analysis chapter8 update

8/6/2019 Analysis Chapter8 Update

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8 Conformal Mappings

Let D be a domain in R2. The problem of finding a function u : D R suchthat

2ux2

+ 2u

y2= 0 and u|D = h where h is given, that is, finding harmonic

function with given boundary values, is known as the Dirichlet problem andis much easier to solve for the disc B1(0) than for a general domain.

IfD is a domain with the property that for each simple closed curve in D,the interior of lies in D then for each harmonic function u on D there existsanother harmonic function v such that u and v satisfy the CR equations. Thenthe function f(x+iy) = u(x, y)+iv(x, y) is holomorphic. IfD is another domainwith a holomorphic bijection g : D D then f g : D C is holomorphicand Re (f g) = u g is harmonic. For suitable D the Dirichel problem may beeasier to solve. In this chapter we study holomorphic maps between domains.

Definition 8.1. Let D be a domain in R2 and f : D R2 a continuously differen-tiable function. The function f is said to be conformal if it preserves angles.

This means that if1, 2 : (, ) R2 are continuously differentiable pathswith 1(0) = 2(0) and

1(0) = 0, 2(0) = 0 then the angle between 1(0) and

2(0) equals the angle between (f 1)(0) and (f 2)(0).

Conformal mappings and holomorphic functions

If we now identify R2 with C in the usual way and put 1(0 = 2(0) = z0, thissays that

arg 1(0) arg 2(0) = arg(f 1)(0) arg(f 2)(0), that is,

arg

1(0)2(0)

= arg (f 1)(0)

(f 2)(0) ,

when all these numbers are defined.

Proposition 8.1. IfD is a domain in C f : D C is holomorphic and such thatf(z) = 0 for all z D then f is conformal.

Proof. Let 1, 2 : (, ) D be continuously differentiable and such that1(0) = 2(0) = x0. Then

arg(f

1)(0)

(f 2)(0) = arg f(1(0))1(0)f(2(0))2(0)

= arg

1(0)2(0)

.

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COMMENT: The converse is also true. This follows from the fact that alinear map

x

y

a b

c d

x

y

preserves angles at 00 ifdet

a b

c d = 0, a = d and b =

c.

Iff(x + iy) = u(x, y) + iv(x, y) is holomorphic then the CR equations giveus that

ux uyvx vy

satisfies this property.

Example 8.1. Let f(z) = z2 then f(0) = 0. Let 1(t) = t, 2(t) = it, t R.Then f(1(t)) = t

2, f(2(t)) = t. The angle between the positive real axis and thepositive imaginary axis is

2and this angle maps to the straight angle, , showing that

the conditionf(0)

= 0

is necessary.

Lecture 22

Last year you studied Mobius transformations, maps of the form

z f(z) = az + bcz + d

where a,b,c,d C, ad bc = 0. This Mobius transformation is defined on thedomain C \ {d

c} and

f(z) = (cz + d)a c(az + b)(cz + d)2 = ad bc(cz + d)2 = 0.

We quickly review the properties of Mobius transformation needed for ourpurposes in this course.

PROPERTIES OF MOBIUS TRANSFORMATIONS

(i) A Mobius transformation

f(z) =az + b

cz + dextends to a bijection of the extended complex plane, C := C{}, withitself. This is done by defining f(d

c) = and

f() = ac

.

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The reasoning behind this choice for f() is thataz + b

cz + d=

a + bz

c + dc

,

ifz

= 0, and as

|z

| ,

a + bz

c +d

c a

c

. This extension gives a homeomor-

phism of the extended complex plane (when identified with the Riemannsphere).

(ii) If we let w =az + b

cz + dthen

(cz + d)w = az + b

(cw a)z = dw + bw =

dw + bcw a .

We see that the inverse of a Mobius transformation is again a Mobiustransformation.

(iii) The composition of two Mobius transformations is a Mobius transfor-mation,and with composition as multiplication Mobius transformationsform a group. If we let GL(2,C) denote the group of invertible 2 2matrices with complex coefficients, the map given by

a b

c d

=

az + b

cz + d,

defines a surjective homomorphism from GL(2,C) onto the group of

Mobius transformations. The kernel of is the subgroup of invertiblescalar matrices. The coefficients of a,b,c,d are only determined up tomultiplication by a non-zero complex number.

(iv) Two complex numbers and are said to be inverse points for a circleof centre a and radius r, |z a| = r, if

( a)( a) = r2.The inverse points lie on a ray coming from a. The equation

z z = ,

, C, = , > 0, = 1, defines a circle with inverse points and; the equation z z

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gives a line. If

w = f(z) =az + b

cz + d,

then the circline given by

z z

=

is mapped to a circline given by an equation of the form

w f()w f() = ,

provided that f() = , f() = .Mobius transformations map circlines to circlines. A circline is de-termined by three points, that is , given three distinct points there is aunique circline containing those points. To find the image of a circline bya Mobius transformation, it is sometimes convenient to keep track of the

image of three points.

Example 8.2. Find a bijective conformal map from upper half of the complex plane

H+ := {z C : Im z > 0}

onto D = {z C : |z| < 1}.

Note that H+ = {z C :

z iz + i

< 1}. Put w = z iz + i

then z H+ iff|w| < 1and the Mobius transformation f(z) = zi

z+isolves the problem. Alternatively,

we note that , 0, 1 all lie on the boundary ofH+ and that 1, 1, i lie on theboundary ofD. We calculate the Mobius transformation f(z) =

az + b

cz + dmap-

ping to 1, 0 to 1 and 1 to i:

1 ac

= 1 a = c,

0 1 bd

= 1 b = d,

1 i a + b

c + d = i.

Soc dc + d

= i.

We know the coefficients a,b,c,d are only determined up to multiplicationby a non-zero constant so for simplicity we take c = 1. This immediately gives

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us that a = 1. Also 1 d = i(1 + d) 1 = i = d(1+i). So d = 1i1+i

= i and andb = i. We now consider the map

w = f(z) =z + i

z i .

We must check whether this maps H+ to D but immediately we see that

i H+ maps to and f(z) = z+izi maps H+ to the outside ofD. This is easilyremedied by composing with the map w 1

w.

In this exercise we used the Mobius transformation z 1z

and it is worth

noting one of its key properties: it maps any circline through 0 to one through, that is, it maps a circline through 0 to a line.Example 8.3. Find the image by the map z 1

zof the disc given by |z 1| < 1 .

Three convenient points on the circle are 0,1 + i and 2. These points map to

, 1i2 and 12 , respectively, which lie on the line given by Re z = 12 . The centreof the circle is at 1 which is mapped to itself. Thus the disc is mapped onto thedomain Re z > 12 .

Note also that z 1z

maps the unit circle given by |z| = 1 to itself but itmaps the inside containing 0 to the outside containing .Example 8.4. Find a conformal map between

H+ D = {z C : Im z > 0, |z| < 1}

and D ={

zC :

|z|

= 1}

.

Notice that the boundary ofH+ D has two right-angled corners and theboundary ofD has none and so the map we seek is not a Mobius transforma-tion which is conformal at all points except possibly at one which it maps to.

However, we might use a Mobius transformation to map one corner, saythat at 1, to , and the other at 1 to 0. As both components of the bound-ary intersect orthogonally at 1, both components being parts of circlines aremapped to circlines through , that is, to lines which intersect orthogonally.The image of the boundary ofH+ D is mapped to a region between two or-thogonal lines intersecting at 0. We know that the map z z

2

will double theangle at 0 and maps a quadrant to a half plane. The previous example tells uswhat to do next. We put this plan into action.

The map f(z) =z 1z + 1

maps 1 to 0 and 1 to . It maps the circle given by|z| = 1 and the real axis to two lines intersecting orthogonally at 0. This map f

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sends 0 to 1 and so maps the real axis to itself. Then f(i) = i1i+1

= i so f sendsthat part of the circle given by |z| = 1 lying in H+ to the positive part of theimaginary axis. From this we see that H+ D maps to {z C :

2< arg z < }.

We rotate this second quadrant back to lie in the first quadrant by multiplying

by i. Then the map z (i)z 1z + 1

maps H+ D to the first quadrant. Next

we compose this map with z z2

to get

z

z 1z + 1

2,

which maps H+ D onto H+. Composing with the Mobius transformationused in the previous exercise we get

z z1z+1

2+ i

z1z+1

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

mappingH+ D onto D. This map is bijective because it is a composition ofmaps which are bijective on the domains described in the construction.

Lecture 23

We explore the exponential maps z ez and z eiz. First we consider z ez.If z = x + iy, then z ez becomes x + iy ex.eiy. If c R is fixed thenc + iy ec.eiy. The line Re z = c is wrapped around the circle |z| = ec. Theline Im z = c, is parametrised by x x + ic and x + ic ex.eic. Thus the lineIm z = c maps to the ray arg z = c.

Example 8.5. Find a conformal mapping from the strip{z C : 0 < Re z < 1}

onto the annulus{z C : 1 < |z| < e}.

On the boundary line Re z = 0, iy eiy lying on the circle |z| = 1; on theboundary line Re = 1, 1 + iy e.eiy lying on the circle |z| = e. So the mapz ez maps the given strip onto the given annulus.Example 8.6. Find a conformal mapping from the strip

{z : 0 < Re z < 1}onto the annulus

{z : a < |z| < b},where 0 < a < b, are given.

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We see that the strip

{z : log a < Re z < log b}

is mapped by z ez onto the annulus

{z : 0 < a < |z| < b}.Then z log a + (log b log a)z = log a + log( b

a)z maps the strip given by

0 < Re z < 1 onto the strip given by log a < Re z < log b. Composing the mapswe get

z elog a+log( ba )z = a

b

a

z

which maps{z : 0 < Re z < 1}

onto the annulus

{z : 0 < a < |z| < b}.Next we explore z eiz.

For the line Re z = c: c + iy ei(c+iy) = eyeic. The line Re z = c maps to theray arg z = c. For the line Im z = c: x + ic eixec. The line Im z = c maps tothe circle given by|z| = ec.Example 8.7. Find a bijective conformal mapping from

{z : 0 < Re z < , Im z > 0}

onto D = {z : |z| < 1}.Notice that the boundary has two corners. Try z eiz. This map is 1-1 on

the given domain. Then x + iy w = eix.ey and ify > 0 and 0 < x < , then|w| < 1 and 0 < arg < . The image is H+ D. We have already solved theproblem of mapping H+ D onto D by taking the map z (z1)2+i(z+1)2

(z1)2+i(z+1)2 . Amap solving this problem given by the composition:

z (eiz 1)2 + i(eiz + 1)2

eiz 1)2 i(eiz + 1)2 .

Example 8.8. Find a conformal mapping from

{z : |z 14| > 1

4, |z| < 1}

onto an annulus.

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Here we are asked to find a conformal mapping from a region between twocircles which are not concentric onto a region between two concentric circles.Mobius transformations map inverse points to inverse points and do not, ingeneral, map centres to centres. So we might try to solve this problem usingMobius transformations.

Any two non-intersecting circles share a common pair of inverse points,

that is, the circles can be written in the formz z = 1 and

z z = 2.

Then the map z w = zz will map these circles to circles given by |w| = 1

and |w| = 2 which are concentric. So our problem has been reduced to findingthe common inverse points for the circles |z 14 | = 14 and |z| = 1. The commoninverse points are collinear with both centres, 1

4and 0, and lie on the real axis.

This simplifies our calculation.If and are inverse points for the circle given by

|z

14

|= 1

4then

1

4

1

4

=

1

4

2=

1

16,

(no conjugation involved because all points are real);if and |beta are inverse points for the circle given by |z| = 1 then (

0)( 0) = 1, that is, = 1 and = 1

. Substituting we get:

( 14

)(1

1

4) =

1

16

( 14)(1 4 ) = 16(4 1)(4 ) = 42 + 17 4 =

2 4 + 1 = 0 =

4 16 42

= 2

3.

Let = 2 3 and = 2 + 3. Then lies inside and lies outside bothcircles.

To find the equation of the form

z (2 3z (2 + 3

= 1for the circle given by |z 1

4| = 1

4, we note that 0 satisfies |z 1

4| = 1

4. Hence

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1 =

0 (2 30 (2 + 3

=2 32 +

3

= 7 4

3.

To find the equation of the form

z

(2

3

z (2 + 3 = 2

for the unit circle given by |z| = 1, we note that 1 satisfies |z| = 1. Hence

2 =

1 (2 31 (2 + 3

=

3 13 + 1

= 2

3.

So z z(23

z(2+3) maps the given domain onto the annulus

{z : 7 4

3 < |z| < 2

3}.

Lecture 24

Example 8.9. Define a branch of

z2 1.We seeking a holomorphic function f such that (f(z))2 = z2 1. For a

particular value of z C, z2 1 is defined to be elog(z21) 12 . We know thatlog is a multivalued function with a branch point at 0 and z2 1 = 0 ifz = 1or z = 1. We should expect our function to have branch points at 1 and 1.We simplify the problem by factorising. Note that z2 1 = (z 1)(z + 1). Weproceed by defining branches of the factors

z 1 and z + 1. First we define

a holomorphic function f1 such that (f1(z))2 = z

1 with domain C

\[1,

) by

taking f1(z) = |z 1| 12 ei arg(z

1)2 where 0 < arg(z 1) < 2. Similarly, we defineholomorphic g1 such that (g1(z))

2 = z + 1 with domain C \ [1, ) by takingg1(z) = |z + 1| 12 ei arg(

z+1)2 where 0arg(z + 1) < 2. We also define holomorphic

g2 such that (g2(z))2 = z + 1, this time with domain C \ (, 1] by taking

g2(z) = |z + 1| 12 ei arg(z+1)2 where < arg(z + 1) < . Then (f1(z)g1(z))2 =z2 1. Both f1 and g1 change by a factor of1 across their cuts, [1, ) and[1, ), respectively, (ei 22 = ei = 1). In the product the factors cancel outacross [1, ) and so the product f1g1 can be extended so as to be continuousacross [1, ). It follows that the product is holomorphic on C \ [1, +1] witha jump of a factor

1 across [

1, 1]. The other product f1g2 is holomorphic on

C \ [(, 1] [1, )] with jumps across , 1] and[1, ).Example 8.10. Let a > 1 and calculate1

1

1

(x a)1 x2 dx.

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For this problem we chose the branch f1g1 of

z2 1 which is homomor-phic on C \ [1, 1]. We exploit the jump across [1, 1]. Let R > a. Then

|z|=R

1

(z a)z2 1 dz

20

Rieit dt

Reit

a)(Re

it)2

1

2

0

R

(R a)R2 1 dt

= O(1

R)

which 0 as R . Hence|z|=R

1

(z a)z2 1 dz = 0.

Further,

|z|=R

1(z a)z2 1 dz =

|z|= a+1

2

1(z a)z2 1 dz+

|za|=a1

2

1(z a)z2 1 dz

by the deformation version of Cauchys Integral Theorem. Now|za|=a1

2

1

(z a)z2 1 dz = 2i1

a2 1 .

Deforming the circle, |z| = a+12

, within the domain {z : |z| < a} \ [1, 1] toflatten it so it runs along the interval [1, 1] from 1 to 1 then back along from1 to 1, we find that|z|= a+1

2

1

(z a)z2 1 dz =

1

1

1

(z a)(z2 1)dz

1

1

1

(z a)(z2 1)+dz

where (

z2 1) is the limiting value of the branch of

z2 1 taken frombelow the real axis and (

z2 1)+ is the limiting value of the branch of

z2 1

taken from above the real axis. The difference between the two branches is afactor 1.At this point we wont describe the limiting value of branch exactly but simplyobserve that it must be one ofi1 x2 because (i1 x2)2 = x2 1 and

|z|= a+1

2

1(z a)z2 1 dz = 211 1(x a)1 x2 i dx.

Therefore,

2i11

1

(x a)1 x2 dx + 2i1

a2 1 = 0

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and so 11

1

(x a)1 x2 dx =

a2 1 ,

the choice of sign being clear because x a < 0 when x [1, 1].

ANOTHER WAY OF LOOKING AT THE MULTIFUNCTION z2

The function z z2 is not one-to-one because (z)2 = z2 and its inversez is a multifunction which, for z = 0, gives two values. The graph of z z2

is{(z, z2) : z C}.

If we switch the factors in the graph ofz2 we get the set

G = {(z2, z) : z C},

the graph of the inverse multifunction

z.

Let 1, 2 : CC Cbe given by 1(z, w) = z and 2(z, w) = w. Then 2 isa well defined function on G and 2|G : G C can be thought of as the functionz and G as the Riemann surface ofz. The complication of a multifunction

has been transferred by defining

z on a more complicated domain. Noticethat except at (0, 0) on G, z is a two-to-one function.

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