Chapter VIIIConformal MappingsInthischapter,weintroduceanddeveloptheconceptofaconformal mapping.Thegeometricinterpretationofafunction of a complex variableas amapping,ortransformation, was introducedin Secs.2.2and 2.3(Chap.2).We saw therehowthenatureof such a functioncanbe displayedgraphically, to some extent, by the manner in which it maps certain curves and regions. In thischapter,weshall seefurther examplesof howvariouscurves andregionsaremappedby someelementaryanalyticfunctions.s8.1.ConformalmappingsDefinition 8.1.1. A transformation w = f(z) is said to be conformal at a point zo if fis analytic there and f(=) 0.Such a transformation is actually conformal at each point in a neighborhood of zo.For fmust be analytic in a neighborhood of zo (Sec. 2.13); and, since f is continuous at zo (Sec.4.13), it follows from Theorem 2.7.2 in Sec. 2.7 that there is also a neighborhood of thatthroughout whichf'()0Definition 8.1.2.A transformation w=f(-) on a domain Dis referred to as aconformaltransformation,orconformal mapping,whenitis conformalateachpointinDEachoftheelementaryfunctionsstudied inChap.3canbeusedtodefineatransformationthat is conformal in somedomain.Example 1.The mapping w=e=is conformal throughout the entire zplane since(e')=e'+O for each z.Consider any two lines x= and y=c, in the zplane, thefirst directed upward and the second directed to the right.According to Sec.2.3, their imagesunder the mapping w=eare a positively oriented circle centered at the origin and a ray fromthe origin, respectively.As illustrated in Fig.2-5 (Sec. 2.3), the angle between the lines at theirpoint of intersection is a right angle in the negative direction, and the same is true of the anglebetween the circleandtherayat the correspondingpoint inthewplane.Example 2. Consider two smooth arcs which are level curves u(x,y)=C andv(x,y)=Cofthereal andimaginarycomponents,respectively,ofafunctionf(z) =u(x,y)+iv(x,y),and suppose that they intersect at a point zo where f is analytic and f'(zo)+0. Thetransformation w= f(z) is conformal at zo and maps the arcs into the lines u=c, andV=C2,which are orthogonal at the point w=f(=)because of the Cauchy-Riemannequations.According to our theory,then, the arcsmust beorthogonal at zo.This has alreadybeenverified and illustrated in Exercises 7 through 11 of Sec. 2.15.A mappingthat preserves the magnitude of the anglebetween two smooth arcsbut notnecessarily the sense is called an isogonal mapping.Example3.Thetransformation w=z,whichisareflection inthereal axis,is isogonal butnot conformal.If it is followed by a conformal transformation, the resulting transformationw=f(-)isalsoisogonalbutnotconformal
Chapter Ⅷ Conformal Mappings In this chapter, we introduce and develop the concept of a conformal mapping. The geometric interpretation of a function of a complex variable as a mapping, or transformation, was introduced in Secs. 2.2 and 2.3 (Chap. 2). We saw there how the nature of such a function can be displayed graphically, to some extent, by the manner in which it maps certain curves and regions. In this chapter, we shall see further examples of how various curves and regions are mapped by some elementary analytic functions. §8.1. Conformal mappings Definition 8.1.1. A transformation = zfw )( is said to be conformal at a point if is analytic there and . 0 z f 0)(′ zf 0 ≠ Such a transformation is actually conformal at each point in a neighborhood of . For must be analytic in a neighborhood of (Sec. 2.13); and, since is continuous at (Sec. 4.13), it follows from Theorem 2.7.2 in Sec. 2.7 that there is also a neighborhood of that throughout which 0 z f 0 z f 0 z ′ zf ≠ 0)( . Definition 8.1.2. A transformation = zfw )( on a domain is referred to as a conformal transformation, or conformal mapping, when it is conformal at each point in . D D Each of the elementary functions studied in Chap. 3 can be used to define a transformation that is conformal in some domain. Example 1. The mapping is conformal throughout the entire plane since for each . Consider any two lines z = ew z ′ ≠= 0)( zz ee z 1 = cx and 2 = cy in the plane, the first directed upward and the second directed to the right. According to Sec. 2.3, their images under the mapping are a positively oriented circle centered at the origin and a ray from the origin, respectively. As illustrated in Fig. 2-5 (Sec. 2.3), the angle between the lines at their point of intersection is a right angle in the negative direction, and the same is true of the angle between the circle and the ray at the corresponding point in the plane. z z = ew w Example 2. Consider two smooth arcs which are level curves and of the real and imaginary components, respectively, of a function 1 ),( = cyxu 2 ),( = cyxv = + yxivyxuzf ),(),()( , and suppose that they intersect at a point where is analytic and . The transformation is conformal at and maps the arcs into the lines and , which are orthogonal at the point 0 z f 0)(′ zf 0 ≠ = zfw )( 0 z 1 = cu 2 = cv )( 0 0 = zfw because of the Cauchy-Riemann equations. According to our theory, then, the arcs must be orthogonal at . This has already been verified and illustrated in Exercises 7 through 11 of Sec. 2.15. 0 z A mapping that preserves the magnitude of the angle between two smooth arcs but not necessarily the sense is called an isogonal mapping. Example 3. The transformation = zw , which is a reflection in the real axis, is isogonal but not conformal. If it is followed by a conformal transformation, the resulting transformation = zfw )( is also isogonal but not conformal
Suppose that f is not a constant function and is analytic at a point zo. If, in addition,f'(=)=0,then zis calleda critical pointofthetransformation w=f(z)Example 4. The point z = 0 is a critical point of the transformationW=1+2?,which is a composition of the mappingsZ=22andw=l+Z.A ray =α from the point z=O is evidently mapped onto the ray from the point w=1whose angleof inclination is 2α.Moreover,the anglebetweenanytworays drawn fromthecritical point z=O isdoubled by thetransformation.More generally, it can be shown that if zo is a critical point ofa transformation w= (z),there is an integer m(m ≥2) such that the angle between any two smooth arcs passing throughzo is multiplied by m under that transformation. The integer m is the smallest positiveinteger such that f(m)(z)+ 0. Verification of these facts is lefto the exercises.Next, let us discuss the angle-preserving property of a conformal mapping. Let C be asmooth arc (Sec. 4.3), represented by the equation z=z(t)(a≤t≤b), and let f(=) be afunction defined at all points z on C. The equation w= f[=(t)] (a≤t≤b) is a parametricrepresentation ofthe image of C under the transformation w= f(2).Suppose that C passes through a point zo=z(to) (a≤t.≤b) at which f isconformal at zo-According tothe chain rule, if w(t)=f[z(t)],thenw(to) = f'[=(to)]-'(to);(8.1.1)and this means that (see Sec. 1.7)(8.1.2)Argw'(to)= Arg f'[z(to)}-(to)+Argz(to)Statement (8.1.2) is useful in relating the directions of C and F at the points zo andW。= f(zo),respectivelyTo be specific, let o denote a value of Argf(z), and let be the angle ofinclination ofa directed line tangent to C at =。 (Fig. 8-1),山VAC10.ZoxoouFig. 8-1AccordingtoSec.4.3,.isavalueoftheargumentofz'(t.);and itfollowsfromstatement(8.1.2) that the quantity中o=Vo+0(8.1.3)is a value of Argw'(to) and is, therefore, the angle of inclination of a directed line tangent toI at the point Wo=f(z)
Suppose that is not a constant function and is analytic at a point . If, in addition, , then is called a critical point of the transformation f 0 z 0)(′ zf 0 = 0 z = zfw )( . Example 4. The point z = 0 is a critical point of the transformation 2 1+= zw , which is a composition of the mappings 2 Z = z and =1+ Zw . A ray θ =α from the point is evidently mapped onto the ray from the point whose angle of inclination is z = 0 w =1 2α . Moreover, the angle between any two rays drawn from the critical point z = 0 is doubled by the transformation. More generally, it can be shown that if is a critical point of a transformation , there is an integer such that the angle between any two smooth arcs passing through is multiplied by m under that transformation. The integer is the smallest positive integer such that . Verification of these facts is left to the exercises. 0 z = zfw )( mm ≥ )2( 0 z m ( 0) 0 ) z ( f ≠ m Next, let us discuss the angle-preserving property of a conformal mapping. Let be a smooth arc (Sec. 4.3), represented by the equation C = tzz )( ≤ ≤ bta )( , and let be a function defined at all points on C . The equation zf )( z = tzfw )]([ ≤ ≤ bta )( is a parametric representation of the image Γ of C under the transformation = zfw )( . Suppose that passes through a point C )( 00 = tzz )( ≤ 0 ≤ bta at which is conformal at . According to the chain rule, if f 0 z = tzftw )]([)( , then )()]([)( 0 00 ′ = ′ ′ tztzftw ; (8.1.1) and this means that (see Sec. 1.7) ′ 0 = ′ ′ 00 + ′ tztztzftw 0 )(Arg)()]([Arg)(Arg . (8.1.2) Statement (8.1.2) is useful in relating the directions of C and Γ at the points and , respectively. 0 z )( 0 0 = zfw To be specific, let ψ 0 denote a value of ′ zf 0 )(Arg , and let θ 0 be the angle of inclination of a directed line tangent to C at (Fig. 8-1). 0 z Fig. 8-1 According to Sec. 4.3, θ 0 is a value of the argument of )( 0 ′ tz ; and it follows from statement (8.1.2) that the quantity φ =ψ +θ 000 (8.1.3) is a value of ′ tw 0 )(Arg and is, therefore, the angle of inclination of a directed line tangent to Γ at the point )( 0 0 = zfw
Now let C and C be two smooth arcs passing through zo, and let , and 3, beangles ofinclination of directed linestangentto C and C2,respectively,at zo.Weknow from(8.1.3) that the quantitiesΦ=+0,andΦ,=+0,are angles of inclination of directed lines tangent to the image curves and ,,respectively, atthe point w。= f(=o). Thus, Φ, -Φ, =0, -,; that is, the angle Φ, -Φ from , to , isthe same in magnitude and sense as the angle , -, from C to C2.Those angles aredenoted by α in Fig. 8-2.山CW/Z0ox可uFig. 8-2Lastly, let us discuss the geormetrical meaning of the derivative f'(zo) of a conformalmapping f. To consider a transformation w = f(-) that is conformal at a point zo. From thedefinition of derivative, we know thatm L(=)- f(zo)f(z)-f(zo))[F(z0) =lim .:(8.1.4)[z- z0]Z-Z0Now |z-zol is the length of a line segment joining Zo and z, and [f(z)- f(zo)l is thelength of the line segment joining the points f(zo) and f(z) in the w plane. Evidently.then, if z is near the point zo, the ratio[F(2) - f(=0)[z - zolofthe two lengths is approximately the numberf'(z).That is[)-(0 [1(c0) o [(2) - (=0) (c0)-=0,[2 zowhenever zzo.Note thatf(zo)represents an expansion if it is greater than unity and acontraction if it is less than unity.Although the angle of rotation arg f'(zo) and the scale factor f'(z)] vary, in general,from point to point, it follows from the continuity of f' that their values are approximatelyargf'(=)and f(=)at point znear =o.Consequently,the image ofa small region in aneighborhood of zo conforms to the original region in the sense that it has approximately thesame shape.A large region may be transformed into a region that bears no resemblance to theoriginal one
Now let and be two smooth arcs passing through , and let C1 C2 0 z θ1 and θ 2 be angles of inclination of directed lines tangent to and , respectively, at . We know from (8.1.3) that the quantities C1 C2 0 z φ =ψ +θ101 and φ =ψ +θ 202 are angles of inclination of directed lines tangent to the image curves Γ1 and , respectively, at the point . Thus， Γ2 )( 0 0 = zfw φ −φ =θ −θ1212 ; that is, the angle φ −φ12 from to is the same in magnitude and sense as the angle Γ1 Γ2 θ −θ12 from to . Those angles are denoted by C1 C2 α in Fig. 8-2. Lastly, let us discuss the geormetrical meaning of the derivative )( 0 ′ zf of a conformal mapping f . To consider a transformation = zfw )( that is conformal at a point . From the definition of derivative, we know that 0 z Fig. 8-2 0 0 0 0 0 )()( lim )()( lim)( 0 0 zz zfzf zz zfzf zf zz zz − − = − − ′ = → → . (8.1.4) Now || is the length of a line segment joining and , and 0 − zz 0 z z )()( 0 − zfzf is the length of the line segment joining the points and in the plane. Evidently, then, if is near the point , the ratio )( 0 zf zf )( w z 0 z 0 0 )()( zz zfzf − − of the two lengths is approximately the number )( 0 ′ zf . That is, ≈ − − 0 0 )()( zz zfzf )( 0 ′ zf , or 0 0 0 ≈− ′ |)(|)()( −⋅ zzzfzfzf , whenever . Note that 0 ≈ zz )( 0 ′ zf represents an expansion if it is greater than unity and a contraction if it is less than unity. Although the angle of rotation ′ zf 0 )(arg and the scale factor ′ zf )( vary, in general, from point to point, it follows from the continuity of f ′ that their values are approximately )(arg and 0 ′ zf )( 0 ′ zf at point near . Consequently, the image of a small region in a neighborhood of conforms to the original region in the sense that it has approximately the same shape. A large region may be transformed into a region that bears no resemblance to the original one. z 0 z 0 z
Example 5. When f()= 22, the transformationW= f(2)=x - J2 +i2xyis conformal at the point z=1+i, where the half lines y= x (x≥0) and x=1(x≥0)intersect.We denote those half lines by C and C,with positive sense upward, and observethat the anglefrom C, to C, is 元/4 at their point of intersection (Fig.8-3). Since the imageofa point z=(x,y) is a point in the w plane whose rectangular coordinates areu=x-y?andv=2xy,the half line C is transformed into the curve with parametric representationu=0,V=2x2(0≤x<00)(8.1.5)Thus is the upper half v≥0 ofthe axis. The halfline C, is transformed into the curveI,representedbytheequationsu=1-y,v=2y((8.1.6)(0≤y<8)Hence , is the upper half of the parabola y=-4(u-1). Note that, in each case, the positivesense of the image curve is upward.vir12元C10ol1Fig. 8-3If u andvarethevaritation (8.1.6)fortheimagecurveF,,thendy dvI dy 272dudu/dy-2yVIn particular, dv/ du=-1 when v= 2. Consequently, the angle from the image curve I, tothe image curve I, at the point w= f(1+i)=2i is π /4, as required by the conformalityof the mapping at z =1+i.As anticipated, the angle of rotation π/4 at the point z =1+i isa value ofArg[(1+i)]= Arg[2(1 +i)] =+2n元(n = 0,±1,±2,...) .4The scalefactoratthatpoint isthenumber[F(1+i)= 2(1+i)=2/2To illustrate howthe angleofrotation and the scale factor can change from point topoint, we notethat they are 0 and 2, respectively, at the point z =1 since f'(I) = 2. See Fig. 8-3, where thecurves C, and I, are the one just discussed and where the nonnegative x axis C, istransformed into the nonnegative u axis I
Example 5. When , the transformation 2 )( = zzf )( 2xyiyxzfw 22 +−== is conformal at the point 1+= iz , where the half lines = xxy ≥ )0( and intersect. We denote those half lines by and , with positive sense upward, and observe that the angle from to is xx ≥= )0(1 C1 C2 C1 C2 π 4/ at their point of intersection (Fig. 8-3). Since the image of a point = yxz ),( is a point in the w plane whose rectangular coordinates are 22 −= yxu and = 2xyv , the half line is transformed into the curve C1 Γ1 with parametric representation u = 0 , 2 = 2xv ≤ x < ∞)0( . (8.1.5) Thus is the upper half of the axis. The half line is transformed into the curve represented by the equations Γ1 v ≥ 0 v C2 Γ2 2 1−= yu , = 2 yv ≤ y < ∞)0( . (8.1.6) Hence is the upper half of the parabola . Note that, in each case, the positive sense of the image curve is upward. Γ2 )1(4 2 uv −−= If u and v are the variables in representation (8.1.6) for the image curve , then Γ2 vydydu dydv du dv 2 2 2 / / −= − == . Fig. 8-3 In particular, dudv −= 1/ when v = 2 . Consequently, the angle from the image curve to the image curve at the point Γ1 Γ2 = + = 2)1( iifw is π 4/ , as required by the conformality of the mapping at =1+ iz . As anticipated, the angle of rotation π 4/ at the point is a value of 1+= iz π+ π ′ if =+=+ 2ni 4 )]1(2[Arg)]1([Arg n = ± ± K),2,1,0( . The scale factor at that point is the number ′ iif =+=+ 22)1(2)1( . To illustrate how the angle of rotation and the scale factor can change from point to point, we note that they are 0 and 2, respectively, at the point z = 1 since f ′ = 2)1( . See Fig. 8-3, where the curves and are the one just discussed and where the nonnegative C2 Γ2 x axis is transformed into the nonnegative axis C3 u Γ3
$8.2.UnilateralFunctionsIn this section, we will discuss a very important class of conformal mappings, called unilateralfunctionsDefinition 8.2.1. Let f be an analytic functiuon on a domain D.If it is an injection onD,thenwecall faunilateral functionon DTodiscuss theproperties of unilateral functions, we need thefollowinglemmaLemma 8.2.1. Suppose that a function f is analytic on a neighbordood N(zo,R) of apoint -o suchthatwo = f(=0), f"(=o)= "(=o)=..= J(m-1)(z0)=0, f(m)(z)±0. (8.1.6.)Then there is a positive number r0 satisfing(l) The point zo is the unique zero of f(2)-wo in N(zo,e) and it is oforder m(2)For each fixed point weN(wo,),thefunctionf()-whas exactlym zeroscounting multiplicity in the deleted neighborhood N°(zo,) and these zeros are all of order 1.Proof. Since f is not a constant in N(zo,r), f'(z) is not identically zeron inN(zo,r). Thus, there is a positive number rwo-w| when [z-zo/-, it follows from the Rouche's theoremthat the function f(z)-whas just m zeros in N°(zo,). Let z be any zero off(z)-w in N°(zo,3), then(f(a)-w)=f(=)0Thus, z must be a zero of order 1 of (=)-w.The proofis completed.Theorem 8.2.1.Every unilateral function f on a domain D is a conformal mapping onthat domain.Proof. Suppose that f'(z)=0 for some o ED, then there exists a positive integerm>1 suchthatf'(=o)= f"(=0) =...= f(m-)(=0) = 0, f(m)(z0)± 0.Take an R>O such that N(zo,R)cD. From Lemma 8.2.1, there is a positive numberrO so that for each fixed point we N°(wo,o), thefunction f(z)-w has exactly m zeros counting multiplicity in the deleted neighborhoodN°(zo,) and they are all of order 1. Since m>1, for each fixed point w e N°(wo,), thereare two distinct points z),z2 in N°(zo,3) such that f(z)= f(z2)=w. This contradictsthe assumptionthatfisunilateral inD.Thisprovesthatfisconformal inDandcompletes the proofNote that a conformal mapping is not necessarily unilateral. For example, the functionf(z)=e is conformal throughout the plane, but it is not unilateral since it is 2mi-periodic.However,wehavethefollowingresult
§8.2. Unilateral Functions In this section, we will discuss a very important class of conformal mappings, called unilateral functions. Definition 8.2.1. Let be an analytic functiuon on a domain . If it is an injection on , then we call a unilateral function on . f D D f D To discuss the properties of unilateral functions, we need the following lemma. Lemma 8.2.1. Suppose that a function is analytic on a neighbordood of a point such that f ),( 0 RzN 0 z )()(),( 0)(,0)( 0 )( 0 )1( 0 = 0 ′ 0 = ′′ 0 == = ≠ − zfzfzfzfzfw L m m . (8.1.6.) Then there is a positive number r 0,0 satisfing (1) The point is the unique zero of 0 z 0 )( − wzf in ),( 0 zN ε and it is of order m (2) For each fixed point ),( , the function wNw 0 δ o ∈ )( − wzf has exactly zeros counting multiplicity, in the deleted neighborhood and these zeros are all of order 1. m ),( 0 zN ε o Proof. Since f is not a constant in ),( , 0 rzN ′ zf )( is not identically zeron in ),( . Thus, there is a positive number 0 rzN r =− wzf zz ε δ . Let ),( , we write wNw 0 δ o ∈ )())(()( − = − 0 + 0 − wwwzfwzf . Since |||)(| 0 δ 0 −>≥− wwwzf when − || = ε 0 zz , it follows from the Rouche’s theorem that the function has just zeros in . Let be any zero of in , then )( − wzf m ),( 0 zN ε o 1z )( − wzf ),( 0 zN ε o ( ) )( 0)( 1 1 = ′ ≠ ′ − = zfwzf zz . Thus, must be a zero of order 1 of 1z )( − wzf . The proof is completed. Theorem 8.2.1. Every unilateral function on a domain is a conformal mapping on that domain. f D Proof. Suppose that 0)( for some ′ zf 0 = ∈ Dz0 , then there exists a positive integer m >1 such that )()( 0)(,0)( 0 )( 0 )1( ′ 0 = ′′ 0 == = ≠ − zfzfzfzf L m m . Take an R > 0 such that ),( ⊂ DRzN . From Lemma 8.2.1, there is a positive number 0 r ∃1 ),(wNw 0 δ o ∈ 21,zz ),( 0 zN ε o = )()( = wzfzf 1 2 . This contradicts the assumption that is unilateral in . This proves that is conformal in and completes the proof. f D f D Note that a conformal mapping is not necessarily unilateral. For example, the function is conformal throughout the plane, but it is not unilateral since it is z )( = ezf 2πi -periodic. However, we have the following result
Theorem 8.2.2. If a finction f is conformal at a point zo, then it is unilateral on someneighborhood of zo.Proof. Since f'(z)+0, using Lemma 8.2.1 for m=1, we know that there exists aneighborhood N(zo,e) of z and a neighborhood N(wo,) of wo such thatza is theuniquezero off()-woinN(zo,)and it is of order 1,and for eachfixedpointweN(wo,), the function f(z)-w has exactly 1 zero in the deleted neighborhoodN(zo,). From the continuity of f(z)-w, we can choose a positive number μ0 so that foreach fixed point weN(wo,),the function f(z)-whas exactly m zeros countingmultiplicity in the deleted neighborhood N°(zo,). Thus,N(wo,8)c f(N(z0,8) cDHence, wo is an interior pointof f(D) and so f(D) is open.For points w,w, E f(D), take z),=2 ED with (z)=w(k=1,2). Since D isconnected, there is a polygonal lineL: z= z(t)(a≤t≤b)contained in D such that z(a)=zi,z(b)=z2. Since f has conitinuous derivative in D,wegetapiecewise smooth curver: w= w(t)= f(z(t)(a≤t<b)contained in f(D) such thatw(a) = w,w(b)= W2The compactness of the I enable us to find a polygonal line 'contained in f(D) andjioning w, and w,:This proves that f(D) is a domain in the w-plane.The proof iscompleted
Theorem 8.2.2. If a function is conformal at a point , then it is unilateral on some neighborhood of . f 0 z 0 z Proof. Since 0)(′ zf 0 ≠ , using Lemma 8.2.1 for m =1, we know that there exists a neighborhood ),( 0 zN ε of and a neighborhood 0 z ),(wN 0 δ of such that is the unique zero of in w0 0 z 0 )( − wzf ),( 0 zN ε and it is of order 1, and for each fixed point ),( , the function wNw 0 δ o ∈ )( − wzf has exactly 1 zero in the deleted neighborhood ),( . From the continuity of 0 zN ε o )( − wzf , we can choose a positive number μ ∃ 0,0 so that for each fixed point ),( , the function wNw 0 δ o ∈ )( − wzf has exactly zeros counting multiplicity in the deleted neighborhood . Thus, m ),( 0 zN ε o 0 δ ⊂ 0 ε )),((),( ⊂ DzNfwN . Hence, is an interior point of and so is open. w0 Df )( Df )( For points )(, , take 21 ∈ Dfww ∈ Dzz 21, with = kwzf = )2,1()( kk . Since is connected, there is a polygonal line D = ≤ ≤ btatzzL ))((: contained in D such that 1 2 = )(,)( = zbzzaz . Since has conitinuous derivative in , we get a piecewise smooth curve f D Γ = = ≤ ≤ btatzftww )))((()(: contained in Df )( such that 1 2 = )(,)( = wbwwaw . The compactness of the Γ enable us to find a polygonal line Γ′ contained in and jioning and . This proves that is a domain in the -plane. The proof is completed. Df )( w1 w2 Df )( w
s8.3.Local InversesA transformation w= f(-) that is conformal at a point zo has a local inverse there. That is,we haveTheorem 8.3.1. If a transformation w= f() is conformal at a zo and w= f(z),then there exists a unique transformation z=g(w),which is defined and analytic in aneighborhood N of wo, such that g(wo)==o and flg(w))=w for all pints w in Nand the derivative of g(w) is, moreover,1g'(w) =(8.3.1)f(2)Proof. As noted in Sec. 8.1, the conformality of the transformation W=f(=) at zoimplies that there is a neighborhood N(zo,) of zo throughout which f is analytic andf'(-)+0.Henceifwewritez=x+iy, zo=xo +iyo,and f(z)=u(x,y)+iv(x,y),we know that there is a neighborhood N(zo,) of the point (xo,yo) throughout which thefunctions u(x,y) and v(x,y) along with their partial derivatives of all orders, are continuous(see Sec. 4.13).Now, the pair ofequationsu=u(x,y), v=v(x,y)(8.3.2)represents a transformation from the neighborhood N(zo,) into the uv plane. Moreover, theJacobiandeterminantluxu,J==uy, -vu, =(u) +(v) =f()Pyxyyis nonzero on N(=o,).Ifwewrite(8.3.3)o = u(xo,yo) and Vo = v(xo,yo)then from the theory of implicit functions in mathematical analysis, we know that there is a uniquecontinuous transformationx=x(u,v), y= y(u,v),(8.3.4)defined on a neighborhood N of the point (uo,vo) such thatu= u(x(u,v),y(u,v), = v(x(u,v),y(u,v), xo = x(uo,vo), yo = y(uo,vo)Also, the functions (8.3.4) have continuous first-order partial derivatives satisfying the equations111(8.3.5)xu=Vy,x,=uysyu=Vx,y=-uJ1Jthroughout N.Ifwewrite w=u+iv and wo=uo+ivo,aswell as(8.3.6)g(w)= x(u, v)+iy(u,v),the transformation z = g(w) is evidently the local inverse of the original transformationw= f(z) at zo.Transformations (8.3.2)and (8.3.4) can be writtenu+iv=u(x,y)+iv(x,y)andx+iy=x(u,v)+iy(u,v);and these last two equations are the same asw= f(=) and z= g(w),where g has the desired properties. From expression (8.3.5), we see that the function (8.3.4)have continuous partial derivatives and that the Cauchy-Riemann equationsx =y,x, =-yu
§8.3. Local Inverses A transformation = zfw )( that is conformal at a point has a local inverse there. That is, we have 0 z Theorem 8.3.1. If a transformation = zfw )( is conformal at a and , then there exists a unique transformation 0 z )( 0 0 = zfw = wgz )( , which is defined and analytic in a neighborhood N of , such that w0 00 )( = zwg and )]([ = wwgf for all pints in and the derivative of is, moreover, w N wg )( )( 1 )( zf wg ′ ′ = . (8.3.1) Proof. As noted in Sec. 8.1, the conformality of the transformation at implies that there is a neighborhood = zfw )( 0 z ),(zN 0 δ of throughout which is analytic and . Hence if we write 0 z f ′ zf ≠ 0)( z = x + iy , 000 = + iyxz , and = + yxivyxuzf ),(),()( , we know that there is a neighborhood ),(zN 0 δ of the point throughout which the functions and along with their partial derivatives of all orders, are continuous (see Sec. 4.13). ),( 00 yx yxu ),( yxv ),( Now, the pair of equations = yxuu ),( , = yxvv ),( (8.3.2) represents a transformation from the neighborhood ),(zN 0 δ into the plane. Moreover, the Jacobian determinant uv yxyx yx yx uvvu vv uu J −== 2 2 2 zfvu )()()( x x =+= ′ is nonzero on ),(zN 0 δ . If we write ),( 0 00 = yxuu and ),( 0 00 = yxvv . (8.3.3) then from the theory of implicit functions in mathematical analysis, we know that there is a unique continuous transformation = vuxx ),( , = vuyy ),( , (8.3.4) defined on a neighborhood N of the point ),( such that 00 vu = vuyvuxuu )),(),(( , = vuyvuxvv )),(),(( , ),( 0 00 = vuxx , ),( . 0 00 = vuyy Also, the functions (8.3.4) have continuous first-order partial derivatives satisfying the equations u y v J x 1 = , v y u J x 1 −= , u x v J y 1 −= , v x u J y 1 = (8.3.5) throughout N . If we write = + ivuw and 000 = + ivuw , as well as = + vuiyvuxwg ),(),()( , (8.3.6) the transformation is evidently the local inverse of the original transformation at . Transformations (8.3.2) and (8.3.4) can be written = wgz )( = zfw )( 0 z +=+ yxivyxuivu ),(),( and + = + vuiyvuxiyx ),(),( ; and these last two equations are the same as = zfw )( and = wgz )( , where has the desired properties. From expression (8.3.5), we see that the function (8.3.4) have continuous partial derivatives and that the Cauchy-Riemann equations g uvvu = , = −yxyx
are satisfied in N.Thus, g is analytic in N.Furthermore, we compute when w= f(z),11-(y,-iv)=_(u-iv)=g(w)=x, +iyu=ux+ivf(=)The proof is completed.Definition 8.3.1. The function g in Theorem 8.3.1 is called the local inverse of thefunction f at oClearly, unilateral function f on a domain D has an inverse function f-l: f(D)→Dthat equals to the local inverse of at every point zo.Thus, from Theorems 8.2.1, 8.2.3 and8.3.1, we haveCorollary 8.3.1. Every unilateral function f on a domain D has a unilateral inversef-1 on thedomain f(D) and1f-l(w)=VW= f(=)E f(D)f'()Corollary 8.3.2.Every unilateral function f on a domain D is a homoemorphism fromD onto f(D), i.e. f : D-→ f(D),f-l: f(D)-→D are continuous.Example 1. We saw in Example 1, Sec. 9.1, that if f()= e", the mapping w = f(2) isconformal everywhere in the z plane and, in particular, at the point zo =2πi. The image ofthis choice of zo is the point w=1.When points in the w plane are expressed in the formw=pexp(i),the local inverse at zo can be obtained bywriting g(w)=logw,wherelogwdenotesthebranchlogw=Inp+ig(p>0,元0,-元<0<元）of the logarithmic function to define g . In this case, g(l) = 0
are satisfied in N . Thus, g is analytic in N . Furthermore, we compute when = zfw )( , )( 11 )( 1 )( 1 )( zfivu ivu J ivv J iyxwg xx uu xy xx ′ = + ′ =−=−=+= . The proof is completed. Definition 8.3.1. The function g in Theorem 8.3.1 is called the local inverse of the function f at . 0 z Clearly, unilateral function on a domain has an inverse function that equals to the local inverse of at every point . Thus, from Theorems 8.2.1, 8.2.3 and 8.3.1, we have f D → DDff − )(: 1 f 0 z Corollary 8.3.1. Every unilateral function on a domain has a unilateral inverse on the domain and f D −1 f Df )( )()(, )( 1 )(1 Dfzfw zf wf ∈=∀ ′ = − . Corollary 8.3.2. Every unilateral function on a domain is a homoemorphism from onto , i.e. are continuous. f D D Df )( → → DDffDfDf − )(:),(: 1 Example 1. We saw in Example 1, Sec. 9.1, that if , the mapping is conformal everywhere in the plane and, in particular, at the point z )( = ezf = zfw )( z 2π iz0 = . The image of this choice of is the point . When points in the plane are expressed in the form 0 z w0 = 1 w = ρ iw φ)exp( , the local inverse at can be obtained by writing , where denotes the branch 0 z = log)( wwg log w = lnlog ρ + iw φ ρ > π ,0( −π < θ < π ) of the logarithmic function to define g . In this case, g = 0)1(
s8.4.AffineTransformationsTo study the mappingW= Az,(8.4.1)where A isa nonzero complexconstantand z+O,we write A and z inexponential form:A=aeia, z=reieThenW= (ar)ei(a+0)(8.4.2)and we see from equation (8.4.2)thattransformation (8.4.1)expands or contracts the radius vectorrepresenting z by the factor a=A and rotates it through an angle α=argA abount theorigin.The imageofa given region is, therefore,geometrically similar to that region.In this sencewe call (8.4.1)a similarity.ThemappingW=z+B,(8.4.3)where B is any complex constant, is a translation by means of the vector representing B.Thatis, if we writew=u+iv, z=x=iy,and B=b, +ib,thentheimageofanypoint (x,y)inthezplaneisthepoint(u.v)=(x+b.y+b,)(8.4.4)in the w plane.Since each point in any given region of the z plane is mapped into the wplane in this manner, the image region is geometrically congruent to the original one.Generally,thetransformationW=Az+B (A+O),(8.4.5)is calledan affinetransformation,which is acomposition of thetransformationsZ=Az (A+0) and w=Z+B.Example. The mapping w =(1+i)z +2 transforms the rectangular region shown in thez plane of Fig. 8-4 into the rectangular region shown in the w plane there. This is seen bywritingitasacompositionofthetransformationsZ=(1+i)z and w=Z+2.Since 1+i = /2exp(iπ /4), the first of these transformations is an expansion by the factorV2and arotation through the angle π/4.The second is a translation two units to the rightyiY-1+3i+311+2iBB'BA1olAXXo2Fig. 8-4
§8.4. Affine Transformations To study the mapping = Azw , (8.4.1) where A is a nonzero complex constant and z ≠ 0, we write A and z in exponential form: iα = aeA , . iθ = rez Then )( )( +θα = i earw , (8.4.2) and we see from equation (8.4.2) that transformation (8.4.1) expands or contracts the radius vector representing z by the factor = Aa || and rotates it through an angle α = arg A abount the origin. The image of a given region is, therefore, geometrically similar to that region. In this sence, we call (8.4.1) a similarity. The mapping = + Bzw , (8.4.3) where B is any complex constant, is a translation by means of the vector representing B . That is, if we write += ivuw , z = x = iy , and 21 = + ibbB , then the image of any point yx ),( in the z plane is the point ),(),( 1 2 = + + bybxvu (8.4.4) in the plane. Since each point in any given region of the plane is mapped into the plane in this manner, the image region is geometrically congruent to the original one. w z w Generally, the transformation = + ABAzw ≠ )0( , (8.4.5) is called an affine transformation, which is a composition of the transformations = AAzZ ≠ )0( and = + BZw . Example. The mapping = + ziw + 2)1( transforms the rectangular region shown in the plane of Fig. 8-4 into the rectangular region shown in the plane there. This is seen by writing it as a composition of the transformations z w = + )1( ziZ and = Zw + 2 . Since =+ ii π )4/exp(21 , the first of these transformations is an expansion by the factor 2 and a rotation through the angle π 4/ . The second is a translation two units to the right. Fig. 8-4
$8.5.TheTransformationw=1/zThe equation1W=(8.5.1)2establishesaonetoonecorrespondencebetweenthenonzeropointsofthezandthewplanesSincezz=z2,themappingcanbedescribedbymeansofthesuccessivetransformations1z,W=Z2.(8.5.2)/=PThe first of these transformations is an inversion with respect to the unit circle [z =1. Thatis,the imageofa nonzero point z is thepoint Zwith the properties1[ZFand argZ =argz[z]Thus thepoints exterior to the circleIz=1 aremapped onto the nonzero points interior to it(Fig.8-5), and conversely.Any point on the circle is mapped onto itself.The second oftransformations(8.5.2)is simplya reflection in thereal axis.y40Fig.8-5If we write transformation (8.5.1)as1T(=) =(= ±0),(8.5.3)Nwe can define T at the origin and at the point at infinity so as to be continuous on the extendedcomplexplane.To do this,we need onlyrefer to Sec.2.6to see that1limT(=) = 0 since lim:0(8.5.4)0 T(2)=→0andlimT(=)=0 sincelimT(8.5.5)→0(2In ordertomake Tcontinuous on theextended plane,wedefineT(0)= 8, T() = 0(8.5.6)More precisely, equations (8.5.6), together with the first of limits (8.5.4) and (8.5.5), show thatlim T(2) = T(=)(8.5.7)→=0for every point zo in the extended plane, including zo = O and =。 = co
§8.5. The Transformation = /1 zw The equation z w 1 = (8.5.1) establishes a one to one correspondence between the nonzero points of the and the planes. Since z w 2 = zzz || , the mapping can be described by means of the successive transformations z z Z 2 || 1 = , = Zw . (8.5.2) The first of these transformations is an inversion with respect to the unit circle . That is, the image of a nonzero point is the point z = 1|| z Z with the properties || 1 || z Z = and Z = argarg z . Thus the points exterior to the circle z = 1|| are mapped onto the nonzero points interior to it (Fig. 8-5), and conversely. Any point on the circle is mapped onto itself. The second of transformations (8.5.2) is simply a reflection in the real axis. Fig. 8-5 If we write transformation (8.5.1) as ( 0) 1 ( ) = z ≠ z T z , (8.5.3) we can define T at the origin and at the point at infinity so as to be continuous on the extended complex plane. To do this, we need only refer to Sec.2.6 to see that = ∞ → lim ( ) 0 T z z since 0 ( ) 1 lim 0 = → T z z (8.5.4) and lim ( ) = 0 →∞ T z z since 0 1 lim 0 ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ → z T z . (8.5.5) In order to make T continuous on the extended plane, we define T (0) = ∞ , T (∞) = 0 . (8.5.6) More precisely, equations (8.5.6), together with the first of limits (8.5.4) and (8.5.5), show that lim ( ) ( ) 0 0 T z T z z z = → (8.5.7) for every point in the extended plane, including 0 z z0 = 0 and z0 = ∞