Solucionario Gamelin

7/24/2019 Solucionario Gamelin

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IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 191 X X X X X X X X X X X2 X X X X X X X X3 X X X X X X X4 X X X X5 X X X X6 X X X X7 X X X X X X X X X X X8 X X X X X X X X

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I.1.1

Identify and sketch the set of points satisfying.(a) jz 1 ij = 1 (f) 0< Im z < (b) 1< j2z 6j


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x < 1; then jx 1j + jx+ 1j = (x 1) (x+ 1) = 2x 2;1 x 1 then jx 1j + jx+ 1j = (x 1) + (x+ 1) = 2 2x >1; then jx 1j + jx+ 1j = (x 1) + (x+ 1) = 2x 2:

(e) Halfplanex >1=2.

jz 1j < jzj , jz 1j2 < jzj2 , jx+iy 1j2 < jx+iyj2 ,, (x 1)2 +y2 < x2 +y2 , x >1=2

(f) Horizontal strip,0 < y < .(g) Vertical strip, < x < .(h) CnR.

jRe zj < jzj , jRe (x+iy)j2 < jx+iyj2 , x2 < x2 +y2 , jyj >0(i) Half plane y 0 , Re (i (x+iy) + 2)>0 , y+ 2 > 0 , y


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I.1.2

Verify from the denitions each of the identities(a) z+w= z+ w (b) zw = zw (c) jzj = jzj (d) jzj2 =z zDraw sketches to illustrate (a) and (c).

SolutionSubstitutez = x+iy andw = u+iv, and use the denitions.(a)

z+w= (x+iy) + (u+iv) = (x+u) + (y+v) i=

= (x+u) (y+v) i= (x iy) + (u iv) = z+ w:(b)

zw = (x+iy) (u+iv) = (xu yv) + (xv+yu) i== (xu yv) (xv+yu) i= (x iy) (u iv) = zw:

(c)

jzj =x+iy= jx iyj = qx2 + (y)2 = px2 +y2 = jx+iyj = jzj :

(d)

jzj2 = jx+iyj2 =p

x2 +y22

=x2 +y2 =

=x2 i2y2 = (x+iy) (x iy) =z z:

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I.1.3

Show that the equationjzj2

2Re(az) + jaj2

=2

represents a circlecentered at a with radius .

SolutionLetz = x+iy anda = +i, we have

jzj2 2Re(az) + jaj2 ==x2 +y2 2Re(( i) (x+iy)) +2 +2 =

=x2 +y2 2Re((x+y) +i (y x)) +2 +2 =

=x2

+y2

2 (x+y) +2

+2

== (x )2 + (y )2 :

Thus the equationjzj2 2Re(az) + jaj2 =2, becomes

(x )2 + (y )2 =2;which is the equation for a circle of radius centered at (; ), which incomplex notation is the point a = +i.

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I.1.4

Show thatjzj jRe zj + jIm zj, and sketch the set of points for whichequality holds.

SolutionApply triangle inequality toz = Re z + i Im z, to obtain jzj 6 jRe zj + jIm zj.Now set z = x+iy, and see then equality holds

jzj = jRe zj+jIm zj ,p

x2 +y2 =p

x2+

py2 ,

p

x2 +y22

=p

x2 +

py22

,

, x2+y2 = px22+px22+2px2py2 , x2y2 = 0 , x= 0ory = 0:Equality holds only whenz is real or pure imaginary, which are all the pointson the real and imaginary axis.

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I.1.5

Show that

jRe zj jzj ; jIm zj jzj :Show that

jz+wj2 = jzj2 + jwj2 + 2 Re (zw) :Use this to prove the triangle inequalityjz+wj jzj + jwj.

SolutionLetz = x+iy. Then since the square root function is monotone, we have

jRe zj = jxj = px2 6p

x2 +y2 = jzj ;jIm zj = jyj =

py2 6

px2 +y2 = jzj :

Now forz ; w we have

jz+wj2 == (z+w) (z+w) = (z+w) (z+ w) =z z+zw+wz+ww=

= jzj2 + 2 Re (zw) + jwj2 ;

where we have used the fact that 2R e (zw) =zw +zw= zw + wz = zw +wz.We use both of the above facts and the trivial identities jzj = jzj and jzwj =jzj jwj to prove the triangle inequality forz ; w. We have

jz+wj2 = jzj2 + 2 Re (zw) + jwj2 jzj2 + 2 jRe (zw)j + jwj2 jzj2 + 2 jzwj + jwj2 = jzj2 + 2 jzj jwj + jwj2 = (jzj + jwj)2 :

The desired inequality now follows by taking square root of both sides.

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I.1.6

For xeda 2 C, show that jz aj = j1 azj = 1 ifjzj = 1 and1az6= 0.SolutionIfjzj = 1, thenjzj = 1 and z z= 1. Use this and get

jz aj = jz aj jzj = jzz azj = j1 azj = j1 azj :We have

jz ajj1 azj = 1;

as was to be shown.

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I.1.7

Fix >0, 6= 1, and xz0; z12 C. Show that the set of z satisfyingjz z0j = jz z1j is a circle. Sketch it for = 12 and = 2, withz0 = 0 andz1= 1. What happens when= 1?

SolutionRecall that a circle in R2 centered at (a; b) with radius r is given by theequation

(x a)2 + (y b)2 =r2:We manipulate the equation

jz z0j = jz z1j :The solutions set of the equation above remains the same if we square bothsides,

jz z0j2 =2 jz z1j2 :Letz = x + iy,z0= x0+ iy0 andz1= x1+ iy1. Thus our equation becomes

(x x0)2 + (y y0)2 =2

(x x1)2 + (y y1)2

:

Expanding the squares, and grouping terms, we have

1 2x22 x0 2x1 x+x20 2x21+1 2 y22 y0 2y1 y+y20 2y21= 0

Dividing both sides by (1 2), we have

x2 2(x0 2x1)

1 2 x+(x20 2x21)

1 2 +y2 2(y0

2y1)

1 2 y+(y20 2y21)

1 2 = 0

Now complete the squares for both the x and y terms. Recall that

x2 2ax+b= (x a)2 a2 +b:So we have

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x x0

2x11 2

2+

(1 2) (x20 2x21) (x0 2x1)2(1 2)2 +

+

y y0

2y11 2

2+

(1 2) (y20 2y21) (y0 2y1)2(1 2)2 = 0:

This becomes

x x0 2x1

1

2 2

+ y y0 2y1

1

2 2

=

=(x0 2x1)2 + (y0 2y1)2 (1 2) (x20+y20 2 (x21+y21))

(1 2)2 =

=2

(x1 x0)2 + (y1 y0)2

(1 2)2 ;

which is the equation for a circle. Ifz0 = 0, andz1= 1, we havex

2

1 22

+y2 =

1 22

:

When= 12 we have a circle of radius 23 centered at 13 ; 0, and when= 2,

we have a circle of radius 23 centered at43 ; 0

. When = 1, we have the

equation

jz z0j = jz z1j ;which is the line bisecting the two points. Whenz0 = 0; z1 = 1, this is theline Re z= 12 .

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-1 1 2

-2

-1

1

2

Re z

Im z

I.1.7

-1 1 2

-2

-1

1

2

Re z

Im z

I.1.7

-1 1 2

-2

-1

1

2

Re z

Im z

I.1.7

= 12 = 1 = 2

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I.1.8

Let p (z) be a polynomial of degree n 1 and let z02 C. Showthat there is a polynomial h (z) of degree n 1 such that p (z) =(z z0) h (z)+p (z0). In particular, ifp (z0) = 0, thenp (z) = (z z0) h (z).

SolutionSet

p (z) =anzn +an1zn1 +an2zn2 + +a2z2 +a1z+a0:

and

h (z) =bn1zn1 +bn2zn2 +bn3zn3 +: : :+b2z2 +b1z+b0:

We equate coecients in the polynomial identity p (z) = (z z0) h (z) +p (z0), and get

anzn +an1zn1 +an2zn2 + +a2z2 +a1z+a0=

= (z z0)

bn1zn1 +bn2zn2 +bn3zn3 +: : :+b2z2 +b1z+b0

+p (z0) =

=bn1zn +bn2zn1 +bn3zn2 +: : :+b2z3 +b1z2 +b0z+p (z0) bn1z0zn1 bn2z0zn2 bn3z0zn3 : : : b2z0z2 b1z0z b0z0 =

bn1zn+(bn2 bn1z0) zn1+(bn3 bn2z0) zn2+(bn4 bn3z0) zn3+ + (b2 b3z0) z3 + (b1 b2z0) z2 + (b0 b1z0) z+p (z0) b0z0:

Equate and solve for thebjs in terms ofajs.

8>>>>>>:

bn1 = an

bk =nk1Pi=0

ak+1+izi0; 0 k n 2

p (z0) =n

Pi=0aiz

i0

Proof by induction on degree n ofp (z), set

p (z) =anzn +an1zn1 : : :+a0;

wherean6= 0.

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Fixz0 and write

p (z) =an(z z0) zn1 +r (z) ;wheredeg r (z) n 1.By using the induction hypothesis, we can assume that

r (z) =q(z) (z z0) +c;wheredeg q(z) deg r (z). Then

p (z) =

anz

n1 +q(z)

(z z0) +c= h (z) (z z0) +c

Since deg q(z)

n

2,deg r (z)

n

1.Plug inz0, getp (z0) =c.

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I.1.9

Find the polynomialh (z)in the preceding exercise for the followingchoices of p (z) andz0(a) p (z) =z2 andz0 = i(b) p (z) =z3 +z2 +z andz0 = 1(c) p (z) = 1 +z+z2 + +zm andz0 = 1

SolutionFrom the preceding exercise we have

p (z) = (z z0) h (z) +p (z0) :We solve forh (z)then

h (z) =p (z) p (z0)

z z0 :

(a)We have that p (z) =z2 andz0 = i, thusp (z0) =p (i) = 1.Thus

h (z) =p (z) p (z0)

z z0 =z2 + 1

z i =z +i;

and

z2 = (z i) (z+i) 1:(b)We have that p (z) =z3 +z2 +z andz0 = 1, thusp (z0) =p (1) = 1.Thus

h (z) =p (z) p (z0)

z z0 =z3 +z2 +z+ 1

z+ 1 =

(z+ 1) (z2 + 1)

z+ 1 =z2 + 1;

and

z3 +z2 +z = (z+ 1)

z2 + 1 1:

(c)We have that p (z) = 1 +z+z2 + +zm andz0= 1, thus

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p (z0) =p (1) = 0; modd1; mevenThus

h (z) =p (z) p (z0)

z z0 =

=

8 N.Ifn > N, then we have

jsn+1 snj = jsn+1 s+s snj jsn+1 sj + jsn sj < "=2 +"=2 =":

Thus

jsn+1 snj ! 0asn ! 1.

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II.1.9

Plot each sequence and determine its lim inf andlimsup.(a) sn= 1 + 1n+ (1)n (c) sn= sin (n=4)

(b) sn= (n)n (d) sn= xn (x 2 R xed)

Solution(a)lim sup sn= 2 lim infsn= 0(b)lim sup sn= 1lim infsn= 1(c)lim sup sn= 1 lim infsn= 1

(d)

lim sup sn=

81;1; jxj = 1;0; jxj >>>>>>:

1; 1 < x < 1;1; x= 1;0; jxj 1:

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II.1.10

At what points are the following function continuous? Justify youranswer.(a) z (c) z2= jzj(b) z= jzj (d) z2= jzj3

Solution(a)continuous everywhere(b)Continuous except at z = 0, where it is not dened(c)

Continuous forjzj 6= 0. It is not dened at z = 0, but it has a limit 0 asz! 0. If we dene the function to be 0 at z = 0, it is contionuous there.(d)Contionuous forjzj 6= 0. It has no limit asz! 0

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II.1.11

At what points does the functionArg z have a limit? Where is Arg zcontinuous? Justify your answer.

SolutionArg zhave a limit at each point ofCn (1; 0]. It is continuous in Cn (1; 0].It is discontinuous at each point of(1; 0]. Values of function! fg atpoints of (1; 0], the function may take any value in intervall [; ] asz! 0. The value is depending in that direktion we approachz = 0.

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II.1.12

Let h (z) be the restriction of the function Arg z to the lower half-planefIm z


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II.1.13

For which complex values of does the principal value of z

havea limit as z tends to 0? Justify your answer.

SolutionSetz = rei =eLog rei =eLog r+i and = Re +i Im , then

z =

eLog r+iRe +i Im

=eLog r Re +i Log r Im +i Re Im =

=eLog r Re e Im ei(Log r Im + Re ) =rRe e Im ei(Log r Im + Re );

thus

jzj =rRe e Im :Becausee Im is bounded we look at rRe in three intervalsIfRe 0,r Re ! 0 as r ! 0 thusjzj have limit 0 asz! 0, andz ! 0asz

!0.

The conclution is z ! 0 ifRe > 0, and z ! 1 if = 0, otherwise wehave no limit at 0.

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II.1.14

Let h (t) be a continuous complex-valued function on the unit in-terval [0; 1], and consider

H(z) =

Z 10

h (t)

t z dt:

Where is H(z) dened? Where is H(z) continuous? Justify youranswer. Hint. Use the fact if jf(t) g (t)j < " for 0 t 1, thenR1

0jf(t) g (t)j dt < ".

Solution

We have that H(z) is dened for z 2 Cn [0; 1]. If h (z1) = 0 for somez12 [0; 1]. ThenH(z)is also dened for z = z1.ThenH(z) =

1R0

h(t)tz dz is continuous forz2 Cn [0; 1]. Ifzn! z2 Cn [0; 1],

then h(t)tzn !

h(t)tz uniformly for 0 t 1, so the used proof shows that

H(zn) ! H(z).

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II.1.15

Which of the following sets are open subsets ofC

?Which are close?Sketch the sets.(a) The punctured plane, Cn f0g .(b) The exterior of the open unitdisk in the plane,fjzj 1g .(c) The exterior of the closed unitdisk in the plane,fjzj >1g .(d) The plane with the open unitinterval removed, Cn (0; 1) .(e) The plane with the closed unitinterval removed, Cn [0; 1] .(f) The semidisk,fjzj


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II.1.16

Show that the slit planeCn (1; 0] is star-shaped but not convex.Show that the slit plane Cn [1; 1] is not star-shaped. Show that a

punctured disk is not star-shaped.

Solution(a) Cn (1; 0] is starshaped, because we can see every point from z = 1.To show that Cn (1; 0] is not convex we may choose any two points suchthat the stright line segment joining them contains a point in (1; 0]. Forexample take1 +i and1 i. These two points are joined by a verticalline which contain1. From this follows that Cn (1; 0] is not convex.(b) Cn [1; 1]is not starshaped. Given any z we have thatz can not beeseen from a stright line betweenz and z that must contain0. We also havethat ifz belong to the domain thenz belongs also to the domain.(c) Same argument as in (b) shows that Cn f0gis not starshaped.

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II.1.17

Show that a set is convex if and only if it is star-shaped with respectto each of its points.

Solution (A. Kumjian)Let D C, since the empty set trivially satises both conditions we willassume thatD 6= ?.Suppose rst that D is convex. We must show thatD is star-shaped withrespect to each of its points. So x z02 D, to show that D is star-shapedwith respect toz0. We must show that for every point z2 Dthe line segmentconnecting z0 to z is contained in D. Letz2 D be given, then since D isconvex it contains the line segment joining z0 andz. Hence,D is star-shapedwith respect to z0, then, since z0 was chosen arbitrarily, D is star-shapedwith respect to each of its points.

Conversely, suppose that D is star-shaped with respect to each of its points.To show that D is convex we must show that given two points in D, theline segment joining them is also contained inD. So letz0; z12 D be given.Then since D is star-shaped with respect to z0,D contains the line segment

joiningz0 andz1. Hence,D is convex.

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II.1.18

Show that the following are equivalent for an open subset Uof thecomplex plane.(a)Any two points of Ucan be joined by a path consisting of straightlinesegments parallel to the coordinate axis.(b)Any continuously dierentiable function h (x; y)on Usuch that rh=0 is constant.(c)If V and W are disjoint open subsets of U such that U = V[ W,then eitherU=V orU=W. Remark. In the context of topological

spaces, this latter property is taken as denition of connectedness.

SolutionShow that the following equivalent for an open subsetUof the complex plane.(a)) (b) Supposerh= 0 inD. Fix z02 D. Ifz12 D, join by polygonalcurve.rh= 0 ) his constant on each segment of curve) h (z0) =h (z1).) his constant in D.(b)) (c) Supposerh= 0, h not constant, say h (z0) = 0 for some z0. LetW =fh= 0g, V =fh 6= 0g. Evidently V is open. Sincerh = 0, h isconstant in a neighborhood of each point, so

fh= 0

gis open andWis open.

W6= ?,V6= ?,W\ V = ?,W[ V =U(c)) (a) Suppose z02 U. Let D = points that can be joined to z0 by apolygonal curves, intervals parallel to coordinate axis. EvidentlyD is open.Also if can connected points near z02 U, then can connect to U, soUnD isopen. By (c), UnD must be empty, someD 6= ?:

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II.1.19

Give a proof of the fundamental theorem of algebra along the fol-lowing lines. Show that if p (z) is a nonconstant polynomial, thenjp (z)j attains its minimum at some point z0 2 C. Assume thatthe minimum is attained at z0 = 0, and that p (z) = 1 +az

m + ,wherem 1 anda 6= 0. Contradict the minimality by showing thatP"ei0< 1 for an appropriate choice of 0.Solution (K. Seip)Set

p (z) =anzn

+an1zn

1

+ +a0;where we assume an6= 0. Since

limjzj!1

jp (z)jjzjn = janj ;

9 R > 0 such thatjp (z)j >ja0j for alljzj > R. Thusjp (z)j attainsits minimum injzj R. We may assume it is attained at z0 = 0.Suppose a06= 0, say a0 = 1. Then p (z) = 1 +azm+ higher orderterms, a 6= 0. Choosez= "

1

a

1

m (any m th root does the job).Then

p (z) = 1 "m +O "m+1 :Thus for" suciently small " we havejp (z)j


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Follow the suggestion.

Let

p (z) =a0+a1z+am+1zm+1 +:::+aNz

N; aN6= 0:Choose R so that

jaNj RN > ja0j + ja1j R+ jam+1j Rm+1 +:::+ jaN1j RN1:Then p (z)6= 0 forjzj = R, the term aNzN dominates and furtherjp (z)j >jp (0)j forjzj =R.Letz0 be a point in the diskjzj 6 R at which the continuous function for

jp (z)

jattains a minimum. Then

jz0

j< R, so that is a disk centred at z0 so

thatjp (z)j >jp (z0)j on the disk. Supposep (z0)6= 0. We can assume thatp (z0) = 1.Write

p (z) = 1 +amP(z z0)m +O

(z z0)m+1

;

wheream6= 0. Suppose am= r0ei0 .Conside

zm = z0+"ei0=mei=m; " >0:

We have

p (zm) = 1 +r0ei0emei0ei +O "m+1= 1 r0"m +O "m+1 :Have p (zm) 1 r0"m < 1 for" > 0 small. Contradiction! We concludethatp (z0) = 0.

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II.2.1

Find the derivatives of the following function.(a) z2 1 (c) (z2 1)n (e) 1= (z2 + 3) (g) (az+b) = (cz+d)(b) zn 1 (d) 1= (1 z) (f) z= (z3 5) (h) 1= (cz+d)2

Solution(a) 2z (c) n (z2 1)n1 2z (e) 2z= (z2 + 3)2 (g) (ad bc) = (cz+(b) nzn1 (d) 1= (1 z)2 (f) (2z3 5) = (z3 5)2 (h) 2c= (cz+d)3

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II.2.2

Show that

1 + 2z+ 3z2 + +nzn1 = 1 zn

(1 z)2 nzn

1 z :

SolutionUse the geometric sum

1 +z+z2 +z3 + +zn =1 zn+1

1 z ; z6= 1:Dierentiate both sides

1 + 2z+ 3z2 + +nzn1 =

=(n+ 1) (1) zn (1 z) + (1 zn+1)

(1 z)2 =

=nzn+1 nzn +zn+1 zn + 1 zn+1

(1 z)2 =

= 1 zn(1 z)2 +

nz n+1 nzn(1 z)2 =

1 zn(1 z)2

nzn

(1 z) :

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II.2.3

Show from the denition that the functions x = Re z and y = Im zare not complex dierentiable at any point.

SolutionDierentiation f(z) =x = Re z (from the denition)

limz!0

f(z+ z) f(z)z

= limz!0

Re (z+ z) Re (z)z

=

= limz!0

Re z

z

= 1; ifz= x0; ifz= iy :Dierentiation g (z) =y = Im z (from the denition)

limz!0

g (z+ z) g (z)z

= limz!0

Im (z+ z) Im (z)z

=

= limz!0

Im z

z =

0; ifz= xi; ifz= iy :

The functions x = Re z andy = Im z are not complex dierentiable at anypoint, because the functions have dierent limit asz

!0through real and

imaginary axis.

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II.2.4 (+ IV.8.2)

Suppose f(z) =az2

+bz z+cz2

, where a, b, and c are xed complexnumbers. By dierentiating f(z) by hand, show that f(z) is com-plex dierentiable at z if and only if bz+ 2cz = 0. Where is f(z)analytic?

SolutionWe will nd d

dzf(z)using the limit denition

f(z+ z) f(z)z

=

= a (z+ z)2

+b (z+ z) (z+ z) +c(z+ z)2

(az2 +bz z+cz2)z

=

=a

z2+2zz+ z2

+b

zz+zz+zz+ zz

+c

z2 + 2zz+ z2

(az2 +bzzz

=2azz+az2 +bzz+bzz+bzz+ 2czz+cz

2

z =

= 2az+az+bzz

z+bz+bz+ 2cz

z

z+c

z2

z =

= 2az+az+bz+bz+czz

z

+ (bz+ 2cz)z

z

We know z is not analytic at any open set ofC and lim4z!0

zz does not ex-

ist. Therefore, unless bz+ 2cz= 0, the derivative ddz f(z) would not exist.Therefore f(z) is analytic on C ifb = c = 0, otherwise is the function notanalytic on any open set.

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II.2.5

Show that if f is analytic on D, theng (z) =f(z) is analytic on thereected domain D = fz : z2 Dg, and g0 (z) =f0 (z).

Solution (A. Kumjian)Letfbe analytic on the domain D and dene g onD as above (note thatD is also a domain). For a complex-valued function ' dened near a pointz0 it is easy to show that

limz!z0

' (z) exists i limz!z0

' (z)exists;

and if either exists, then the two limits are complex conjugates of each other.Let z2 D be given. We rst show that g is dierentiable at z and thatg0 (z) =f0 (z).

limh!0

1

h(g (z+h) g (z)) = lim

h!01

h

f

z+h f(z) = lim

k!01

k(f(z+k) f(z)) = f0 (z)

where equation () follows using the substitution k =h and basic algebraicproperties of conjugation, note that conjugation is a continuous function soh! 0 i h! 0. Since f is analytic on the domain D, its derivative f0is continuous on D. Further, since g0 (z) = f0 (z) for all z2 D, g0 is thecomposite of continuous functions and therefore is itself continuous. Hence,g is analytic on D.

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II.2.6

Let h (t) be a continuous complex-valued function on the unit in-terval [0; 1], and dene

H(z) =

Z 10

h (t)

t z dt; z2 Cn [0; 1] :

Show that H(z) is analytic and compute its derivative. Hint. Dif-ferentiate by hand, that is, use the dening identity (2.4) of com-plex derivative.

Solution

We use denition (2.4) of derivative to nd H0 (z).

H0 (z) =

= limz!0

H(z+ z) H(z)z

= limz!0

1

z

Z 10

h (t)

t z z dt Z 1

0

h (t)

t z dt

=

= limz!0

1

z

Z 10

h (t)

1

t z z 1

t z

=

= limz!0Z

1

0

h (t)

(t z z) (t z)= Z 1

0

h (t)

(t z) (t z)dt=

=

Z 10

h (t)

(t z)2 dt:

H0 (z)is continuous in z , also by uniform convergence of integrandH(z)isanalytic forz2 Cn [0; 1].

29


116/717

II.3.1

Find the derivatives of the following functions.(a) tan z = sin zcos z

(b) tanh z = sinh zcosh z

(c) sec z = 1= cos z

Solution(a)

d

dztan z=

d

dz

sin z

cos z =

cos z cos z sin z ( sin z)cos2 z

=cos2 z+ sin2 z

cos2 z =

1

cos2 z:

(b)

d

dztanh z=

d

dz

sinh z

cosh z =

cosh z cosh z sinh z sinh zcosh2 z

=cosh2 z sinh2 z

cosh2 z=

1

cosh2 z:

(c)

d

dzsec z=

d

dz

1

cos z = 1

cos2 z( sin z) = sin z

cos2 z= tan z sec z:

30


117/717

II.3.2

Show that u = sin x sinh y and v = cos x cosh y satisfy the Cauchy-Riemann equations. Do you recognize the analytic function f =u+iv? (Determine its complex form)

SolutionWe have

@u

@x =

@

@xsin x sinh y= cos x sinh y=

@

@ycos x cosh y=

@v

@y;

@u@y = @@y sin x sinh y= sin x cosh y= @@x cos x cosh y= @vdx :

Hence, u and v satisfy Cauchy-Riemann equations, and now we calculatef(z).

f(z) =

=u+iv=

eix eix

2i

ey ey

2

+i

eix +eix

2

ey +ey

2

=

= ieix+y

eixy

eix+y +eixiy

4+ie

ix+y +eixy +eix+y +eixy

4=

=ieixy +eix+y

2 =i

ei(x+iy) +ei(x+iy)

2 =i

eiz +eiz

2 =

=i cos z:

31


118/717

II.3.31 2 3 P L K

LLL

Show that if f and fare both analytic on a domain D, then f isconstant.

Solution (A. Kumjian)LetD be a domain and let fbe a complex valued function such that both fand fare analytic onD. Then bothRe f= 12

f+ f

andIm f= 12i

f f

must also be analytic on D. Since Re f andIm fare also both real-valued,it follows by the theorem on page 50, that both are constant. Hence,f =

Re f+i Im f is constant.

32


119/717

II.3.4

Show that if f is analytic on a domain D, and if jfj is constant,then f is constant.Hint. Write f= jfj2 =f.

Solution (with use of hint)Iff(z) = 0 for somez2 D. Thenf 0on D, since jfj is constant. Assumethat f(z)6= 0 for every z2 D. Since f is analytic in D we have that1=f and f =jfj2 =f (wherejfj is constant) are analytic. From this followsthat Re f andIm fare analytic and real-valued and therefore constant. Sof= Re f+i Im f is constant.

33


120/717

II.3.4

Show that if f is analytic on a domain D, and if jfj is constant,then f is constant.Hint. Write f= jfj2 =f.

SolutionSet f =u+iv, where u andv are realvalued functions. Now suppose thatjfj =k is constant. We have that u2 + v2 =k2. Dierentiate both sides withrespect tox respectivelyy

uu0

x+vv0

x= 0;

uu

0

y+vv

0

y = 0:

(1)

We use Cauchy-Riemanns equations u0

x= v0

y andu0

y = v0x in (1) and have uu

0

x vu0y = 0;uu

0

y+vu0

x= 0: (2)

In the simultaneusly systems of equations in (2) we rst rst multiply therst row with u and second row with v in (2) and add them together, andhave (u2 +v2) u

0

x = 0. And if we in the same system multiply the rst rowwith v and the second row withu and add them we have (u2 +v2) u0y = 0.Thus

(u2 +v2) u0

x = 0;(u2 +v2) u

0

y = 0: (3)

If in (3) u2 +v2 = 0, and we have that u = v = 0 and thus f is constant.Suppose that u2 +v2 6= 0. We have that (3) gives thatu0x = u0y = 0, withsays that u is constant. Cauchy-Riemanns equations gives

0 =u0

x= v0

y;0 =u

0

y = v0y; (4)

thus by (4) f is constant.

34


121/717

II.3.5

If f=u+iv is analytic, thenjruj = jrvj = jf0j.

Solution (K. Seip)Because f=u+iv is analytic. Then

ru=

@u

@x;@u

@y

=

@u

@x; @v

@x

=

@v

@y; @v

@x

from which it follows that

jru

j=

jf0 (z)

j=

jrv

j:

35


122/717

II.3.6

If f =u+iv is analytic on D, thenrv is obtained by rotatingruby90. In Particular,ru andrv are orthogonal.

Solution (A. Kumjian)Let 2 R then rotation of a vector (x; y)2 R2 by the angle (about theorigin) is given by matrix multiplication:

xy

7! A

xy

=

cos sin sin cos

xy

=

x cos y sin x sin +y cos

:

Hence, rotation by 90 is given by the transformation (x; y)7!(y; x). Bythe Cauchy-Riemann equations we have

rv=

@v

@x;@v

@y

=

@u

@y;@u

@x

=A=2ru:

Thus, rv is obtained by rotating ruby90. It follows that ruand rv areorthogonal.

Solution (D. Jakobsson)Given since fis analytic, then

(0.6) @u@x

= @v@y

and @u@y

= @v@x

Sinceru =

@u@x

; @ u@y

andrv =

@v@x

; @ v@y

all we need to check is whether

rv rv= 0. To prove that we use (0.6).

ru rv= @v@x

@u

@x+

@ v

@y

@u

@y =

@v

@x

@v

@y @ v

@y

@v

@x= 0

We conclude thatruandrv are orthogonal.Notice that the operation that sends ru ! rvis given by the counter-clock-

wise rotation matrix with =

2 . Keeping in mind (0.6), one can write thefollowing.

rv=

vxvy

=

0 11 0

vyvx

=

cos

2 sin

2

sin 2 cos2

vyvx

= R

2

ru

36


123/717

Remark.

We have that

vx = @v

@x; vy =

@v

@y; ux =

@u

@x; uy =

@u

@y;

and

R (=2) =A=2:

37


124/717

II.3.7

Sketch the vector eldsru andrv for the following functionsf=u+iv.(a) iz (b) z2 (c) 1=zSolution

-4 -2 2 4

-4

-2

2

4

x

y

II.3.7a

-4 -2 2 4

-4

-2

2

4

x

y

II.3.7a

-4 -2 2 4

-4

-2

2

4

x

y

II.3.7a

(a)

f(z) =iz = y+ix )

u= yv= x

)ru= (0; 1)

rv= (1; 0)(b)

f(z) =z2

=x2

y2

+ 2ixy) u=x2 y2v= 2xy ) ru= (2x; 2y)rv= (2y; 2x)(c)

f(z) =1

z =

z

jzj2 = x iyx2 +y2

)

u= xx2+y2 = cos

r

v= yx2+y2

= sin r8>>>>>>>:

@u@x

= y2x2

(x2+y2)2 = sin

2 cos2 r2

= cos2r2

@u@y =

2xy(x2+y2)2

= 2cos sin r2 = sin 2

r2

@v@x =

2xy

(x2+y2)2 = 2cos sin r2 =

sin2r2

@v

@y =

y2x2(x2+y2)

2

=

sin2

cos2

r2 = cos2

r2ru= 1r2

( cos2; sin2)rv= 1r2( sin2; cos2)

38


125/717

II.3.8 (see III.4.2) (36)

Derive the polar form of the Cauchy-Riemann equations for u andv,

@u

@r = 1r

@v@ ;

@u@ = r @v@r :

Check that for any integer m, the functions u

rei

= rm cos(m)andv

rei

= rm sin(m)satisfy the Cauchy-Riemann equations.

SolutionSetx = r cos andy = r sin , then take the partial derivative,

@x@r = cos

@x@ = r sin

@y@r = sin

@y@ =r cos

Use Cauchy-Riemanns equations and the derivative to get the rst equation

@u

@r =

@u

@x

@x

@r +

@ u

@y

@y

@r =

@u

@xcos +

@ u

@ysin =

@v

@ycos @v

@xsin =

=1

r

r @v

@xsin +r

@v

@ycos

=

1

r

@v

@x

@x

@+

@ v

@y

@y

@

=

1

r

@v

@;

Use Cauchy-Riemanns equations and the derivative to get the second equa-tion

@u

@ =

@u

@x

@x

@+

@u

@y

@y

@ = r @u

@xsin + r

@u

@ycos = r @v

@ysin r @v

@xcos =

= r

@v

@xcos +

@ v

@ysin

= r

@v

@x

@x

@r+

@ v

@y

@y

@r

= r @v

@r:

we get

@u@r = 1r @v@ ;@u@

= r @v@r

:

Set u

rei

= rm cos(m) and v

rei

= rm sin(m) and use it in theCauchy-Riemanns equations,

39


126/717

@u@r

= @@r

(rm cos(m)) =mrm1 cos(m) =1r

rmm cos(m) =1r

@@

(rm sin m) =1r

@v@

;

@u

@ =

@

@(rm cos(m)) = mrm sin(m) = rmrm1 sin(m) = r@

@r(rm sin(m)) =

40


127/717

II.4.1

Sketch the gradient vector eldsru andrv for(a) u+iv= ez (b) u+iv= Log z

Solutiona)We have

u+iv= ez =ex cos y+iex sin y

and get

ru= (ex cos y;

ex sin y) =ex (cos y;

sin y)

(in black), and

rv= (ex sin y; ex cos y) =ex (sin y; cos y)(in red).b)We have

u+iv= Log z = log r+i

and get

ru = 1r!ur ;

rv = 1r!u:

FIGURE II.4.1a NF FIGURE II.4.1b NF

41


128/717

II.4.2

Letabe a complex numbera 6= 0, and letf(z)be an analytic branchof za on Cn (1; 0]. Show thatf0 (z) =af(z) =z. (Thusf0 (z) =aza1,where we pick the branch of za1 that corresponds to the originalbranch of za divided by z.)

SolutionWe may write the branch as follows

f(z) =za =ea log z =ea(Log z+2im) =ea Log z+2iam =cea Log z;

wherec = e2iam for somem2Z. It follows by the chain rule that

f0 (z) =cea Log za

z =

af(z)

z :

So, the desired result follows.

42


129/717

II.4.3

Consider the branch of f(z) = pz (1 z) on Cn [0; 1] that has posi-tive imaginary part at z= 2. What is f0 (z)? Be sure to specify thebranch of the expression for f0 (z).

SolutionSetf(z) =

pz (1 z)(principal branch) and dierentiate

f0 (z) =1

2

1pz (1 z)

d

dz(z (1 z)) =1

2

1 2zpz (1 z) :

Branch off(z) is i

increasing for x > 0 large, thus f0 (z) is i

positive for

x >0 large.Use branch off0 (z) =

pz (1 z)that it is i positive, i.e., same on f(z).

f0 (z) =1

2

1 2zf(z)

:

Solution (D. Jakobsson)Givenf(z) =

pz (1 z), we can dene w = f(z)and get w2 =z (1 z).

d

dz(1 z) z = d

dzw2

1 2z = 2w dwdz

One gets

d

dz

pz (1 z) = 1 2z

2p

z (1 z)and is dened on Cn [0; 1]. Take the principal branch with the positive imag-inary part.

43


130/717

II.4.4

Recall that the principal branch of the inverse tangent functionwas dened on the complex plane with two slits on the imaginaryaxis by

Tan1 = 12iLog

1+iz1iz

; z62 (i1; i] [ [i; i1) :

Find the derivative of Tan1 z. Find the derivative of tan1 z forany analytic branch of the function dened on a domain D.

SolutionWe have that

d

dzTan1 z =

1

2i

1

1 +izi 1

1 iz(i)

=1

2

1 iz+ 1 +iz(1 +iz) (1 iz)

=

1

1 +z2:

Any two branches oftan1 z dier by a constant, so derivatives are same.

44


131/717

II.4.5

Recall that cos1

(z) =i log z pz2 1. Suppose g (z) is an ana-lytic branch of cos1 (z), dened on a domain D. Find g0 (z). Dodierent branches of cos1 (z) have the same derivative?


d

dzcos1 z=

= iz pz2 1

1 12z2 11=2 2z=

= i

z pz2 1

1 zp

z2 1

=

ipz2 1 =

= 1p

1 z2 :

Derivatives of branches ofcos1 z are not always the same.

45


132/717

II.4.6

Supposeh (z)is an analytic branch of sin1

(z), dened on a domainD. Find h0 (z). Do dierent branches of sin1 (z) have the samederivative?

Solution (A. Kumjian)Regard h (z) as a local inverse off(z) where f(z) = sin z. Then f0 (z) =cos z, so using the formula given in the statement of theorem on page 51 wehave

h0 (z) = 1

f0 (h (z))=

1

cos h (z)

ifcos h (z) 6= 0.One local inverse off satises h1(0) = 0, while another local inverse offsatisesh2(0) =. Using the formula above we see

h0

1(0) = 1

cos0= 1 and h

0

2(0) = 1

cos = 1:

Hence, dierent branches ofsin1 (z)do not necessarily have the same deriv-ative.

46


133/717

II.4.7

Let f(z) be a bounded analytic function, dened on a boundeddomain D in the complex plane, and suppose that f(z) is one-one-one. Show that the area of f(D) is given by

Area (f(D)) =

ZZD

jf0 (z)j2 dxdy:

Solution (K. Seip)We have f :D! C,jf(z)j M for some M < 1 when z2 D. Since f isassumed to be one-to-one, we may compute

A (f(D)) = ZZf(D)

dudv

by the change of variables (u (x; y) ; v (x; y))! (x; y), and since det Jf =jf0 (z)j2, we get

Area(f(D)) =

ZZD

jf0 (z)j2 dxdy:

47


134/717

II.4.8

Sketch the image of the circlefjz 1j 1g under the map w = z2

.Compute the area of the image.

Solution

-4 -2 2 4

-4

-2

2

4

x

y

II.4.8 z-plane

-4 -2 2 4

-4

-2

2

4

x

y

II.4.8 w-plane

We have that f(z) =z2, that gives f0 (z) = 2z andJf(z) = 4 jzj2.The area of the image isZZ

Jfdxdy= 4

ZZ(x1)2+y261

x2 +y2

dxdy:

Sett =x 1, get

4

ZZt2+y261

t2 + 2t+ 1 +y2

dtdy=

= 4

264 ZZt2+y261

t2 +y2

dtdy+ 0 +

ZZt2+y261

dtdy

375=

= 4 h

2

+i= 6:Solution (D. Jakobsson)Given w = f(z) = z2, one can dierentiate f(z) and get d

dzf(z) = 2z =

2 (x+iy), we can use maple to integrate and get the following

48


135/717

Area (f(D)) =

=

ZZD

jf0 (z)j2 dxdy=Z 2

0

Zp1(x1)2p

1(x1)24 jx+iyj2 dx dy=

=

Z 20

Zp1(x1)2p

1(x1)24

x2 +y2

dx dy =

Z 20

8x2p

2x x2+8=3 2x x23=2 dx== 6

Change of variables x= 1 + sin t

dx= cos tdt

Z =2=2

8 (1 + sin t)2p

1 sin2 t+83

1 sin23=2 cos tdt=

=

Z =2=2

8

1 + 2sin t

|{z}odd+ sin2 t

!cos2 +

8

3cos4 tdtdt=

= 16Z =2

0

2cos2 23

cos4 tdt=

= 8

Z =20

2 (1 + cos 2t) 13

1 + 2 cos2t+ cos2 2t

dt=

= 8

Z =20

5

3+

4

3cos 2t 1

3(1 + cos 4t) dt=

= 8 2

5

31

6

= 4

9

6= 6:

49


136/717

II.4.9

Compute ZZD

jf0 (z)j2 dxdy;

for f(z) = z2 and D the open unit diskfjzj


137/717

II.4.101 2 3 P L K

For smooth functionsg andh dened on a bounded domain U, we dene theDirichlet formDU(g; h)by

DU(g; h) =

ZZU

@g

@x

@h

@x+

@ g

@y

@h

@y

dxdy

Show that ifz = f() is a one-to-one analytic function from the boundeddomain V onto U, then

DU(g; h) =DV(g

f; h

f) :

Remark. This shows that the Dirichlet form is a "conformal invariant".

SolutionDu(g; h) =

RRU

rgrhdxdy Assumeh; g ????Du(g; h) =

RRU

r (g f) r (h f)dudv=RRU

g (u (x; y) ; v (x; y)) h (u (x; y) ; v (x; y)) dudv

@g(u(x;y);v(x;y))@x

@h(u(x;y);v(x;y))@x

+@g(u(x;y);v(x;y))

@y@h(u(x;y);v(x;y))

@y =

@g@u @u@x + @g@v @v@x @h@u @u@x + @h@v @v@x + @g@u @u@y + @g@v @v@y@h@u @u@y + @h@v @v@y=@g@u

@h@u

@u@x

2+ @g

@u@u@x

@h@v

@v@x

+ @g@v

@v@x

@h@u

@u@x

+ @g@v

@h@v

@v@x

2+

@g@u

@h@u

@u@y

2+ @g

@u@u@y

@h@v

@v@y

+ @g@v

@v@y

@h@u

@u@y

+ @g@v

@h@v

@v@y

2=

@g@u

@h@u

@u@x

2+ @g@v

@h@v

@v@x

2+ @g@u

@h@u

@u@y

2+ @g@v

@h@v

@v@y

2=

@g@u

@h@u

@u

@x

2+

@u

@y

2!

| {z }jf0(z)j2

+ @g@v@h@v

@v

@x

2+

@v

@y

2!

| {z }jf0(z)j2

+

@g

@u

@u

@x

@h

@v

@v

@x +

@ g

@v

@v

@x

@h

@u

@u

@x +

@g

@u

@u

@y

@h

@v

@v

@y +

@ g

@v

@v

@y

@h

@u

@u

@y| {z }Use C-R equationsRR

U

rgrh jf0 (z)j2dxdy= RRU

rgrhdudv

51


138/717

II.5.1

Show that the following functions are harmonic, and nd its har-monic conjugates:(a) x2 y2 (c) sinh x sin y (e) tan1 (y=x) ; x >0(b) xy+ 3x2y y3 (d) ex2y2 cos(2xy) (f) x= (x2 +y2)

Solution(a)Set

u (x; y) =x2 y2 ) u0

x= 2xu0

y =

2y

) u00

xx = 2u00

yy =

2

) 4u= 0;

thus is the function u (x; y)is harmonic. Cauchy Riemann give us v0

x= u0y = 2y) v (x; y) = 2xy+g (y) ) v0y = 2x+g0 (y)v0

y =u0

x= 2x

Thus we have that

g0 (y) = 0 ) g (y) =C;thus

v (x; y) = 2xy+C:(b)Set

u (x; y) =xy+3x2yy3 )

u0

x = y + 6xyu0

y = x+ 3x2 3y2 )

u00

xx= 6yu0

yy = 6y ) 4u= 0;

thus is the function u (x; y)is harmonic. Cauchy Riemann give us

v0x= u0y = x 3x2 + 3y2 ) v (x; y) = x22 x3 + 3xy2 +g (y) ) v0y = 6xy+g0 (y)v0

y =u0

x= y + 6xy

Thus we have that

52


139/717

g0 (x) =y) g (y) =y2

2 +C;

thus

v (x; y) =y2

2 + 3xy2 x

2

2 x3 +C:

In paranthesis we remark that the analytic function f(z) =u +iv is givenby

f(z) =xy +3x2y

y3 +i

y2

2

+ 3xy2

x2

2 x3 +C= iz

2 (z+ 1=2)+iC:

(c)Set

u (x; y) = sinh x sin y)

u0

x = cosh x sin yu0

y = sinh x cos y )

u00

xx = sinh x sin yu0

yy = sinh x sin y ) 4u= 0;

thus the functionu (x; y)is harmonic. Cauchy Riemann give us

v0x= u0y = sinh x cos y) v (x; y) = cosh x cos y+g (y) ) v0y = cosh x sin y+g0 (y)v0

y =u0

x= cosh x sin y

Thus we have that

g0 (y) = 0 ) g (y) =C;thus

v (x; y) = cosh x cos y+CIn paranthesis we remark that the analytic function f(z) =u +iv is given

by

53


140/717

f(z) = sinh x sin y i (cosh x cos y+C) =

=

ex ex

2

eiy eiy

2i

i

ex +ex

2

eiy +eiy

2

+iC=

= i4

ex ex eiy eiy + ex +ex eiy +eiy +iC=

= i

ex+iy +e(x+iy)

2

+iC= i cosh z+iC:

(d)Set

u (x; y) =ex2y2 cos (2xy) )

u0

x = 2xex2y2 cos(2xy) 2yex2y2 sin(2xy)

u0

y = 2yex2y2 cos(2xy) 2xex2y2 sin(2xy) )

)

u00

xx = 2ex2y2 (cos2xy+ 2x2 cos2xy 2y2 cos2xy 4xy sin2xy)

u0

yy = 2ex2y2 (cos2xy+ 2x2 cos2xy 2y2 cos2xy 4xy sin2xy) ) 4u= 0;


8


141/717

Set

u (x; y) = tan1 (y=x) ; x >0 )

=)

8>:

u0

x= 1

1 + (y=x)2

y

x2

=

yx2 +y2

u0

y = x

x2 +y2

=)

)

8>>>:

u00

xx = 2xy

(x2 +y2)2

u0

yy = 2xy(x2 +y2)2

) 4u= 0;


8>:

v0

x = u0y = xx2 +y2

) v (x; y) =a 12log (x2 +y2) +g (y) ) v0y =

yx2 +y2

+g0 (y)

v0

y =u0

x= yx2 +y2

Thus we have that

g0 (y) = 0)

g (y) =C;

thus

v (x; y) = 12

log

x2 +y2

+C:

(f)Set

u (x; y) = x

x2 +y2)

8>>>:

u0

x = x2 y2(x2 +y2)2

u0y = 2xy(x2 +y2)2

)

8>>>:

u00

xx = 2x 3y2 x2(x2 +y2)3

u0yy = 2x 3y2

x2

(x2 +y2)3) 4

u= 0;

thus is the function u (x; y)is harmonic. Cauchy Riemann give us

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8>>>:

v0

x = u0y = 2xy(x2 +y2)2

) v (x; y) = y(x2 +y2)

+g (y) ) v0y = x2 y2

(x2 +y2)2+g0 (y)

v0

y =u0

x= x2 y2(x2 +y2)2

Thus we have that

g0 (y) = 0 ) g (y) =C;thus

v (x; y) = y(x2 +y2)

+C

In paranthesis we remark that the analytic funktion f(z) =u +iv is givenby

f(z) = x

x2 +y2+i

y(x2 +y2)

+C

=

z

jzj2 +iC= z

zz+iC=

1

z+iC:

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II.5.2

Show that if v is a harmonic conjugate for u, then uis a harmonicconjugate for v.

SolutionIff(z) =u (x; y) +iv (x; y) is analytic thenif(z) =v (x; y) iu (x; y) isanalytic. Thusu is a harmonic conjugate ofv .

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II.5.3

Dene u (z) = Im (1=z2

) forz6= 0, and set u (0) = 0.(a) Show that all partial derivatives of u with respect to x existat all points of the plane C, as do all partial derivative of u withrespect to y.(b) Show that @

2u@x2 +

@2u@y2 = 0.

(c) Show that u is not harmonic on C.(d) Show that @

2u@x@y does not exist at (0; 0).

Solution(a)

We rst dene

u (z) =

81,f(z)maps the circlefjzj =rg onto an ellipse, and that f(z) maps the circlefjzj = 1=rgonto the same ellipse. Show thatf(z)is one-to-one on the exteriordomain D= fjzj >1g. Determine the image ofDunder f(z). Sketchthe images under f(z) of the circlesfjzj =rg for r >1, and sketchalso the images of the parts of the raysfarg z = g lying in D.

Solution

f(z) =z + 1=z,f0 (z) = 1 1=z2,f0 (z) = 0at z 2 = 1,z = 1. Not denedatz = 0, not conformal atz = 1. Iff(z) =, thenz =

p2 4

=2.

Ifz =ei, then w = u+iv = cos +i sin + cos = i sin =, whichcan be written u2= (+ 1=)2 +v2= ( 1=)2 = 1, i.e. image is an ellipse,replace by1= and get same ellipse. Sincef(D) =f(D), and f(z) is atmost twotoone, f(z) is onetoone on D. Since f(z) maps @D onto theinterval[2; 2], andf(z)maps onto C, the image ofD is Cn [2; 2].

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II.6.7

For the function f(z) =z + 1=z =u + iv, sketch the families of levelcurves of u and v. Determine the images under f(z) of the tophalf of the unit disk, the bottom half of the unit disk, the partof the upper half-plane outside the unit disk, and the part of thelower half-plane outside the unit disk. Hint. Start by locatingthe images of the curves where u = 0, where v = 0, and wherev= 1. Note that the level curves are symmetric with respect to thereal and imaginary axis, and they are invariant under the inversionz7! 1=z in the unit circle.

Solution

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II.6.8 ( From Hints and Solutions)

1 2 3 P L K

Consider f(z) =z +ei=z, where 0< < . Determine where f(z)is conformal and where it is not conformal and where it is notconformal. Sketch the images underf(z)of the unit circlefjzj = 1gand the intervals(1; 1]and[+1; +1)on the real axis. Show thatw =f(z) mapsfjzj >1g conformally onto the complement of a slitplane in the w plane. Sketch roughly the images of the segmentsof rays outside the unit circlefarg z= ; jzj 1g under f(z). Atwhat angels do they meet the slit, and at what angles do theyapproach

1?

Solutionf(z) = z +ei=z , f0 (z) = 1 ei=z2, f0 (z) = 0 at z2 = ei. Not de-ned at z = 0, not conformal at z = ei=2. The expression f(z) =ei=2

ei=2z+ 1=

ei=2z

, shows that f(z) is a composition of the ro-

tation by=2, the function of Exercises 6 and 7, and a rotation by =2.Thusf(z)maps fjzj >1g onetoone onto the complement rotate of[2; 2]by=2.

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II.6.9

Letf=u + iv be a continuously dierentiable complex-valued func-tion on a domain D such that the Jacobian matrix of f does notvanish at any point ofD. Show that if fmaps orthogonal curves toorthogonal curves, then eitherf or fis analytic, with nonvanishingderivative.

SolutionSupposef(0) = 0andux6= 0 at0. The tangents in the orthogonal directions(1; t) and (t; 1) are mapped to the tangents in the directions (ux; vx) +t (uy; vy) andt (ux; vx) + (uy; vy).The orthogonality of these directions for all t gives

h(ux; vx) +t (uy; vy) ; t (ux; vx) + (uy; vy)i = 0;can be rewritten

h(ux; vx) ; (uy; vy)i+t (h(uy; vy) ; (uy; vy)i h(ux; vx) ; (ux; vx)i)t2 h(uy; vy) ; (ux; vx)i = 0:The systems of equations

(1)

(2) uxuy+vxvy = 0

u2

y+v2

y (u2

x+v2

x) = 0most hold to get orthogonality. Put (1) into (2) and get

u2x

u2y+v2y u2x v2x

= v2xv

2y+u

2xv

2y u4x u2xv2x=

=v2x

v2y u2x

+u2x

v2y u2x

=

v2x+u2x

v2y u2x

= 0:

Ifux6= 0we are lead to ux= vy anduy = vx.Tangent to curves ! (s;ts)is tangent to image curves + (u (s;ts) ; v (s;ts))(1; 0)! (ux; vx), (0; 1)! (uy; vy), orthogonal) (uy; vy) = C(vx; uy) forsome. A ????? gives C=

1, same sign holds in ?????.

Eitherux =vy, uy =vx) conformal in neighborhood of0, orux =vy,uy =vx) anti conformal in neighborhood of0. ux= vy anduy = vx.So Cauchy Riemann is satised for either f or f.

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II.7.1

Compute explicitly the fractional linear transformations determinedby the following correspondences of triples(a) (1 +i; 2; 0) 7! (0; 1; i 1) (e) (1; 2; 1) 7! (0; 1; 1)(b) (0; 1; 1) 7! (1; 1 +i; 2) (f) (0; 1; i) 7! (0; 1; 1)(c) (1; 1 +i; 2) 7! (0; 1; 1) (g) (0; 1; 1) 7! (0; 1; i)(d) (2; i; 2) 7! (1 2i; 0; 1 + 2i) (h) (1; i; 1) 7! (1; 0; 1)

Solution

(w w0) (w1 w2)(w w2) (w1 w0)=

(z z0) (z1 z2)(z z2) (z1 z0)

(a)We have the points

z0= 1 +i w0 = 0z1= 2 w1 = 1z2= 0 w2 = i 1

The mapping is

(w w0) (1 w2=w1)(w w2) (1 w0=w1) =

(z z0) (z1 z2)(z z2) (z1 z0) ;

take limit as w1! 1, to obtainw 0

w (i 1)=(z (1 +i))(2 0)(z 0)(2 (1 +i))) w=

2iz+ (2 2i)z 2 :

(b)We have the points

z0= 0 w0= 1z1= 1 w1= 1 +iz2= 1 w2= 2

The mapping is(w w0) (w1 w2)(w w2) (w1 w0)=

(z z0) (z1=z2 1)(z=z2 1) (z1 z0) ;

take limit as z2! 1, to obtain

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(w

1) ((1 +i)

2)

(w 2) ((1 +i) 1)=z

0

1 0) w=2z

(1 +i)

z (1 +i) :(c)We have the points

z0= 1 w0 = 0z1= 1 +i w1 = 1z2= 2 w2 = 1

The mapping is

(w

w0) (w1=w2

1)

(w=w2 1) (w1 w0)=(z=z0

1) (z1

z2)

(z z2) (z1=z0 1) ;take limit as z0; w2! 1, to obtain

w 01 0 =

(1 +i) 2z 2 ) w=

i 1z 2 :

(d)We have the points

z0= 2 w0 = 1 2iz1= i w1 = 0

z2= 2 w2 = 1 + 2iThe mapping is

(w w0) (w1 w2)(w w2) (w1 w0)=

(z z0) (z1 z2)(z z2) (z1 z0) ;

obtain

(w (1 2i))(0 (1 + 2i))(w (1 + 2i))(0 (1 2i))=

(z (2))(i 2)(z 2) (i (2))) w= iz + 1:

(e)

We have the points

z0 = 1 w0= 0z1 = 2 w1= 1z2 = 1 w2= 1

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The mapping is

(w w0) (w1=w2 1)(w=w2 1) (w1 w0)=

(z z0) (z1=z2 1)(z=z2 1) (z1 z0) ;

take limit as z2; w2! 1, to obtainw 01 0 =

z 12 1) w= z 1:

(f)We have the points

z0 = 0 w0= 0

z1 = 1 w1= 1z2 = i w2= 1

The mapping is

(w w0) (w1=w2 1)(w=w2 1) (w1 w0)=

(z z0) (1 z2=z1)(z z2) (1 z0=z1) ;

take limit as z1; w2! 1, to obtainw 01 0 =

z 0z i) w=

z

z i(g)We have the points

z0 = 0 w0= 0z1 = 1 w1= 1z2 = 1 w2= i

The mapping is

(w w0) (1 w2=w1)(w w2) (1 w0=w1)=

(z z0) (z1=z2 1)(z=z2 1) (z1 z0) ;

take limit as z2; w1! 1, to obtainw 0w i =

z 01 0) w=

iz

z 1 :

(h)

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We have the points

z0= 1 w0= 1z1= i w1= 0z2= 1 w2= 1

The mapping is

(w w0) (w1 w2)(w w2) (w1 w0)=

(z z0) (z1 z2)(z z2) (z1 z0) ;

obtain

(w

1)(0

(

1))

(w (1))(0 1)=(z

1) (i

(

1))

(z (1))(i 1)) w= iz+ 1

z+i :

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II.7.2

Consider the fractional linear transformation in Exercise 1a above,which maps 1 +i to0, 2 to1, and 0 to i 1. Without referring toan explicit formula, determine the image of the circlefjz 1j = 1g,the image of the diskfjz 1j


168/717

II.7.3

Consider the fractional linear transformation that maps 1 to i, 0to 1 +i,1 to 1. Determine the image of the unit circlefjzj = 1g,the image of the open unit diskfjzj


169/717

II.7.3 z-plane II.7.3 w-plane

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II.7.4

Consider the fractional linear transformation that maps1 toi,1 to2i, and i to0. Determine the image of the unit circlefjzj = 1g,the image of the open unit diskfjzj


171/717


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II.7.5

What is the image of the horizontal line through i under the frac-tional linear transformation that interchanges0 and1 and maps 1to 1 +i? Illustrate with a sketch.

SolutionWe have the points

z0 = 0 w0 = 1z1 = 1 w1 = 0z2 = 1 w2 = 1 +i

The mapping is

(w w0) (w1 w2)(w w2) (w1 w0)=

(z z0) (z1 z2)(z z2) (z1 z0) ;

obtain

(w 1)(0 (1 +i))(w (1 +i))(0 1)=

(z 0)(1 (1))(z (1))(1 0)) w=

iz iz i

The tripple(0; 1; 1)on the real axis is mapped to the points(1; 0; 1 +i)inthew plane, it must be so that the real axis in the z plane is mapped ontothe circle

w 12 + 12 i= p22 in thew plane, since tree points determinesthe "circle". The real axis and the line through i have one common point atinnity. The common point at innity for both real axis and the line throughimust be mapped to the same point in the w plane, we have that the pointisw (1) =i. The two lines in the z plane is parallell, thus the images ofthe lines must go through i in the w plane and be parallell in that point,thus the mapping of the horizontal line through i must be the line throughi, which is tangent line to circle at i, i.e. Im z = Re z+ 1. Remark that theimage can not be a circle because w (i) = 1, so the image of the horizontalline must contain the point at innity.

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II.7.6

Show that the image of a straight line under the inversion z7! 1=zis a straight line or circle, depending on whether the line passesthrough the origin.

SolutionThe image of the straight line ax+by= cis (kontrollera xiyi lsningshftet)

w=1

z =

1

zz =

z

jzj2 = (x iy) 1

r2

whereu = x=r2 andv =

y=r2. Get

au bv= ax+byr2

= c

r2 =

c

x2 +y2 =c

u2 +v2

The image is the solution ofc (u2 +v2) au+bv= 0.Ifc = 0we have that

au+bv= 0;this is a straight line through 0.Ifc 6= 0we have

c u2 +v2 au+bv= 0;which can be rewritten as

u a

2c

2+

v+

b

2c

2=

0@sa

2c

2+

b

2c

21A2 ;it is a circle through 0.

Solution (K. Seip)Setf(z) = 1z . Then

1 2f(S)

,0

2S, which proves the claim.

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II.7.7

Show that the fractional linear transformation f(z) = (az+b) = (cz+d)is the identity mapping z if and only if b= c = 0 anda= d 6= 0.


az+b

cz+d=z, az+b= cz2 +dz, cz2 + (d a) z b= 0:

Put in values onz for intance 0 and1the expession becomes

(1)(2)(3)

c (d a) = 0; z= 1:b= 0 z= 0c+ (d a) = 0; z= 1:

From (2) we haveb = 0 and if we add (1) and (3) we have c = 0.Thus the fractional linear transformationf(z)is the identity mapping if andonly ifb =c = 0,a = d 6= 0.

SolutionSet R= R[f1g. Supposef(z) = az+bcz+d maps R to R. Then0 7! bd2 R,1 7! a

c2 R, 0 - ba2 R,1 - dc2 R. At least one of theese rations

is dierent from0 and1

, and since one ofa; b; c; dcan be chosen freely, theresult follows.

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II.7.8

Show that any fractional linear transformation can be representedin the form f(z) = (az+b) (cz+d), where ad bc= 1. Is this repre-sentation unique?

SolutionIf

w= (z+) (z +)

divide each coecient by the square root of

to obtain representation with

ad bc= 1:Representation is not unique, as can multiply all coecients by1.

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II.7.9

Show that the fractional linear transformations that are real onthe real axis are precisely those that can be expressed in the form(az+b) = (cz+d), where a;b; c;andd are real.

Solution (K. Seip)Set R = R[f1g. Supposef(z) = az+b

cz+dmaps Rto R. Then0 7! b

d2 R,

1 7! ac2 R, 0- ba 2 R,1 - dc 2 R. At least one of these is

dierent from 0 and1, and since one ofa;b;c;d can be chosen freely, theresult follows.

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II.7.9

Show that the fractional linear transformations that are real onthe real axis are precisely those that can be expressed in the form(az+b) = (cz+d), where a;b; c;andd are real.

Solution (A. Kumjian)Set R= R [f1g, note that a fractional linear transformation f : C! Cis real on the real axis if(R) R (sincef is continuous). ForA = a bc dwithdet A6= 0 we dene the fractional linear transformation fA : C! Cby the formula fA(z) = (az+b) = (cz+d). It remains to prove that given afractional linear transformation f, we have f(R) R if =fA for some2 2 matrix A with real entries.Letf : C! C be a fractional linear transformation. Suppose that f=fAwhere A is a 2 2 matrix with real entries a; b; c andd as above. Then forx 2 R we have

f(x) =fA(x) =ax+b

cx+d2 R:

This is clear forx 2 Rbut requires a little work for x = 1. We have

f(1) = limz!1

az+b

cz+d=

a=c if c 6= 0;1 if c= 0:

Hence,f(R) R.Conversely, suppose thatf(R) R, we must show thatf=fAfor some22matrix with real entries. IfAis a 22matrix (with non-zero determinant),thenfA1 = (fA)

1. Hence,f=fA for some2 2matrixA with real entriesi its inversef1 has the same property.By the unlikeness part of the theorem on page 64,fis completely determinedby the three extended real numbers x0 =f(0), x1 =f(1) andx1 =f(1)(recall that we have assumed f(R) R). So it suces to show thatthe fractional linear transformation g for which g (x0) = 0, g (x1) = 1 andg (x1) =1 has the requisite properties, because we must have g = f1

(again by the uniqueness part of the theorem).There are four cases to consider. The rst case is when x0; x1; x16= 1 andthe remaining three cases are x0 =1; x1 =1andx1 =1. The formulaforg (z)in the four cases is given as follows (in the rst case k = x1x1

x1x0 2 R):

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k

x

x0

x x1 if x0; x1; x16= 1; x1

x1x x1 if x0= 1;

x x0x x1 if x1 = 1;

x x0x1 x0 if x1= 1:

In each caseg (z)has the desired form. Hence,f=fAfor some2 2matrixA with real entries as required.

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II.7.10

Suppose the fractional linear transformation (az+b) = (cz+d)mapsR to R, and ad bc= 1. Show thata;b; c; andd are real or they areall pure imaginary.

SolutionBecause R is mapped on R and the orientation is perserved, f(x) will beeither increasing or decreasing.

Case 1 : Supposef(x) is increasing, then f0 (x) = 1= (cx+d)2 > 0 for allx ) c; dare real. f(z) = (az+b) = (cz+d), thenax + bis real for all x 2 R,soa; bare also real.

Case 2 : Supposef(x) is decreasing, then f0 (z) = 1= (cz+d)2 < 0 for allx ) c; dare pure imaginary. f(z) = (az+b) = (cz+d), thenax + bare pureimaginary for all x 2 R, soa; bare also pure imaginary.

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II.7.11

Two mapsf andg are conjugate if there is h such thatg = hfh1

.Here the conjugating map h is assumed to be one-to-one, withappropriate domain and range. We can think of f and g as the"same" map, after the change of variable w = h (z). A point z0 isa xed point of f if f(z0) = z0. Show the following. (a) If f isconjugate to g, then g is conjugate to f.(b) If f1 is conjugate to f2 andf2 tof3, then f1 is conjugate to f3.(c) If f is conjugate to g, then f f is conjugate to g g, and moregenerally, the mfold compositionf f (m times) is conjugateto g g (m times).(d) If f andg are conjugate, then the conjugating function h maps

the xed points of f to xed points of g. In particular, f and ghave the same number of xed points.

Solution(a)Iffis conjugate to g then

g= h f h1 ) f=h1 g hthusg is conjugate to f

(b)Iff1 = h f2 h1 andf2= g f3 g1 we get

f1= h g f3 g1 h1 = (h g) f3 (h g)1 :(c)

g= h f h1thenm timesz }| {

g g : : : g==h f h1 h f h1 : : : h f h1 =

=h m timesz }| {

f f : : : f h1

(d)g= h f h1,f(z0) =z0, andw0 = h (z0), then

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g (w0) =h fh1 (w0)= h (f(z0)) = h (z0) =w0:We have thatfandg have the same number of xed points since every pointofg is mapped to a xed point offby the functionh1.

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II.7.12

Classify the conjugacy classes of fractional linear transformationsby establishing the following(a)A fractional linear transformation that is not the identity has either1 or 2 xed points, that is, points satisfying f(z0) =z0.(b)If a fractional linear transformationf(z)has two xed points, thenit is conjugate to the dilationz7! az witha 6= 0,a 6= 1, that is, thereis a fractional linear transformation h (z)such thath (f(z)) =ah (z).Isa unique? Hint. Consider a fractional linear transformation thatmaps the xed points to 0 and

1.

(c)If a fractional linear transformation f(z) has exactly one xedpoint, then it is conjugate to the translation 7! + 1. In otherwords, there is a fractional linear transformation h (z) such thath (f(h1 ())) =+ 1, or equivalently, such that h (f(z)) = h (z) + 1.Hint. Consider a fractional linear transformation that maps thexed point to1.

Solutiona) az+bcz+d =z, az+b =cz 2 +dz, cz2 + (d a) z b = 0. If this is = 0for every z, we get w (z) = z. It is not identically equall to zero, so it haseither 1 or 2 nite solutions. Ifc6= 0, it has 2 solutions possibly one withmultiplicity and1 is not a xed point. Ifc = 0, it has one nite solutionz1 = b= (d a), andz2 = 1 is a xed point, so they are two.b) Make a change of variables h (z) = (z+) = (z +) that maps xedpoints to 0 and1 . Then h f h1 is FLT with xed points at 0 and1.(h f h1) (w) =aw for some a6= 0, a6= 1. f is conjugate to a dilationSuppose aw is conjugate to Aw, i.e. 9h, (h (ah1)) (w) = Aw, h (az) =Ah (z). h must map xed points to xed points, so either h (z) = cz orh (z) =c=z . So eitherh (z) =cz) h (az) =caz =Ah (z) =Acz) A = aorh (z) =c=z) h (az) =c= (az) =Ah (z) =Ac=z) A = 1=a. Thusa isnot unique.c) Supposeh maps the xed points to 1, thenh f h1 has only one xedpoint at 1, so(h f h1) (w) =aw+b. Since this has no nite xed points,aw+b= w has no solutions anda= 1,b 6= 0. Thus(h f h1) (w) =w +b.

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Now w+b is conjugate to w+ 1, by a dilation g (w) = Aw. w! w=A!w=A+b ! w+Ab, takeA = 1=b. So

g h f h1 g1 (w) =w+ 1 ()() (g h) f (g h)1 (w) =w+ 1 ()

() ((g h) f) (z) = (g h) (z) + 1:

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III 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1912345678

1


186/717

III.1.1

Evaluate Ry2dx+ x2dy along the following paths from (0; 0) to(2; 4),(a) the arc of the parabola y= x2,(b) the horizontal interval from (0; 0) to (2; 0), followed by the ver-tical interval from(2; 0) to(2; 4), (c) the vertical interval from (0; 0)to (0; 4), followed by the horizontal interval from (0; 4) to(2; 4).

Solution(a)

Z

y2dx+x2dy=

x= t dx=dt; 0 t 2y= t2 dy= 2t dt

=

Z 20

t22

dt+

Z 20

t22t dt=72

5:

(b)

Z1

y2dx+x2dy=

x=t dx=dt 0 t 2y= 0 dy= 0 dt

=

Z 20

02dt+

Z 20

t2 0 dt= 0

Z2

y2dx+x2dy=

x= 2 dx= 0 dt 0 t 4y= t dy= dt

=

Z 40

t2 0 dt+

Z 40

22dt= 16

Z

y2dx+x2dy=Z

1

y2dx+x2dy+Z

2

y2dx+x2dy= 16:

(c)

Z1

y2dx+x2dy=

x= 0 dx= 0 dx 0 t 4y= t dy= dt

=

Z 40

y2 0 dt+

Z 40

02dt= 0

Z2

y2dx+x2dy=

x=t dx=dt 0 t 2y= 4 dy= 0dt

=

Z 20

42dt+

Z 20

t2 0 dt= 32

Z

y2dx+x2dy=Z

1

y2dx+x2dy+Z

2

y2dx+x2dy= 32:

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187/717

III.1.2

Evaluate Rxy dx both directly and using Greens theorem, whereis the boundary of the square with vertices at (0; 0), (1; 0), (1; 1),and(0; 1).

Solution (A. Kumjian)

1

1

1234

VII.1.2

Denote the square by D and note that xy dx = P dx+Q dy where P =xyandQ = 0. ThenP andQare continuously dierentiable on Dand=@D,hence by Greens Theorem we have,

Z

xy dx= Z@D

P dx+Q dy= ZZD

@Q@x

@ Pdy

dxdy= Z 10

Z 10

xdxdy =12

:

Observe that = 1 + 2+ 3+ 4 where 1 and 3 are the bottom andtop of the square while 2 and4 are the last two sides taken in the orderindicated by the order of the vertices in the statement of the problem (so theboundary is oriented counter clockwise). Note that the path integrals on 2and4 are zero because the edges are vertical.

Z xy dx= Z1 xy dx+ Z3 xy dx= Z 1

0

x 0 dx Z 1

0

x 1 dx= 0 1

2

= 1

2

:

3


188/717

III.1.3

Evaluate R@Dx2dy both directly and using Greens theorem, whereD is the quarter-disk in the rst quadrant bounded by the unitcircle an the two coordinate axes.

SolutionEvaluate

R@D

x2dy directly, set @ D= 1+2+3.

Z1

x2dy=

x= t dx= dt 0 t 1y= 0 dy= 0 dt

=

Z 10

t2 0 dt= 0

Z2

x2dy= x= cos t dx= sin t dt 0 t =2

y= sin t dy= cos t dt

=

Z =20

cos2t cos t dt= 23Z

3

x2dy=

x= 0 dx= 0 dt 0 t 1y= 1 t dy= dt

=

Z 10

02dt= 0

Z@D

x2dy=

Z1

x2dy+

Z2

x2dy+

Z3

x2dy=2

3

Now we evaluateR@D

x2dy, this time using Greens theorem. In this case,P(x; y) = 0andQ (x; y) =x2.

Z@D

x2dy=

ZZD

2xdxdy=

= 2

ZZD

xdxdy=

x=r cos dx dy= r dr d 0 r 1y= r sin 0 =2

= 2

Z 10

Z =20

r cos rdrd

= 2

Z 10

r2dr

Z =20

cos d =2

3:

4


189/717

III.1.4

Evaluate Ry dx both directly and using Greens theorem, where is the semicircle in the upper half-plane from R toR.

SolutionEvaluate

Ry dxdirectly, set =1+2.

Z1

y dx=

x= t dx=dt R t Ry= 0 dy= 0 dt

=

Z RR

0 dt = 0

Z2 y dx= x= R cos t dx= R sin t dt 0 t

y= R sin t dy= R cos t dt = Z

0

R sin t R sin t dt=

Z@D

x2dy=

Z1

x2dy+

Z2

x2dy= R2

2

Now we evaluateRy dx, this time using Greens theorem. In this case,

P(x; y) =y andQ (x; y) = 0.

Z@D

y dx=

ZZD

dxdy=

= ZZD

dxdy= x=r cos dx dy= r dr d 0 r Ry = r sin 0

= Z R0

Z 0

rdrd=

=

Z R0

r dr

Z 0

d= R2

2 :

5


190/717

III.1.5

Show that R@Dx dy is the area of D, while R@Dy dx is minus the areaof D.

SolutionUsing in the two casesFor

R@D

x dy, we have P = 0 and Q = x, using Greens theorem we haveZ@D

x dx=

ZD

Z dxdy= Area D:

For R@Dy dx, we have P =y andQ = 0, using Greens theorem we haveZ@D

y dx=

ZD

Z dxdy= Area D:

6


191/717

III.1.6

Show that if P andQ are continuous complex-valued functions ona curve , then

R

P dx

z w+

R

Qdy

z w; (z= x+iy)

is analytic for w 2 Cn. ExpressF0 (w) as a line integral over .

SolutionDierentiate by hand, use uniform convergence of

1

w

1

z (w+ w)

1

z w

=

1

(z w) (z (w+ w))!

1

(z w)2

asw ! 0, uniformly forz 2 . Get

F0 (w) =R

P dx

(z w)2+

R

Qdx

(z w)2; z= x+iy

7


192/717

III.1.7

Show that the formula in Greens theorem is invariant under coor-dinate changes, in the sense that if the theorem holds for a boundeddomain U with piecewise smooth boundary, and if F(x; y) is asmooth function that mapsUone-to-one onto another such domainVand that maps the boundary of U one-to-one smoothly onto theboundary ofV, then Greens theorem holds for V. Hint. First notethe change of variable formulae for line and area integrals, givenby

Z@V P d = Z@U (P F)@

@x

dx+@

@y

dy ;ZZV

R d d =

ZZU

(R F)detJFdxdy;

where F(x; y) = ((x; y) ; (x; y)), and where JF is the Jacobian ma-trix of F. Use these formulae, with R = @P=@. The summandR

Q d is treated similarly.

Solution

ZZV

@P

@d d=ZZU

@P@ ((x; y) ; (x; y))det(JF) dxdy=F(x; y) = ((x; y) ; (x; y))

@F@x =

@@x +

@@x

@F@y =

@@y +

@@y

det JF(x; y) = @

@x

@

@y

@

@y

@

@x

=

=

ZZU

@P

@ ((x; y) ; (x; y))

@

@x

@

@y

@

@y

@

@x

dxdy

Using Greens theorem

8


193/717

Z@V

P d=

=

Z@U

(P F)

@

@xdx+

@

@ydy

=

Z@U

P((x; y) ; (x; y))

@

@xdx+

@

@ydy

=

=

ZZU

@

@y

P

@

@x

+

@

@x

P

@

@y

dxdy=

=

ZZU

@

@yP((x; y) ; (x; y))

@

@x P

@2

@y@x+

@

@xP((x; y) ; (x; y))

@

@y+P

@2

@x@y

dxd

= ZZU

@P@@

@y @ P

@

@

@y@

@x +@P

@

@

@x +@ P

@

@

@x@

@y dxdy=

=

ZZU

@P

@

@

@y

@

@x+

@ P

@

@

@x

@

@y

dxdy=

=

ZZU

@P

@ JFdxdy:

Similar argument as above hold forR

QdHere we use that

Z@V

Qd=

Z@U

(Q F)

@

@xdx+

@

@ydy

and replace R = @P=@ by R=@Q=@to conclude thatZ@V

Qd=

ZZV

@Q

@dd:

9


194/717

III.1.8

Prove Greens theorem for the rectangle dened by x0< x < x1 andy0 < y < y1 (a) directly, and (b) using the result for triangles.

Solution

I

II

III

IV R

VII.1.8a

I

II

III

IV ST

VII.1.8b

(a)Make gure and set the linesegment between (x0; y0) and(x1; y0) to1 andproceed in this way in counterclockwise direction. We haveZ

1

Q dy=

Z2

P dx=

Z3

Q dy=

Z4

P dx= 0:

Integrate around the rectangle

10


195/717

ZZR

@Q

@x

@ P

@y

dxdy=

=

Z y1y0

Z x1x0

@Q

@xdx

dy

Z x1x0

Z y1y0

@P

@ydy

dx=

=

Z y1y0

[Q (x1; y) Q (x0; y)] dy

Z x1x0

[P(x; y1) P(x; y0)] dx =

=

Z y1y0

Q (x1; y)dy

Z y1y0

Q (x0; y)dy

Z x1x0

P(x; y1)dx +

Z x1x0

P(x; y0) dx =

= Z x1

x0

P(x; y0)dx+Z

y1

y0

Q (x1; y)dy Z

x1

x0

P(x; y1) dxZ

y1

y0

Q (x0; y)dy =

=

Z1

P dx+

Z2

Q dy+

Z3

P dx+

Z4

Q dy=

=

Z@R

(P dx+Q dy) :

(b)Write R = S[ T. The integrals on the diagonal are in dierent directionsand will cancel

Z@R

(P dx+Q dy) =

=

Z@S

(P dx+Q dy) +

Z@T

(P dx+Q dy) =

=

ZZS

@Q

@x

@ P

@y

dxdy+

ZZT

@Q

@x

@ P

@y

dxdy=

=

ZZR

@Q

@x

@ P

@y

dxdy:

11


196/717

III.2.1

Determine whether each of the following line integrals is indepen-dent of path. If it is, nd a function h such that dh= P dx+Q dy.If it is not, nd a closed path around which the integral is notzero. (a)x dx+y dy, (b) x2dx+y5dy, (c) y dx+x dy, (d) y dx x dy.

SolutionWe compute the partial derivatve because if h exist we have that dh =P dx+Qdy where @P=@y= @Q=@x.(a)

P =x Q=y@P@y = 0

@Q@x = 0R

P dx= x2

2 +g (y)

RQdy= y

2

2 +f(x)

) h= x2+y2

2

(b)

P =x2 Q=y5@P@y = 0

@Q@x = 0R

P dx= x3

3 +g (y)

RQdy= y

6

6 +f(x)

) h= 2x3+y6

6

(c)

P =y Q= x@P@y = 1

@Q@x = 1R

P dx=xy +g (y)R

Qdy= xy +f(x)) h= xy

(d)

P =y Q= x@P@y = 1

@Q@x = 1

The line integral is not independent of path, we use Greens theoremZjzj=1

(y dx x dy) =

ZZjzj


197/717

III.2.2

Show that the dierentialydx+xdy

x2 +y2 ; (x; y) 6= (0; 0) ;

is closed. Show that it is not independent of path on any annuluscentered at 0.

Solution

P = yx2+y2

Q= xx2+y2

@P@y =

(x2+y2

)+y(2y)

(x2+y2)2 = y2

x2

(x2+y2)2 @Q@x =(x2+y2

)x(2x)

(x2+y2)2 = y2

x2

(x2+y2)2

Because @P=@y = @Q=@x;the dierential is closed and we can use GreensTheorem

Ijzj=r

P dx+Q dy=

=

Ijzj=r

y

x2 +y2dx+

x

x2 +y2dy=

1

r2

Ijzj=r

y dx+x dy=

= 1

r2

ZZjzj


198/717

III.2.3

SupposeP andQ are smooth functions on the annulus fa < jzj < bgthat satisfy @P=@y =@Q=dx. Show directly using Greens theoremthat

Hjzj=r

P dx+Qdy is independent of the radius r, for a < r < b.


1

2 D

III.2.3

Letr1; r2 be given so thata < r1 < r2 < b. We must prove thatIjzj=r1

P dx+Qdy =

Ijzj=r2

P dx+Qdy:

Let D denote the annulus fz 2 C :r1 < jzj < r2g. Observe that D is a

bounded domain with piecewise smooth boundary @D = 1 [ 2 where 1denotes the circle fz 2 C : jzj =r1g with clockwise orientation while 2 de-notes the circlefz2 C : jzj =r2gwith counter clockwise orientation. More-over, both P andQ are continuously dierentiable on D= D [ @D. Hence,Greens Theorem applies and we obtain

Z@D

P dx+Qdy=

ZZD

@Q

@x

@ P

@y

dxdy= 0;

and since, as noted above, @D consists of the two circles with opposite ori-entation,

Z@D

P dx+Qdy=Ijzj=r2

P dx+Qdy Ijzj=r1

P dx+Qdy:

Hence, equation()holds andHjzj=rP dx + Qdy is independent of the radius

r, fora < r < b.

14


199/717

III.2.4

Let P andQ be smooth functions on D satisfying @P=@y =@[email protected] and1 be two closed paths in D such that the straight linesegment from 0(t) to 1(t) lies in D for every parameter valuet. Then

R0

P dx+Q dy =R1

P dx+Q dy. Use this to give another

solution to the preceding exercise.

SolutionUse the theorem on page 81. Use the straight lines to deform0to1. Denes(t) =s1(t) + ( 1 s) 0(t),0 6 s 6 1,a1 6 t 6 b1. The theorem on page81 applies. Obtain the result for the annulus above by parameterizing the

circles jzj =r0and jzj =r1by0(t) =r0e2it,1(t) =r1e2it,0 6 t 6 1. Thestraight line segments joining 0(t) to1(t) are radical are in the annulus,so the rst part applies.

15


200/717

III.2.5

Let 0(t)and 1(t),0 t 1, be paths in the slit annulus fa < jzj < bg n (b; a)from A to B. Write down explicitly a family of pathss(t) from Ato B in the slit annulus that deforms continuously to 1.Suggestion. Deform separately the modulus and the principal valueof the argument.

SolutionWrite 0(t) = r0e

i0(t), 0 6 t 6 1 , when < 0(t) < , r0(t), 0(t)continuous. (Use the fact that (t) = Arg 0(t) is continuous on the slitannulus) Also1(t) =r1e

i1(t).

Then consider

s(t) = [sr1(t) + (1 s) r0(t)] ei[s1(t)+(1s)0(t)]; 0 s 1:

This does the trick, it deform 0 to 1 continuously in D, and each s is apath in D fromA to B .We see that s(t) belongs to the slit annulus for 0 s 1 and0 t 1since for every0 5 1 we have

a


201/717

III.2.6

Show that any closed path (t),0 t 1, in the annulus fa < jzj < bgcan be deformed continuously to the circular path (t) =(0) e2imt,0 t 1, for some integer m. Hint. Reduce to the case wherej(t)j j(0)j is constant. Then start by nding a subdivision0 = t0 < t1


202/717

III.2.7

Show that if 0 and1 lie in dierent connected components of thecomplement CnD of D in the extended complex plane, then thereis a closed path in D such that

Rd 6= 0. Hint. The hypothesis

means that there are > 0and a bounded subset E of CnD suchthat0 2 E; and every point of Ehas distance at least5from everypoint of CnD not in E. Lay down a grid of squares in the planewith side length ;and let Fbe the union of the closed squares inthe grid that meet Eor that border on a square meeting E. Showthat@Fis a nite union of a closed paths inD, and that

R@Fd = 2.

SolutionProceed as in the hint. Assume that 0 is the centre of one of the squaresS0 in the grid. Then

R@S0

d = 2, whileR@S d = 0 for any other square in

the grid. ThusPR

@Sjd= 2, where we sum over squares. S0; S1; : : : ; S n

in F. If two squares in Fare ?????, the corresponding integrals along theedge cancel. Then this is the sum over the integrals over the edges of theSj s that have Fon one side and CnFon the other, that is the edges thatform@F. We have thus

R@F

d = 2. Note that by construction, @F D.Now note how edges @F can meet. Either a vertex in @F have one edgecoming in and one out, or it have two coming in and two out. By starting on

following edges, and making a left turn at vertices where 4 edges meet, wemust eventually end where we started, with a closed path. Three paths mustbe disjoint, except for vertices with four edges. Call the paths1; : : : ; m.Since

PRj

d= 2, we must haveRd 6= 0 for one of the j s. (In fact,

we will haveRj

d= 2 on thej s we get by moving from0 to@ F.)

18


203/717

III.3.1

For each of the following harmonic functions u, nddu, nddv, andndv, the conjugate harmonic functions of u.(a) u (x; y) =x y (c) u (x; y) = sinh x cos y(b) u (x; y) =x3 3xy2 (d) u (x; y) = yx2+y2

Solution(a)Use Cauchy-Riemanns equations page 83.

u (x; y) =x ydu=dx dy

dv= dx+dyv (x; y) =x+y

(b)Use Cauchy-Riemanns equations page 83.

u (x; y) =x3 3xy2

du= (3x2 3y2)dx 6xy dydv= 6xy dx+ (3x2 3y2) dyv (x; y) = 3x2y y3

(c)

Use Cauchy-Riemanns equations page 83.

u (x; y) = sinh x cos ydu= cosh x cos y dx sinh x sin y dydv= sinh x sin y dx+ cosh x cos y dyv (x; y) = cosh x sin y

(d)Use Cauchy-Riemanns equations page 83.

u (x; y) = yx2+y2

du= 2xy

(x2+y2)2 dx+ x2+y2y(2y)

(x2+y2)2 dy= 2xy

(x2+y2)2 dx+ x2y2

(x2+y2)2 dydv= y

2x2

(x2+y2)2dx 2xy

(x2+y2)2dy

v (x; y) = xx2+y2

19


204/717

III.3.2

Show that a complex-valued functionh (z)on a star-shaped domainD is harmonic if and only if h (z) =f(z) + g (z), wheref(z) and g (z)are analytic on D.

SolutionWrite h = u+iw where u, w are harmonic. By the Theorem on page 83both u andw have harmonic conjugate in D, (since D is star-shaped), sothus there are ', analytic such that u = Re ',w = Re .

u= ('+') =2; w= + =2;givesh= 1

2

'+'+i+i

= 1

2['+i] + 1

2

'+i

:

Take

f= 12['+i] ; g= 12[' i] ;

thenh (z) =f(z) +g (z).

To show the oposite direction. Assume thath (z) =f(z) + g (z)wheref(z)

and g (z) are analytic on D. Then h = Re f+ Re g+ i (Im f+ Im g) andfrom Cauchy-Riemanns equations follows that

@2 (Re f)

@x2 +

@2 (Re g)

@x2 +i

@2 (Im f)

dx2

@2 (Im g)

@x2

+

+@2 (Re f)

@y2 +

@2 (Re g)

@y2 +i

@2 (Im f)

@y2

@2 (Im g)

@y2

=

=@2 Im f

@x@y +

@2 Im g

@x@y +i

@2 Re f

@x@y +

@2 Re g

@x@y

@2 Im f

@y@x

@2 Im g

@y@x +

+i@2 Re f@y@x

@2

Re g@y@x

= 0soh (z)is harmonic.

20


205/717

III.3.3

Let D = fa < jzj < bg n (b; a), an annulus slit along the negativereal axis. Show that any harmonic function on D has a harmonicconjugate on D.Suggestion.Fix c betweena andb, and denev (z)explicitly as a lineintegral along the path consisting of the straight line from c to jzjfollowed by the circular arc from jzj toz. Or map the slit annulusto a rectangle by w= Log z.

SolutionIntegrate the CR equations we have

v (z) =

Z zz0

1

r

@u

@dr+r

@u

@rd=

Z @u

@ydx+

Z @u

@xdy

integrate alongr interval, then interval. Perception is same as gettingfromz0 toz1, then fromz1 to little disk. Integral is well-dened, continuous,harmonic and is harmonic conjugate for u.

21


206/717

III.3.4 (From Hints and Solutions)

Letu (z)be harmonic on the annulus fa < jzj < bg. Show that thereis a constant C such that u (z) Clog jzj has a harmonic conjugateon the annulus. Show that C is given by

C= r

2

Z 20

@u

@r

rei

d;

where r is any xed radius, a < r < b.

Solutionu has harmonic conjugate v1 on the annulus slit along (b; a), and also aharmonic conjugate v2 on the annulus slit (a; b). Since v1 v2 is constant

above the slit (b; a), and also constant below the slit, v1 jumps by aconstant across the slit. Arg z also jumps by a constant across the slit. Byappropriate choice ofC,v1 CArg z is continuous across the slit (b; a),andu Clog jzj has a harmonic conjugatev1 Clog jzj on the annulus. Forthe identity, use the polar form of the Cauchy-Riemann equations to converttherderivative ofu to aderivative ofv .

22


207/717

III.3.5

The ux of a function u across a curve is dened to beZ

@u

@nds=

Z

ru nds;

where n is the unit normal vector to andds is arc length. Showthat if a harmonic function u on a domain D has a conjugate har-monic function v on D, then the integral giving the ux is inde-pendent of path in D. Further, the ux across a path in D fromA toB is v (B) v (A).

Solution!t =

dsdx

; dsdy

,!n =

dyds

; dxds

,R

ru!n ds=

R

@u@x

dyds

@u@y

dxds

ds=

R

@u@y dx+ @u@x dy =|{z}

R

@v@x

dx+ @v@y

dy=Rdv

()CR equations. Ifgoes fromP toQ, this isv (Q) v (P), which showsthat the integral is independent of path in D.

23


208/717

III.4.1

Let f(z) be a continuous function on a domain D. Show that iff(z)has the mean value property with respect to circles, as denedabove, thenf(z)has the mean value property with respect to disks,that is if z0 2 D and D0 is a disk centered at z0 with area A andcontained in D, then f(z0) =

1A

RRD0

f(z)dxdy.

Note: This MVP for area is not exactly what is dened as the MVP fordisks. What is then is that if h (z0) is the average of h (z) for any circlefjz z0j =rg, 0 < r < 1. Thenh (z) is the average ofh (z) over any diskfjz z0j =r0g, 0 < r < r0. Express dxdy in polar coordinates at z0 and

integrate rst with respect to .

SolutionSuppose that fsatises the mean value property,

f(z0) = 1

2

Z 20

f

z0+rei

d

for allz02 Dand r >0such that Br(z0) D. Let D0 = fz 2 C : jz z0j Rg D. We have,A= R2 and parametrizing in polar coordinates, and use themean value property with respect to circles,

1

A

ZZD0

f(z)dxdy= 1

R2

ZZD0

f(z)dxdy=

x= x0+r cos dx dy= r dr d 0 ry= y0+r sin 0

= 1

R2

Z 20

Z R0

f

z0+rei

r dr d= 1

R2

Z R0

Z 20

f

z0+rei

d

r dr=

1

R2

Z R0

2f

=2f(z0)

R2

Z R0

r dr= 1

R2

r2

2

R0

f(z0) =f(z0) :

Another note: Its better to use something other than1=Aas A was alreadyused as the circle average.

24


209/717

III.4.2

Derive (4.2) from the polar form of the Cauchy-Riemann equations(Exercise II.3.8).

SolutionThe polar form of Cauchy-Riemanns equations (from Exercise II.3.8)

@u

@r =

1

r

@v

@;

@u

@ = r

@v

@r:

Now we have

r Z 2

0@u@r

z0+rei d= Z 2

0@v@

z0+rei d= 0

25


210/717

III.4.3

A functionf(t)on an intervalI= (a; b)has the mean value propertyif

f

s+t

2

=

f(s) +f(t)

2 ; s; t 2 I:

Show that any ane function f(t) = At+ B has the mean valueproperty. Show that any continuous function on I with the meanvalue property is ane.


LetI := (a; b) and let f :I! R be given. Suppose rst thatfis an anefunction, that is, there are A; B 2 R such that f(t) =At +B for all t 2 I.Now let s; t 2 Ibe given, then

f

s+t

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