Topic 10 Notes Jeremy Orlo - MIT Mathematics

1 Topic 10 NotesJeremy Orloff10 conformal IntroductionIn this Topic we will look at the geometric notion of conformal maps. It will turn outthat analytic functions are automatically conformal . Once we have understood the generalnotion, we will look at a specific family of conformal maps called fractional linear trans-formations and, in particular at their geometric properties. As an application we will usefractional linear transformations to solve the Dirichlet problem for harmonic functions onthe unit disk with specified values on the unit circle. At the end we will return to somequestions of fluid Geometric definition of conformal mappingsWe start with a somewhat hand-wavy definition:Informal maps are functions onCthat preserve the angles precisely: Supposef(z) is differentiable atz0and (t) is a smooth curve be concrete, let s suppose (t0) =z0.

2 The function maps the pointz0tow0=f(z0) andthe curve to (t) =f( (t)).Under this map, the tangent vector (t0) atz0is mapped to the tangent vector (t0) = (f ) (t0)atw0. With these notations we have the following functionf(z) is conformal atz0if there is an angle and a scalea >0such that for any smooth curve (t) throughz0the mapfrotates the tangent vector atz0by and scales it bya. That is, for any , the tangent vector (f ) (t0) is found byrotating (t0) by and scaling it (z) is defined on a regionA, we say it is a conformal map onAif it is conformal at scale factoraand rotation angle depends on the pointz, but not on any ofthe curves figure below shows a conformal mapf(z) mapping two curves throughz0to two curves throughw0=f(z0).

3 The tangent vectors to each of the original curves areboth rotated and scaled by the same conformal TRANSFORMATIONS2xy 1(t0) 1(t) 2(t0) 2(t)z0uv(f 1) (t0)f( 1(t))(f 2) (t0)f( 2(t))w0z7 w=f(z)w0=f(z0)A conformal map rotates and scales all tangent vectors atz0by the same is a local phenomenon. At a different pointz1the rotation angleand scale factor might be rotations preserve the angles between vectors, a key property of conformalmaps is that they preserve the angles between that way back in Topic 1 we saw thatf(z) =z2maps horizontaland vertical grid lines to mutually orthogonal parabolas.

4 We will see thatf(z) is , the orthogonality of the parabolas is no accident. The conformal map preserves the rightangles between the grid w= Tangent vectors as complex numbersIn , you used parametrized curves (t) = (x(t),y(t)) in thexy-plane. Considered thisway, the tangent vector is just the derivative: (t) = (x (t),y (t)).10 conformal TRANSFORMATIONS3 Note, as a vector, (x ,y ) represents adisplacement. If the vector starts at the origin, thenthe endpoint is at (x ,y ). More typically we draw the vector starting at the point (t).In , we use parametrized curves (t) =x(t) +iy(t) in the complex plane.

5 Consideredthis way, the tangent vector is just the derivative: (t) =x (t) +iy (t).It should be clear that these representations are equivalent. The vector (x ,y ) and thecomplex numberx +iy both represent the same displacement. Also, the length of a vectorand the angle between two vectors is the same in both of tangent vectors to curves as complex numbers allows us to recast conformalityin terms of complex (z) is conformal atz0then there is a complex numberc=aei suchthat the mapfmultiplies tangent vectors atz0byc. Conversely, if the mapfmultipliesall tangent vectors atz0byc=aei thenfis conformal definitionfis conformal atz0means that there is an angle and a scalara >0such that the mapfrotates tangent vectors atz0by and scales them bya.

6 This is exactlythe effect of multiplication byc=aei . Analytic functions are conformalTheorem (Operational definition of conformal ) Iffis analytic on the regionAandf (z0)6= 0, thenfis conformal atz0. Furthermore, the mapfmultiplies tangent vectorsatz0byf (z0). proof is a quick computation. Supposez= (t) is curve throughz0with (t0) =z0. The curve (t) is transformed byfto the curvew=f( (t)). By the chain rulewe havedf( (t))dt t0=f ( (t0)) (t0) =f (z0) (t0).The theorem now follows from Theorem (Basic example) Supposec=aei and consider the mapf(z) = , this map rotates every point by and scales it bya.

7 Therefore, it musthave the same effect on all tangent vectors to curves. Indeed,fis analytic andf (z) = (z) =z2. Sof (z) = 2z. Thus the mapfhas a different affect ontangent vectors at different (Linear approximation) Supposef(z) is analytic atz= 0. The linearapproximation (first two terms of the Taylor series) isf(z) f(0) +f (0) (t) is a curve with (t0) = 0 then, neart0,f( (t)) f(0) +f (0) (t).10 conformal TRANSFORMATIONS4 That is, near 0,flooks like our basic example plus a shift byf(0).Example mapf(z) =zhas lots of nice geometric properties, but it is notconformal.

8 It preserves the length of tangent vectors and the angle between tangent reason it isn t conformal is that is does not rotate tangent vectors. Instead, it reflectsthem across other words, it reverses the orientation of a pair of vectors. Our definition of conformalmaps requires that it preserves Digression to harmonic functionsTheorem harmonic conjugates andg=u+ivhasg (z0)6= 0, thenthe level curves ofuandvthroughz0are proved this in an earlier Topic using the Cauchy-Riemann equations. Here willmake an argument involving conformal we ll examine howgmaps the level curveu(x,y) =a.

9 Sinceg=u+iv, theimage of the level curve isw=a+iv, it s (contained in) a vertical line in , the level curvev(x,y) =bis mapped to the horizontal linew=u+ , the images of the two level curves are orthogonal. Sincegis conformal it preservesthe angle between the level curves, so they must be w=g(z) =u(x, y) +iv(x, y)w0=g(z0)g=u+ivmaps level curves ofuandvto grid Riemann mapping theoremThe Riemann mapping theorem is a major theorem on conformal maps. The proof is fairlytechnical and we will skip it. In practice, we will write down explicit conformal mapsbetween (Riemann mapping theorem) IfAis simply connected and not the wholeplane, then there is a bijective conformal map fromAto the unit any two such regions there is a bijective conformal map from one to theother.

10 We say they are conformally conformal Fractional linear fractional linear transformation is a function of the formT(z) =az+bcz+d,wherea, b, c, dare complex constants andad bc6= 0 These are also called Mobius transforms or bilinear transforms. We will abbreviate fractionallinear transformation as bc= 0 thenT(z) is a constant full proof requires that we deal with all the cases where some of the coefficientsare 0. We ll give the proof assumingc6= 0 and leave the casec= 0 to you. Assumingc6= 0,the conditionad bc= 0 impliesac(c,d) = (a,b).So,T(z) =(a/c)(cz+d)cz+d= is,T(z) is to.