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 r<R such that V0<<r,3S>0 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 r< R such that functions f(=)-Wo and f'(2)have no zeros in D=N(zo,r) except z=zo.Thus, for each O<<r, it is clear that zo isthe unique zero of f(z)-wo in N(zo,) and it is of order m.Clearly,8= min 1f(z)-w0Let weN(wo,),we writef(2)-w=(f(=)-wo)+(w。-w).Since [f(=)-wo≥>wo-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 numberr<R such that VO<<r,38>O 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 < R such that ∀ < ε < r ∃δ > 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 < R such that functions and have no zeros in 0 )( − wzf ′ zf )( ),( 0 = rzND except 0 = zz . Thus, for each 0 < ε < r, it is clear that is the unique zero of in 0 z 0 )( − wzf ),( 0 zN ε and it is of order m . Clearly, 0|)(|min: 0 || 0 = − > =− 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 < R such that ∀ ε r δ >∃<< 0,0 so that for each fixed point , the function has exactly zeros counting multiplicity in the deleted neighborhood and they are all of order 1. Since , for each fixed point , there are two distinct points in such that ),(wNw 0 δ o ∈ )( − wzf m ),( 0 zN ε o m >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