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

1.

a.

to=

b.

tD

(3z2

-

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

= z2 : y2 + i ( 1

z2

! y2)

z

.

-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

f.

2.

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

c \ {i,-i}

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

c

Rew> 5 b. Imw > 0 a.

c.

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

=

1

., = !.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

'

0

to=

0

-

-

re'"o

r

0

i

theray Argw = -90â&#x20AC;˘ Taking values O < r < oo shows that this mapping goes onto the ray Argw = -Bo.

-----·---------·-----------·-

...

� 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.

4 (c) lz-�

5.

= I 2� 0�

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

1 1 b. f(-z)=Cz=-=•

ez

c. circle

f(z)

lw I = e

d. ray Arg w =

1r /

4

e. infinite sector O � Arg w � 6.

a.

1r /

4

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

Z

Z

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

ei

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

i

= rei8

+;)

on the circle

ii

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

Q

b.

-2

J. - l.

-l

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

c.

8.

a.

b.

.s

c.

9.

10.

a.

b.

a. translate by i, rotate 1r /4

b. reduce by 1/2, rotate

1r

c. translate by i, reduce by 1/2

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

1r /4,

-l

-o.s

/4

o.s

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

'). - L,(

0.25

o.s o.

5

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- v2 /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.

1

The

0.

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

20

10

10

15

;;:i -

b

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

4.

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

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,

> 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 -) =

Hi

a. -Si

-

i= 0

b.

7 2i . c.

6i

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

£

1

ti. No: Observe that although -

J (!)

i

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

n

n

.

- 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

+i

L/

d. 1

d.l,

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 ) � w0 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

. f(zo I rm

Az-o

given

e

+ �z) - f(z0) =L� �Z

> 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)

3.

--+

J -J

o

4.

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

Ii

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

A

)

[g(zo

+ �z) - g(zo)]

ixz [f(zo

A

uz

+ �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

dz

z2n

z"

+ 16z + i

b. -12z(z2

-iz4

-

3i)-7

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

c.

+ 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

uz

I-

If'( z )I 0•

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

z-zo -

'1.- l

1.

arg[f'(zo)]

9.

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

z

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 � ....o( 0 + 0 + ll.z) 11.

= 0.

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) Q(z)

=

P'(z) P(z)

1

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

1 -

z - Z1

1 + z - Z2

1

+ ... + 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, Exercise 18). However, f(z2) - f(zi) = 0 but f'(w) = 3w2 # 0 on the line in question.

z = -�

+ 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

1.

2.

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

a.

2

2

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 =

av

-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

=

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

IJx

�o

ll.x

l}y

At,-0

fl.y

lJu = lim u(0,6.y)- u(O,O)

= =

lim _!_ = O � t:J,.z

_!_ = O.

Jim Ay-0

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)

1

- 2· Therefore

5. :

=

f is not differentiable a.t

z

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

8u

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)

OU

+ fJyrcosB

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

ao

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 = --

or

fJv

-

- m

oy ox Bu . Bu = - cosB+ s B

-

or

oy

r {)() 1 ou

= --•

- m

r 80

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 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). .

av

av

OU

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

10. lmf(z) .

11. Ref(z) =

1

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.

=

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-

_

oy

8y

1

81! 1 2

1

olf 1 2

2

2

av

- -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

0

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 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-

ox

ax au

0V = -m2 lJy.

It follows that

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

ou

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 =

aJ

a,,

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

BJ ay

= ax a,, + By a,,

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

2

ax

ay

aJ

21

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.

2

2

au

au 2

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

y

-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

=0

3.

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

d.

= 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)

e.

=

1

x2 OU

01J

Y

+y

2

8v

x 11

= z2 + y2 = - oz = 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

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.

2

=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 -

2

-

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)

ao

--

+

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

2

-

Gz

8

= -(0) = 0. 8x

+

82¢ 8y

By2 88 (r cos 8)

+

2

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

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

8y

+

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

Combining these partial derivatives, one gets

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

+y

2

=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

f

is

analytic. 13. 4,(x,y)

= Imz4 = r4sin49 = -4xy3 + 4x3y

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

1

2

2 In(u OU

8</, 8x

+ v2)

av

= 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.

2

,

the sum

6.</> = 0.

· ts· harmonic for I �lzl�2 Consider cp(z) - Re(A z O + Bz -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

+

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( 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

av

= ay

2

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.

av

au

[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

ax

Thus,

+ 1/J' ( x)

OU

(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.

b.

a..

100°

100°

scr

800

80°

Mr

6()°

60° 40°

40°

40°

2()°

20°

20°

80° 60° 40° 20,

()° ()° ()°

d.

c. 2(1'

40° 60° 8<>°

500 I 00° 50°

100°

goo 60° 40°

oo

2()°

er

er

e. 100° ..._ 80°

80° 60°........__

._ .

60°

40°

40°

20°

20"

oo

oo

2. This does not violate the maximum principle.

oo

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

100Â°

Exercises 2. 7 1.

2.

3.

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.

3

4.

5. 6.

7.

8.

9. 10.

s1