Fundamentals of complex analysis with applications to engineering and science 3rd edition saff sl by Kroenke

Fundamentals of Complex Analysis with Applications to Engineering and Science 3rd Edition Saff SOLUTIONS MANUAL Full download: https://testbanklive.com/download/fundamentals-of-complexanalysis-with-applications-to-engineering-and-science-3rd-editionsaff-solutions-manual/

CHAPTER 2: Analytic Functions EXERCISES 2.1: Functions of a Complex Variable

to=

(3z2

31,2 + 5% + 1) + i(6zt, +Sy+ 1)

= z2 : y2 + i ( 1

! y2)

-y + 1

c. to=-= 2 + ,----z - i z + (y - 1 )2 z 2 + (y - 1 )2 d 2%2 - 2y2 + 3 . 4zy • tD = J(z-1)2 + y2 + a J(z-1)2 + y2

= e3z: cos 3y + ii- sin 3y w = (ec + e-c) cosy+ i(� - e-c) siny = 2 cosh z cosy + i2 sinh z sin y

e. w

a. C b. C \ {O} c.

c \ {i,-i}

d. C \ {1} e. C f. 3.

Rew> 5 b. Imw > 0 a.

lwl > 1

d. The intersection of 4.

fwl < 2 and

-,r

< Arg w < r/2

a. Taking IJ from O to 2r, the points % rei' traveise the circle lzl r exactly once in the oounterclockwise direction. For the

., = !.e-il traverse the circle re• r lw I = .!:.r exactly once in the clockwise direction, hence the mapping . 1s onto. � 1 � . 1 b. For z = re on the ray Argz = B , w = s on /J- = e-

same values of 9 the points

to=

•

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

-----·---------·-----------·-

...

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.

a. domain: C range: C \ {O}

1 1 b. f(-z)=Cz=-=•

c. circle

f(z)

lw I = e

d. ray Arg w =

1r /

e. infinite sector O � Arg w � 6.

1r /

1(!)=!(!+)=!(z+.!.)=J(z) 2 lz 2 Z-

i (i = i (re;e + r:;s) -

8 b. For z = 8 on the unit circle lzl = 1, J(z) = + 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)

(r cos O + (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. 7.

a.::j

-2

J. - l.

-l

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

10.

a. translate by i, rotate 1r /4

b. reduce by 1/2, rotate

c. translate by i, reduce by 1/2

d. reduce by 1/2, rotate translate by i

1r /4,

-l

-o.s

o.s

-1 -o.1s-o.s-o.2s

'). - L,(

0.25

o.s o.

11.

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

b.magnify by 2, rotate -1r/2

c. translate by -3, magnify by 2

d. magnify by 2,· rotate -r /2, translate by -3

-1

12. Let a =

pe•, F(z) = = az + b.

pz, G(z)

e•z,

a.nd H(z)

= z + b.

Then

H(G(F(z)))

w = u + iv=z2 = (1 + iy)2 = 1 -y2+ i2y u = 1-y2, v = 2y �y v/2 � u 1- v 2/4 a parabola in thew-plane. 2 (b) w = u + iv=z2 = (x + iy)2 = (x + i!x)2 =x2 - l/x + 2i

13. (a)

u = x2 - l/x2, 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)ei9 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 l2 + 1), Xi= (xsincp+ycoscp)/(lz l2 + 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. 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, ,!3) = (-x3, x2, X1) so that Cx1, Xi, XJ).is obtained by a 90° counterclockwise rotation about the x2 axis.

15.

.!2,

2-5

1.6.

17.

w = (1 - iz)/(1 + iz) = (1- ix + y)/(1 + ix - y) == (1-lzl2 + i2x)/(1- 2y + lzi2) lwi2 = (1 + 2y + lzl2)/(l- 2y + lzl2). (!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. 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

i, - -i, 16, and 32i . 1

1. The first five terms are, respectively, , 4 sequence converges to O in a spiral-like fashion.

The

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

-o.

2. 2i, -4, -Bi, 16, 32i; divergent because terms grow in modulus without bound. 3

;;:i -

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. 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 if l.zol > 1, then 1�1-+ oo as

1%31-+ 0 as n-+ oo. On the other hand, n -+ oo so z: diverges. 1,

a. converges to O b. does not converge c. converges to

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,

> 0, choose 6 = 1 : e' that lzl > 1 - 6 and Given e

Whenever O

< lz -

(-i)I < 6 notice

1;-i1 = 1(-;) (i+z)j = 1!11z- (-i)I < (A) 6 = e �-7

10. Given that f and g aie mntinuoua a.t Zo, ..l_i.m ...., /(z) ±g(z) = lim /(z) :l:. lim g(z) = /(Zo) ±g(.ro)

.......

........

==> /(z) ± g(z) is

continuous at Zo·

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

..1-uo

= /(Zo)g(Zo)

====> /(z)g(z) is continuous at Zo·

f(z) }i� f(z) f(zo) . r ( ) = -(-, provided g(z0) z-zo g z im g z g zo z-zo f(z) . . ==> g( z) is continuous at z0• .

lim -) =

a. -Si

i= 0

7 2i . c.

d. -1/2

e. 2zo f. I�.

4v'2

Clearly Argz is discontinuous at z = 0. Let a> 0 be any real number and consider the sequence Zn=

For each n,

-a - i/n

-1r

n = 1, 2, ... , which converges to - a.

< Arg Zn <

13, z-zo lim f ( z) exists for all z

-1r /2,

but Arg (-a) =

1r.

i= -1; f is continuous for all z i= 0, -1;

f has

a removable discontinuity at z = 0. 1 �. let z0 be any complex number. whenever lz - zol < S,

Given e

> 0 choose S = c. Then

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 Then, whenever lz - zo I < & a. lf(z) - f(zo)I = lf(z) - f(zo)I

< e whenever lz - zc,I < S.

= 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

> 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. · 16 Gmn

ti. No: Observe that although -

J (!)

-+ 0 and - -+ 0 as n --+ oo,

- 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 I 'l- -- - i, since lim 2 a-1

.-1

+ 31/

1 .:_ -- and· lim :&fl= -1. e-1

.--1

10, For any Zo in the complex plane,

..l.i.m .....· <

'l I,

= ·l-im.. es 'COS r + i -li-m.. .....

t!' sin r

W-.0

= e"' 008 ,..v. + ii"' sin � = -� . IIA

a. 1

b. 0 c. -1r/2

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, n-oo lim Zn= z 0 but lim f(z ) � w 0 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. (b) 3 (c) oo (d) 00 (a) oo (f) the limit does not exist.

25.

...

········--····--·-··-·

-·

EXERCISES 2.3: Analyticity. I. Let �z = z - zo so that �z--+ 0 � z--+ z0• Then

I.rm f(zo

+ �z) - f(z0) �Z

Az-o

given

=L�

> 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 A'( z ) =

f ( z) - f ( zo) -· f, (zo), z-�

then A(z)

0 as z--+ z and o

+ f'(zo)(z - zo) + A(z)(z - z0) = f(z). }i..� f ( z) = }i.� [f( zo) + !' ( zo) ( z - z0) + A ( z) ( z - z0) J = f(zo) + 0 + 0 = f(zo). f(zo)

--+

J -J

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) !l.z

a.-o

c. Case 1, z

Case 2, z

� tl.z -i, if !l.z

0,.

= �x

= illy

= 0. tl.z

tu-o

lim Im(!l.z) tl.z

tu-o

IO+ !l.zl - IOI =

lim

2 lim ./(ll.z) Az-o

+ (!l.y )2 = {

!l.x + i!l.y

±!,

if Llz = !l.x

-i,

if !l.z = ±illy

# O.

lz + !l.zl - lzl

lim A.s-o

!l.z Jim V-x_+_/l.x_)

!l._y_2 - Jx2 + y2

tl.x+i!l.y lim (z + ll.z)2 + (y + ll.y)2 - (x2 + Az-0 (!:ix+ itl.y)(./(x + &;)2 + (y + !l.y)2 + Az-o

= =

lim Az-0

rr)

2xll.x + (ll.x)2 + 2y!l.y + (ll.y)2 (fl.x + ifl.y)( ./(x + fl.z:)2 + (y + fl.y)2 + ,V-2-+-y--.).

_ { ,/ /: 3

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

if !l.z

+ y2 ,

. 'Y , i,vx""+y""

= i!l.y, z # 0

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

fun

v-2-+-y--)-

[/(Zo + ll.z) liz

Az-O

�) - (! ± g)(Zo) z

- f(Zo) ± g(zo + tl.z) - g(Zo)] liz

- /'(Zo) ± g'(zo)

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

J_-\\

= lim { f(zo

+ �z)g(zo + �z) - f(zo + �z)g(zo) �z

.6.z-o

f(zo

+ . = 1im

{!(

+ zo

.6.z-o

+ �z)g(zo) - f(zo)g(zo)}

+ 9 ( zo )

�z

)

[g(zo

+ �z) - g(zo)]

ixz [f(zo

+ �z) - f(zo)]} �z

= f(zo)g'(zo) + g(zo)f'(zo) 6. Let n

> 0 be d

Then

an integer. d ( 1 ) -nzn- I z-n = - - = (using Rule 8) = -nz-n-1 •

dz 7.

a. 18z2

z2n

+ 16z + i

b. -12z(z2

-iz4

3i)-7

+ (2 + 27i)z2 + 21rz + 18 ( iz3

+ 2z + 1r )2

-(z + 2)2(5z2 + (16 + i)z - 3 + Si) (z2 + iz + 1)

d. --------------5------e. 24i(z3

8. Let z = z0

1)3(z2

+ �z.

+ iz)99(53z4 + 28iz3 - 50z - 25i)

Then

1.Im lf(z) - f(zo)I _ -I 1. m f(zo

z-zo

lz - zol

+ �z) - f(zo) �z

.6.z-o

lim arg[f(z) - f(zo)] - arg(z - z 0)

z-+zo

arg [ 1 Im .6.z-+0

f(zo + �z) - f(zo)] A

I-

If'( z )I 0•

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

z-zo -

'1.- l

arg[f'(zo)]

a.. 2 - 3i b. ±i -1 ± i"'15 c. 2 1 d. 2'1 1:_

10. al.l-l-llo

2 2 IZo + ll.zl a - l.zol

fun (Zo + az)(zo + az}- ZoZo ll.z

4%-+0

Jim az-o

(zo+�ZZo+L\z)={Zo+Zo Zo - Zo �z

ifaz=dz if l::iz ill.y

H Zo = 0, then the difference quotient is lim ( 0 + 0 + ll.z) �....o 11.

= 0.

a.. nowhere analytic b. nowhere analytic c. analytic except a.t z

d. everywhere analytic

e. nowhere analytic f. analytic except at z

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) Q(z)

P'(z) P(z)

+ z- Z.+1 2 - I '3

1 -

z - Z1

1 + z - Z2

+ ... + z - Zn+t

13. a, b, d, f, and g are always true 14

lim f ( z) · z-zo

g(z)

lim [f ( z) - f ( zo)] / ( z - zo)

= !' ( zo)

[g(z) - g(z0)]/(z - z0)

g'(zo)

z-zo

15 . �5 16. Any point on the line through z1 and z2 has the form

+ iv'3 (� - t), t real (see Section 1.3, f(z2) - f(zi) = 0 but f'(w) = 3w2 # 0 on the

z = -�

Exercise 18). However, line in question.

+ 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)

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

EXERCISES 2.4: The Cauchy-Riemann Equations

av ax = 1 # oy = -1 OU av b. ax = 1 # ay = 0 OU av c. oy = 2 # - ax = 1 au av au av - = 3x + 3y - 3 = -, but - = 6xy = -. ax ay ay ax OU

au = -• av ax ay

Therefore -

only 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 = - ox.

Si h . ld . . ce t ese part a er vat ves m 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

�o

l}y

At,-0

lJu = lim

=O u(0,6.y)- u(O,O) = Jim _!_ = O. ll.x

lim _!_ � t:J,.z

fl.y

Ay-0

lim u(ax,O) - u(O,O)

IJx

fl.y

Similarly : = O and : = O. However, when 6.z-+ 0 through real values (!:,,.z

= 6.x)

lim /(0 + ax) - /(0) = 0, fl.z

�-+O

while along the real line y

= x (ll.z = ll.z + i�) (Az)'''(Az)5/3+t(Az)5/3(Az)'/3

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

lim

11z

Az-+O

A.a-+0

2(�A-f

ll.x(l

+ i)

- 2· Therefore

5. :

f is not differentiable a.t

2F-r[zcos(2zy)-ysin(2zy)J

= 0.

: = -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) = 6. z = re18

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

==> x

fJr

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

or 8y 8r "'2-\ S"'

IJx

lJy

ez2 .)

Similar applications of the chain rule yield OU OU . oB = fJx(-rsmB)

+ fJyrcosB

ov fJv ov . - = -cosB+-smB or ox oy ov av . av - = -(-rsmO) +-rcosB ax oy

Replace the partial derivatives on the right sides of the equations for :� and :� by their Cauchy- Riemann counterparts to obtain: 1 ov ov ov . OU - = cos() - s B = --

fJv

- m

oy ox Bu . Bu = - cosB+ s B

r {)() 1 ou

= --•

- m

r 80

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 ox fJy ox

f'(z) = ::

i::

= 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). .

= 0 m D ===} == 0 � - =0 ax ay ax O f . B u '( ) . fJ v � is constant i. n D . ===} ! z = ox + i ox =

10. lmf(z) .

11. Ref(z) =

[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.

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, 2u-+2v-=0

ox -

a1112

av ox

ou ox

a1112

ov 8u = -2u- + 2v- = 0 8x ox lJy

lJu 8v 2u- + 2v-

81! 1 2

olf 1 2

- -u- = 0 = (u +v )-==> -v2 IJx 2 lJy 8x 8v 8u ov ou ==> ax =O===> ax= oy = 8y =0 ďż˝ f'(z)

=:: +i:: =

==> f is constant in

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

8v /Ju 8v -=m-=m8y' OV lJv 0% IJy = m8y = -m-

ax au

0V = -m2 lJy.

It follows that

8v _ 0 _ 8u _ IJv _ fJy - - 0% - ax -

l)y

a.nd /'(z) _

Hence f(z) is constant.

15. J(zo,Yo)

auav 8ulJv =-811 8y 0% o:e

(zo,,o)

[::czo,Yof + [:(zo,Yo>J2

ou + /Jv _

- ox â&#x20AC;˘ ox - 0.

= l/'(Zo)l2

':l-,,

(using Equation (1))

16.

aJ 8f8x 8f8y a. ae = ax ae + aJae = (au+ iav � + (au+ iav) � ax ax 2 8y 8y 2i =

a,,

H::+ :) +H:> :;) aJ ax

BJ ay

= ax a,, + By a,,

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

b. a,,, = 0 <=> 0 =

+i 2

+ iav) ay

(au ay

(-1) 2i

av) ax

(au fJv) 1 (au av) fJx - 8y and O = 2 ay + ax

au fJv fJu fJv <=> ax = ay and {)y - - fJx

EXERCISES 2.5: Harmonic Functions 1.

au 2

ax2 = 2 = - ay2 ==} �u = 0 a2v fJ2v v(x, y) = 2xy + 2y, fJx2 = 0 = - {)y2 ==} �v = 0 x a2u 2x(x2 - 3y2) a2 u b.u(x,y)= x 2 +y 2' a2 x = (2 x +y 2)3 =-a2 y ==}�u=O a. u(x, y) = x - y

+ 2x + 1,

av 2

-2y(3x2

y2 )

av 2

v(x,y)=- x 2 +y 2' a x 2= ( x 2 +y 2)3 =-a2 y ==}�v=O fJ2u a2u c. u(x, y) = ex cosy, ax2 = ex cosy= - fJy2 ==} �u = 0 v(x,y)

= exsiny,

2. h(x,y) = ax2

fJ2v fJx2 = exsiny

+ bxy - ay2

fJ2v -ay2 ==> �v

a. u

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

b. u = Re(-ie"), v =-�cosy+ a 2 · ) 1( 2 2) ( ) = R. e (-i 2z - 1z - z , v = - 2 z - 11 - x + y + a It is straightforward to verify that �u = 0. 8u lJv ax= cosxcoshy = 8y

c. u

= j cosxcoshydy = cosxsinhy + tf,(x) : = sin:i:sinhy = -: sinzsinhy + tp'(z) :::> tp(z) = a Thus, v(x,y) = cosxsinhy + a. 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) => v(x,y)

x2 OU

01J

x 11

= z2 + y2 = - oz = z2 + y2

, - t/J ( x)

=> tf,(x)

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

2 6. Since f(z) = u+iv is analytic,! [f(z)] = !(u2 -v2) + iuv is analytic.

2 Thus uv = Im� [J(z)]2is harmonic.

7. </>(x, y) 8.

=x+1

a. Yes, because �(u

+ v) = �u + �v = 0.

b. No. Take u = x, v = x2 c. Yes, because �(ux)

y2 as an example.

= Uxxx + Uxyy = Uxxx + Uyyx 8

= -(�u) 8x

9. ¢,(x, y) = xy - 1 (this is Im 10. Let x

= rcos8 and

8</> _ ar 82 </> 8r2 -

8</> n au 82 </> 802 -

i))

y = rsinB.

8</> ax + 8</> 8y = 8</> cos B + 8</> sin 8 8x 8r ay 8r 8x 8y 2 2 8 </> ax 8 </> 8y 8x2 8r cos8 + 8y8x 8r cosO 82</> 8x . 82</> 8y . 8 + 8x8y 8r sin + 8y2 ar sm 8 a2 </> 2 a2 </> • Bx2 cos 8 + ByBx2sm8cos8

a2 </> . 2 By2 sm 8

8</> ax 8</> 8y 8¢ . 8</> +-= -(-rsm8) + -rcos8 8x 8y 88 ax ay 2 2 8 </> ax . 8 </> 8y . 8</> 8x2 80 (-r sin B) + 8y8x 88 (-r sin fJ) + 8x (-r cos fJ)

82</> 8x axay 88 (r cos 0)

= -(0) = 0. 8x

82¢ 8y

By2 88 (r cos 8)

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

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

8</> . By (-r sin 8) 2 2 8 </> 2 By2 r cos 8

Combining these partial derivatives, one gets

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

=0=>yx 2+y 3-y=y(x 2+y 2-1)=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

analytic. 13. 4,(x,y)

= Imz4 = r4sin49 = -4xy3 + 4x3y

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

2 In(u OU

8</, 8x

+ v2)

= 8</> OU + 8</> lJv = ua; + va; 8u 8x av 8x u + v2 2

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

a2 </> +. 8y IP</>

the Cauchy-Riemann equations of u and v to {Jz'l simplifies to reveal that 15.

the sum

6.</> = 0.

· ts· harmonic for I �lzl�2 Consider cp(z) -- Re(A zO + B z-n) + c which . 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 10) . e e + . 3 3 <p ( re = Ar cos30 + Br- cos30 + C = (Ar3 Br s30 .

-3)co

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( re 18) = (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 2

( z2 ) 2x(x2+4y+y ) b. </>(x, y) = 2Re z + 2i = x2 + y2 + 4y + 4 18. Let u = <Px, v = -<Py· Then

au ax

19.

2._o.

= <Pxx =

-</>yy

au_</> ay - xy -

av ax

= ay

cos 0 = (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.

[Yau

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 au - - ay (x, y) Inor d er t h at av

Thus,

+ 1/J' ( x)

(because u is harmonic) I

+ oy (x,0) + VJ (x).

. = - au t oy, i

must be true that VJ x)

= - au (x, 0). oy

r au

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

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..

100°

scr

800

80°

6()°

60° 40°

40°

2()°

20°

80° 60° 40° 20,

()° ()° ()°

c. 2(1'

40° 60° 8<>°

500 I 00° 50°

100°

goo 60° 40°

2()°

e. 100° ..._ 80°

80° 60°........__

._ .

60°

40°

20°

20"

2. This does not violate the maximum principle.

3. This does not violate the maximum principle. 100Â°

100Â°

Exercises 2. 7 1.

f(z) = z2 + 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. 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. (a) Fixed points are SI = i, s2 = -i. Both are repellors. (b) Fixed points are St= 1/2, s2 = -1/2, s = -1. Fixed points and SJ are repellors, but fixed point s2 is an attractor. zo = ei2rra with a an irrational real number. Zn = ei2rra 2"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. Fixed points are SI= -1/2 + i"5/2 (an attractor) and s2 = -1/2 - i"5/2 (a repellor). 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 = ei2rr. Solving gives Zo = eA(i27t/(2n- l )). 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. 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) If lal s; 1 the whole complex plane is the filled Julia set. If lal 2:: I the origin is the filled Julia set. 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.

5. 6.

9. 10.

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