinhy + a.

e. It is straightforward to verify that � u = 0. 8u x 8v

ox z2 + y2 {Jy

v(x,y) = j

x dy = tan-1 (

) + tf,(x)1

x2 +y2 x OU Y 8v 11 ,

01J z2 + y2 z2 + y2 - t/J (x) => tf,(x) = a

Thus, v(z,y) = tan-1

(!)+a.

f. u = Re (-iez2) , v = -er-,r cos(2zy) + a.

4. Suppose v and w are both harmonic conjugates of u, and consider

4,(z,y) = w(x,y)-v(x,y). Then (using the Cauchy-Riemann equations for v and w),

::=Z-!;=-:-(-:)=o

and similarly :: = O. Hence iKz, y) = a, from which it follows that

w(x,y) = v(x,y) + a.

5. H f(z) = u(z,y) + iv(x,y) is analytic then -i/(z) = v(x,y) - iu(x,y)

is analytic. Thus -u is a harmonic conjugate of e.

2 - I 't

n - -- +-- = -(-rsm8) + -rcos8

8 8

+ +

6. Since f(z) = u+iv is analytic,! [f(z)]2

= !(u2 -v2) + iuv is analytic. 2 2

Thus uv = Im� [J(z)]2is harmonic.

7. </>(x, y) = x + 1

8. a. Yes, because � (u + v) = �u + �v = 0.

b. No. Take u = x, v = x2 - y2 as an example.

c. Yes, because �(ux) = Uxxx + Uxyy = Uxxx + Uyyx

= -(�u) = -(0) = 0. 8x 8x

9. ¢,(x, y) = xy - 1 (this is Im Gz2 - i))

10. Let x = rcos8 and y = rsinB.

8</> _ 8</> ax + 8</> 8y = 8</> cos B + 8</> sin 8 ar 8x 8r ay 8r 8x 8y

82 </> 82 </> ax 82 </> 8y

8r2 - 8x2 8r cos8 + 8y8x 8r cosO

82</> 8x . 82</> 8y .

+ 8x8y 8r sin 8 + 8y2 ar sm

8

a2 </> 2 a2 </> • a2 </> . 2

- Bx2 cos 8 + ByBx2sm8cos8 + By2 sm 8

8</> 8</> ax 8</> 8y 8¢ . 8</>

au 8x ao 8y 88 ax ay

82 </> 82 </> ax . 82 </> 8y . 8</> 802 - 8x2 80 (-r sin B) + 8y8x 88 (-r sin fJ) + 8x (-r cos fJ)

82</> 8x

+ axay 88

(r cos 0)

82¢ 8y By2 88

(r cos 8)

8</> . By (-r sin 8)

28

2 </> 2 • 2 8

2 </> ( 2 • ) 8

</> 2 2- axz r sm 8 + ByBx -2r sin 8 cos() + By2 r cos 8

+ o</> (-r cos 0) + B</> (-r sin 0). ox 8y

1

2

=" c · ·

Combining these partial derivatives, one gets

2 3 2 2

11. lm/(z)=y- z21+/ y2

=0=>yx +y -y=y(x +y -1)=0.

The points satisfying x2 + y2 - 1 = 0 lie on the circle f z I = 1. The

points (other than z = 0) satisfying y = 0 lie on the real axis.

12. f(z) = z" = rn(cos 9 + i sin 9)" = r"(cos n9 + i sin n9) �

Ref( z) = r" cos n6 and Im f(z) = r" sin n8 a.re harmonic since f is

analytic.

13. 4,(x,y) = Imz4 = r4sin49 = -4xy3 + 4x3y

14. Let <l>(x, y) = ln 1/(z)J = 2

In(u2 + v2

)

OU av 8</, = 8</> OU + 8</> lJv = ua; + va;

8x 8u 8x av 8x u2 + v2

A similar calculation yields :�- By applying Laplace's equation and

a2 </> . IP</>the Cauchy-Riemann equations of u and v to {Jz'l + 8y , the sum

15.

simplifies to reveal that 6.</> = 0.

Consider cp(z) - Re(A O B -n) . - z + z + which ts harmonic for I �lzl�2

Co�stder the polar form for z. z = and select n=3 to agree with. the

cos1�e argument. cp(re19

) = Ar3Re(e130) + Br-3R ( -i30) C

<p(re

10) = Ar

3 cos30 + Br-

3 cos30 + C = (Ar3

.

Br e e + .

s30 .+ -3)co + c

r=l=:> (A+B)cos30 + C = o =:>A+ B = O, c = O.

r=2=:=> (A *8+B/8)cos30 = 5cos30. A = 40/63 B - -40/63 cp(

re18)

= (40/63

)(r3 -�-

3 )cos30=(40/63)Re

' z3 -

-\

( -z

16. �e, !f) = �3

In lzl - 1 or �z, !f) = In Ii I are two possibilities.

:t - 21

+

+

+ +

17. a. </>(x,y) = Re(z2 + 5z + 1) = x2 - y2 + 5x + 1

- ( z2

) 2x(x2+4y+y

2)

b. </>(x, y) = 2Re z + 2i = x2 + y2 + 4y + 4

18. Let u = <Px, v = -<Py· Then

au av ax = <Pxx = -</>yy = ay

au_</> av ay - xy - ax

19. cos

20 = (l/2)cos20 + 1/2 = cp(z) = ARe(r-2e-i20)n + B = Ar-2cos20 + B. In

the limit as r�oo cp(z) = 1h =:> B = 1h. On the circlelzle I, r = 1. =:> A = 1/2.

cp(z) = {l/2)r-2cos20 + 1/2 = Re [11(2z2

)] + 1/2.

\ av au [Yau 2._o. In order that ay =ax, let v(x,y) = lo ax (x, TJ)dTJ 1/J(x). Then

av t' a2u

ax - lo ax2 (x, 'T/ )dTJ + 1/J'(x)

t" fPu - - lo ay2 ( x, TJ )dTJ 1/J' ( x) (because u is harmonic)

au OU I - - ay (x, y) oy (x,0) VJ (x).

Inorder t h at av

= - au .

t must be true that VJ '(

x) = - au

(x, 0).

Thus,

ax oy, i oy

r au

and

VJ(x) = - lo ay ((, O)d( + a

fY au fx au v(x, y) = lo ax (x, TJ)dTJ - lo ay ((, O)d( + a.

2 f. It is easily verified that u = In lzl satisfies Laplace's equation on

C\ {O} and that u+iv = ln lzl+iArg(z) satisfies the Cauchy-Riemann

equations on the domain D = C\ { nonpositive real axis}, so that

Arg (z) is a harmonic conjugate of u on D. By Problem 4, any harmonic

conjugate of u has to be of the form Arg( z) + a in D. It is impossible to

have a. harmonic ..conjugate of this form that is continuous on C \ { 0}.

EXERCISES 2.6: Steady-State Temperature as a Harmonic Func• tion.

1. a.. b.

100° 100°

800 80° scr

80° 60°

Mr 6()° 40° 60°

40° 40° 2()° 40°

20° 20° ()° 20, ()°

()°

c. d.

2(1' 40° 60° 8<>° 500 I 00° 50°

100°

goo

60° oo 40°

2()°

er er

e.

80°

100°

..._ 80°

60°........__

40°

20°

oo

._ .

oo

60°

40°

20"

2. This does not violate the maximum principle.

oo

3. This does not violate the maximum principle.

100°

100°

2

Exercises 2. 7

1. f(z) = z

2 + c where c is a real constant.

St= (I + "(1-4c))/2, s2 = (1 - "(1-4c))/2

Only s2 is an attractor for -3/4 < c < 1.4.

2. f(<;) =<;and f'(<;) > 1 Therefore we can pick a real number p between 1

and If(<;) I such that lf(z) - <;I= plz-<;I for all z in a sufficiently small disk

around q, If any point z0 in this disk is the seed for an orbit z1 = f(zo), z2 =

F(z.), ... Zn= f(Zn-t), then we have lzn-<;1 � plzn-t-<;I � ... � pnlZn-t-<;I.

Because p>l, the point Zn moves away from <; until the

magnitude of the derivative becomes 1 or less. The orbit is out of the disk.

3. (a) Fixed points are SI = i, s2 = -i. Both are repellors.

(b) Fixed points are St= 1/2, s2 = -1/2, s3 = -1. Fixed points s1 and SJ are repellors, but fixed point s2 is an attractor.

4. zo = ei2

rra with a an irrational real number. Zn = ei2

rra "n. Because lznl = 1,

the trajectory will follow the unit circle. If iterations p and q coincide,

21ta2P - 21ta2q = 21ta(2P - 2q) = 21tk for some integer k. But because (2P -

2q) is an integer that can be represented by m, the equation 21tam=21tk is

satisfied only if k=am or a= k/m. Because a is irrational it cannot be

represented by a rational number and no iterations repeat.

5. Fixed points are SI= -1/2 + i"5/2 (an attractor) and s2 = -1/2 - i"5/2 (a

repellor).

6. f(z) = z2. The seed is zo. ZI = zl, z2 = z04, ... Zn= zoA(2n). To have an n

cycle Zn= zo = zoA(2n). Or Znlzo = zoA(2n -1) = 1 = ei2

rr. Solving gives

Zo = eA(i27t/(2n- l )).

7. The cycle is 4. 2\27tlp) = 21t mod p => 24

= 1 mod p. p=3,5, 15. 3 will

give repeated cycles of length 2. 5 and 15 will give the desired cycles of

length 4.

8. Student Matlab: n=lOO;c=.253; zo=O;y(l)=zo;

for k= 1 :n-1,y(k+ 1 )=y(k)A2+c;end

plot(y)

9. If lal s; 1 the whole complex plane is the filled Julia set. If lal 2:: I the

origin is the filled Julia set.

10. f(z) = z - F(z)/F'(z). f(s) = s - F(s)/F'(s) = s=> F(s)/F'(s) = 0 =>F(s)=O

with the possible exception of the points where F'(s) = 0.

f'(z) = 1 - F'(z)/F'(z) + F(z)F"(z)/(F'(z))2 = F(z)F"(z)/(F'(z))2

f(s) = F(s)F"(s)/(F'(s))2 = 0 where F'(s) #0 and every zero of F(z) is an

attractor as long as F'(s) # 0.

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Fundamentals of Complex Analysis with Applications to Engineering

Edward B. Saff and Arthur

Arthur David Snider Fundamentals of complex

Fundamentals of Complex Analysis with Applications

Fundamentals of Complex Analysis

Edward B. Saff

Saff & Snider

Saff E.B