MAT665:ANALYTIC FUNCTION THEORY

MAT665:ANALYTIC FUNCTION THEORY DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR Contents 1. About 2 2. Complex Numbers 2 3. Fundamental inequalities 2 4. Continuously differentiable functions 3 5. Holomorphic function 3 5.1. Partial derivative form 4 5.2. The complex derivative 4 6. Harmonic functions 4 6.1. Use of Harmonic mappings 5 6.2. Harmonic functions and holomorphicity 5 6.3. Harmonic Conjugate 5 7. Conformal mapping 6 8. Inverse functions of analytic functions 6 8.1. Continuity of Logarithm function 7 8.2. Analyticity of Logarithm function 7 9. Derivative of analytic function 7 10. Conformal Mapping 7 10.1. The angle between curves 8 11. Analytic functions and conformality 9 12. Möbius transformations 10 12.1. Properties of Möbius transformation 10 12.2. Conformality of Möbius transformation 10 12.3. Unique Möbius transformation 12 12.4. Facts about Möbius transformation 12 12.5. Example of Möbius transformation 13 12.6. Composition of Möbius transformation 13 13. Conformal Self-maps 13 14. Local geometry of Holomorphic functions 14 15. Schwarz s Lemma 14 15.1. Application of Schwarz Lemma 15 15.2. Schwarz-Pick Lemma 16 References 16 Date: August 12, 2016. Email:ragarwal.maths@mnit.ac.in. 1

2 DR. RITU AGARWAL 1. About Complex variables originally conceived in the pursuit of solutions of polynomial equations (such as x 2 + 1 = 0). This subject blossomed in the hands of Euler, Argand and others. Purpose Main purpose of this subject is to study the analytic and geometric properties of complex functions. Applications It has found applications in various fields: (1) Eigenvalues that arise in mechanical vibrations are complex numbers; (2) Incompressible fluid flow can be modelled using complex functions; (3) Solutions of variety of differential equations are complex functions; (4) Fourier transform is possible due to complex variable;. and many more... 2. Complex Numbers Definition 1. The Complex Numbers The complex number C consist of R 2 with some special algebraic operations (x + y) + (x, y ) = (x + x, y + y ) (x, y) (x, y ) = (xx yy, xy + yx ) These operations of + and are commutative and associative. Remark 1. C is a complete field but not ordered. The complex number (x, y) is usually written as z = x + iy where i = 1. Complex conjugate of a complex number z = x+iy is z = x iy and zz = z 2 = x 2 +y 2. z denotes the modulus (norm) of z. Polar form of complex number z is z = r e iθ where r = z = x 2 + y 2 and θ = arg z = tan 1 y/x. If p is a complex number and r > 0 then D(p, r) = {z C : z p < r represents an open disc with center p and radius r. D(p, r) = {z C : z p r represents a closed disc with center p and radius r. C is the smallest field that contains R and this is called algebraic closure property. 3. Fundamental inequalities We next record a few inequalities Triangle inequality: If z, w C, then z + w z + w (3.1)

MAT665:ANALYTIC FUNCTION THEORY 3 Cauchy Schwarz Inequality: If z 1,..., z n and w 1,..., w n are complex numbers then [ n n ] [ n ] z j w j z j 2 w j 2 j=1 j=1 j=1 (3.2) Exercise 1. Q.1 Prove that the set of complex numbers C is a field. Q.2 Consider the polynomial p(z) = z 3 z 2 + 2z 2. How many real roots and complex roots does p have? Explain. Q.3 Let z = 4 6i, w = 2 + 7i. Calculate z/w, w/z and 1/w. Q.4 Sketch these discs on the same set of axes: D(2+3i, 4), D(1 2i, 2), D(i, 5) and D(6 2i, 5). Q.5 Calculate that e z = e x. Find cos z 2 and sin z 2 in terms of x and y. Q.6 If w 2 = z 3, then how their polar forms are related? Q.7 If z and w are any complex numbers then explain why z + w z w? Definition 2. Holomorphic functions Holomorphic functions are generalizations of complex polynomials. The collection of polynomials is closed under addition and multiplication. However, the collection of all holomorphic functions is closed under reciprocals, division, inverse, exponentiation, logarithm, square roots and many other operations as well. We shall discuss here the different approaches to define Holomorphic functions. 4. Continuously differentiable functions Definition 3. C 1 function If U R 2 is a region and f : U R is a continuous function, then f is called C 1 (or continuously differentiable) on U if f/ x and f/ y exist and continuous on U. We write f C 1 (U) for short. Definition 4. C k function If U R 2 is a region and f : U R is a continuous function, then f is called C k, k {0, 1, 2,...} (or k-times continuously differentiable) on U if all partial derivatives of f up to and including order k exist and are continuous on U. We write f C k (U) for short. Remarks: A C 0 function is continuous. A function is C if it is C k for every k. Such a function is called infinitely differentiable function. A function f = u + iv : U C is C k if both u and v are C k. 5. Holomorphic function Definition 5. C-R Equations Let U C be a region and f : U C is a C 1 function written as f(z) = u(x, y)+iv(x, y), z = x+iy, u, v being real valued functions and satisfy the equations u x = v y, u y = v (5.1) x at every point of U, then f is said to be Holomorphic on U. Equations ( 5.1) are known as Cauchy-Riemann (C-R) equations.

4 DR. RITU AGARWAL Example 5.1. (i) Show that the function f defined as f(z) = z 2 z, is holomorphic in C. Here, f(z) = (x 2 y 2 x) + i(2xy y). (ii) Find out whether the function g(z) = z 2 4z + 2z is holomorphic or not? Here, g(z) = (x 2 + y 2 2x) + i( 6y). 5.1. Partial derivative form. Define f = u + iv : U C a C 1 function. Also, x = (z + z)/2 and x = (z z)/2i. We may write, = x. + y. and similarly for z. Hence, z z x z y f z 1 ( 2 x i ) f = 1 ( u y 2 x + v ) + i ( v y 2 x u ) y f z 1 ( 2 x + i ) f = 1 ( u y 2 x v ) + i ( v y 2 x + u ) y (5.2) (5.3) Definition 6. If f is holomorphic, it satisfies C-R equations, then from (5.2) and (5.3), we get f = 0. z Another condition for f to be holomorphic. That is, If f is holomorphic, if f is independent of z. Further, if f f = 0 and = 0, then all directional derivatives of f are identically zero. Hence z z f is constant. Example 5.2. Review examples 1 and 2 by finding partial derivative with respect to z. Certainly, any polynomial in z (without z) is holomorphic. 5.2. The complex derivative. Definition 7. If a function f possesses the complex derivative at every point of its open domain U, then f is holomorphic. Goursat s theorem says that if a function f possesses the complex derivative at each point in the open region U C, then f is in fact continuously differentiable on U and Cauchy- Riemann equations can be verified for f and hence f is holomorphic. Alternative terminology for holomorphic functions Analytic: The use of term analytic derives from the fact that a holomorphic function has a local power series expansion about each point of its domain. Conformal: Conformality is an important geometric property of the holomorphic functions 6. Harmonic functions A C 2 (twice continuously differentiable) real valued function u is said to be harmonic if it satisfies the Laplace equation u = 0 that is, ( ) 2 x 2 + 2 y 2 u = 0 (6.1)

MAT665:ANALYTIC FUNCTION THEORY 5 Example 6.1. 1.The function u(x, y) = x 2 y 2 is harmonic. 2. The function u(x, y) = x 3 is not harmonic. 3. The function u(x, y) = sin x cos y is not harmonic. 6.1. Use of Harmonic mappings. If a function is harmonic (that is, it satisfies Laplace s equation 2 f = 0) over a plane domain (which is two-dimensional), and is transformed via a conformal map to another plane domain, the transformation is also harmonic. For this reason, any function which is defined by a potential can be transformed by a conformal map and still remain governed by a potential. Examples in physics of equations defined by a potential include the electromagnetic field, the gravitational field, and, in fluid dynamics, potential flow, which is an approximation to fluid flow assuming constant density, zero viscosity, and irrotational flow. One example of a fluid dynamic application of a conformal map is the Joukowsky transform. 6.2. Harmonic functions and holomorphicity. If f = u + iv is a holomorphic function then both u and v are harmonic. To see, apply z f = 0 (6.2) Further, take partial derivative w.r.to z Making use of (5.2) and (5.3), we get ( 2 z z f = 0 (6.3) ) [u + iv] = 0 (6.4) x 2 + 2 y 2 Equating real and imaginary parts to zero, we conclude that u and v are harmonic. Example 6.2. The function f(z) = (z + z 2 ) 2 is holomorphic and hence real and imaginary parts of f are each harmonic. 6.3. Harmonic Conjugate. If U is a simply connected region (with no holes) and u is harmonic on U, then there is a holomorphic function F on U such that ReF = u, that is, there exists a harmonic function v defined on U such that F = u + iv is holomorphic on U. v is called harmonic conjugate for u. Example 6.3 (5). Consider the function u(x, y) = x 2 y 2 x on the square U = {(x, y) : x < 1, y < 1}. Find harmonic conjugate for u. Hint: Make use of C-R equations. Ans. v(x, y) = 2xy y + C, F = z 2 z + ic. Remark 2. If a function possesses the complex derivative at every point of its open domain U, then f is holomorphic. Holomorhic is another word for the term complex differentiable functions. The use of the term analytic derives from the fact that a holomorphic function has a local power series expansion about each point of its domain. In fact, power series property is a complete characterization of holomorphic functions. Conformality is an important geometric property of holomorphic functions that make these functions useful for modelling incompressible fluid flow.

6 DR. RITU AGARWAL Exercise 2. Q.1 Find out whether the given functions are holomorphic or not: (i) f(z) = e 2z z3 z 2 (ii) g(z) = cos z z 2 +1 (iii) h(z) = z 4 z 2 (iv) k(z) = z z 2 +1 Q.2 Verify the equations: z = 1, z = 0, z = 0 and z = 1. z z z z Q.3 Find a function g such that g = z zz2 sin z. Q.4 Find out whether the following functions are harmonic or not: (i) f(z) = x 3 3xy 2 (ii)g(z) = z 2 2x 2 (iii)h(z) = z 4 z 2 (iv)k(z) = z sin z 7. Conformal mapping Conformal mappings are invaluable for solving problems in engineering and physics that can be expressed in terms of functions of a complex variable but that exhibit inconvenient geometries. By choosing an appropriate mapping, the analyst can transform the inconvenient geometry into a much more convenient one. For example, one may wish to calculate the electric field, E(z), arising from a point charge located near the corner of two conducting planes separated by a certain angle (where z is the complex coordinate of a point in 2-space). This problem per se is quite clumsy to solve in closed form. However, by employing a very simple conformal mapping, the inconvenient angle is mapped to one of precisely π radians, meaning that the corner of two planes is transformed to a straight line. In this new domain, the problem (that of calculating the electric field impressed by a point charge located near a conducting wall) is quite easy to solve. The solution is obtained in this domain, E(w), and then mapped back to the original domain by noting that w was obtained as a function (viz., the composition of E and w) of z, whence E(w) can be viewed as E(w(z)), which is a function of z, the original coordinate basis. Note that this application is not a contradiction to the fact that conformal mappings preserve angles, they do so only for points in the interior of their domain, and not at the boundary. 8. Inverse functions of analytic functions Logarithm function Given z C\{0}, find w C such that e w = z. Write z = z e iθ then e w = z e iθ. Next write w = u + iv then e u e iv = z e iθ. Thus e u = z and e iv = e iθ. So, u = ln z and v = arg θ. Definition 8. Logarithm function For z 0, we define the principal branch of the logarithm and a multivalued function. Log z = ln z + i arg z log z = ln z + i arg z = Log z + 2kπi Example 8.1. Log 1 = ln 1 + i arg 1 = 0. Log i = ln i + i arg π 2 = iπ/2. Log ( 1) = ln 1 + iπ = iπ. k Z

Log (1 + i) = ln 2 + i arg π 4 = 0. MAT665:ANALYTIC FUNCTION THEORY 7 8.1. Continuity of Logarithm function. Log z = ln z + i arg z is continuous in C\(, 0] z z is continuous in C z ln z is continuous in C\{0}. z arg z is continuous in C\(, 0] as z x (, 0) from above Logz ln x + iπ and z x (, 0) from below Logz ln x iπ Thus Log z = ln z + i arg z is not continuous in (, 0) and not defined at origin (z=0). 8.2. Analyticity of Logarithm function. Principal branch of logarithm, Log z, is analytic in C\(, 0]. Derivative of logarithmic function: e Log z = z e Log z. d dz Log z = 1 d dz Log z = 1 z Example 8.2. Calculate the principal branch of the logarithm of each of the following complex numbers: (i) 2 + 2i (ii) 3 3 3i (iii) i (iv) 2 + 2 3i 9. Derivative of analytic function Theorem 9.1. Let f : U C is an analytic function and there exists a continuous function g : D U from some domain D C into U such that f(g(z)) = z for all z D. Then g is analytic in D and g 1 (z) = for z D (9.1) f (g(z)) 10. Conformal Mapping Path: A path in complex plane from a point A to point B is a continuous function γ : [a, b] C such that γ(a) = A and γ(b) = B. Example 10.1. Draw the following curves in z-plane. (i) γ(t) = (2 + i) + e it, 0 t π. (ii) γ(t) = (2 + i) + t( 3 5i) = (2 3t) + i(1 5t), 0 t 1. (iii) γ(t) = te it = t cos t + it sin t, 0 t 3π. (iv) t(1 + i), 0 t 1 γ(t) = t + i, 1 < t 2 2 + i(3 t), 2 < t 3

8 DR. RITU AGARWAL Solution: (i) 2. z = (2 + i) + e it 1. 4. 3. 2. 1. 0 1. 2. 3. 1. (ii) 2. Definition 9. Smooth Curve A path γ : [a, b] C is smooth if the functions x(t) and y(t) in the representation γ(t) = x(t) + iy(t) are smooth, that is, have as many derivatives as desired. In the above example 10.1, (i), (ii) and (iii) are smooth whereas (iv) is piecewise smooth path. The term curve is typically used for a smooth or piecewise smooth path. If γ : [a, b] C is a smooth curve and t 0 (a, b), then γ (t 0 ) = x (t 0 ) + iy (t 0 ) is a tangent vector to γ at z 0 = γ(t 0 ). 10.1. The angle between curves. Definition 10. Let γ 1 and γ 2 be two smooth curves, intersecting at a point z 0. The angle between two curves at z 0 is defined as the angle bwteen the two tangent vectors at z 0. Figure 1. Angle between two curves

MAT665:ANALYTIC FUNCTION THEORY 9 Figure 2. Mapping of curves Figure 3. Angle between curves in z and w planes Example 10.2. Let γ 1 : [0, π] C, γ 1 (t) = e it and γ 2 : [ π Then γ 1 (π/2) = γ 2 (π) = i. Furthermore, γ 1(t) = ie it and γ 2(t) = 2ie it. γ 1(π/2) = ie iπ/2 = 1 and γ 2(π) = 2ie iπ = 2i. The angle between two curves is thus π/2. 2, 3π 2 ) C, γ2 (t) = 2 + i + 2e it. Definition 11. A function is conformal if it preserves angles between curves. More precisely, a smooth complex valued function g is conformal at z 0 if whenever two curves intersect at z 0 with non-zero tangents, the g γ 1 and g γ 2 have non-zero tangents at g(z 0 ) that intersect at the same angles. A conformal mapping of a domain D onto V is continuously differentiable mapping that is conformal at each point in D and maps D one-to-one onto V. A mapping is isogonal if magnitude of angle is preserved but not the orientation. 11. Analytic functions and conformality Theorem 11.1. If f : U C is analytic and if z 0 U, such that f (z 0 ) 0, then f is conformal at z 0. Reason: If γ : [a, b] U is a curve in U with γ(t 0 ) = z 0 for some t 0 (a, b), then (f γ) (t 0 ) = f (γ(t 0 )).γ (t 0 ) = f (z 0 )γ (t 0 ) (11.1) Thus (f γ) (t 0 ) is obtained from γ (t 0 ) via multiplication by f (z 0 ) (=rotation and stretching). If γ 1, γ 2 are two curves in U through z 0 with tangent vectors γ (t 1 ) and γ (t 2 ), then (f γ) (t 0 ) and (f γ) (t 0 ) are both obtained from γ (t 1 ) and γ (t 2 ), respectively, via multiplication by f (z 0 ). The angle between them is thus preserved.

10 DR. RITU AGARWAL Figure 4. Examples of conformal mappings: Mapping of circles with center at origin and lines passing through origin 12. Möbius transformations Definition 12. A Möbius transformation (also called fractional linear transformation) is a function of the form f(z) = az + b (12.1) cz + d where a, b, c, d C such that ad bc 0. Remark 3. As z, f(z) a c Remark 4. Similarly, f ( d c ) = if c 0. if c 0 and f(z) if c = 0. Remark 5. We thus, regard f as a mapping from the extended complex plane Ĉ = C { } to the extended plane Ĉ. 12.1. Properties of Möbius transformation. f (cz + d)za (az + b)c ad bc (z) = =. The condition ad bc 0 thus simply (cz + d) 2 (cz + d) 2 guarantees that f is non-constant. If we multiply each of the parameters by a, b, c, d by a constant k 0, we obtain the same mapping. SO a given mapping does not uniquely determine a, b, c, d. A Möbius transformation is one-to-one and onto from Ĉ to Ĉ. To prove, pick w Ĉ f(z) = w az + b = w(cz + d) z(a wc) = wd b z = wd b wc+a. 12.2. Conformality of Möbius transformation. Möbius transformation are the only conformal mappings from Ĉ to Ĉ Example 12.1. (i) c = 0, d = 1: Then f(z) = az+b, a 0 are also called affine transformations. They map to and therefore map C to C. In fact, these are the only conformal mappings from C to C. In particular,

MAT665:ANALYTIC FUNCTION THEORY 11 if f(z) = az, b = 0 is a rotation and dilation. f(z) = z + b, (a = 1) is a translation. (ii) a = 0, b = 1 c = 1, d = 0 then f(z) = 1/z. This is an inversion. f interchanges outside and inside of the unit circle. How about other circles? A circle centered at 0, is clearly mapped to a circle centered at 0 of reciprocal radius. Example 12.2. Let K = {z : z 3 = 1} be the circle of radius 1, centered at 3. What is the image of f(k)? w f(k) 1 w K 1 w 3 = 1 1 3w 2 = w 2 1 3w 3w + 9 w 2 = w 2 8 w 2 3w 3w = 1 (w 3/8) (w 3/8) = 9 64 1 8 w 3 8 = 1 8 Thus, the image of the circle K = {z : z 3 = 1} under f(z) = 1/z is another circle of radius 1/8, centered at 3/8. A circle centered at 0, is clearly mapped to a circle centered at 0 of reciprocal radius. Example 12.3. Let K = {z : z 1 = 1} be the circle of radius 1, centered at 1. What is the iamge of f(k)? w f(k) 1 w K 1 w 1 = 1 1 w 2 = w 2 1 w w + w 2 = w 2 w + w = 1 R(w) = 1/2 Thus, the image of the circle K = {z : z 1 = 1} under f(z) = 1/z is the vertical line through 1/2. Mapping of lines in inversion Since f(f(z)) = f(1/z) = z, we also find f maps the line {z : Rz = 1/2} to the circle {z : z 1 = 1}. f maps the circle {z : z 3/8 = 1/8} to the circle {z : z 3 = 1}. Let now L be the line {z : z = t + it, < t < }. Then f(z) = 1/z implies f(z) = 1 i12t = s is. 2t A line could be viewed as circle through infinity.

12 DR. RITU AGARWAL Exercise 3. Show that Every Möbius transformation maps circles and lines to circles or lines. 12.3. Unique Möbius transformation. Given three distinct points z 1, z 2, z 3 Ĉ, there exists a unique Möbius transformation f such that f(z 1 ) = 0, f(z 2 ) = 1 and f(z 3 ) = as follows: Let the corresponding Möbius transformation f : Ĉ Ĉ be f(z) = z z 1 z z 3. z 2 z 3 z 2 z 1 (12.2) w = f(z) = k z z 1 z z 3 regardless of the choice of constant k. f sends z 1 0, z 3. Adjusting the constant k so that f(z 2 ) = 1, we get 12.2. Note that AD BC = (z 2 z 1 )(z 2 z 3 )(z 1 z 3 ) 0. 12.4. Facts about Möbius transformation. The composition of two Möbius transformations is a Möbius transformation and so is the inverse. If f(z) = az + b ez + f, g(z) =, f(g(z)) is a Möbius transformation. cz + d gz + h Given three distinct points z 1, z 2, z 3 and w 1, w 2, w 3, there exists a unique Möbius transformation f : Ĉ Ĉ that maps z j to w j, j = 1, 2, 3. Proof: Let f 1 be Möbius transformation that maps z 1, z 2, z 3 to 0, 1,. Let f 2 be Möbius transformation that maps w 1, w 2, w 3 to 0, 1,. Then f 1 2 f 1 maps z 1, z 2, z 3 to w 1, w 2, w 3 respectively. Theorem 12.4 (Cross Ratio Theorem). If {z 1, z 2, z 3 } and {w 1, w 2, w 3 } are two sets of triplets of distinct points in Ĉ, then there exists a unique Möbius transformation taking z j to w j (j=1,2,3) given by w w 1. w 2 w 3 = z z 1. z 2 z 3 (12.3) w w 3 w 2 w 1 z z 3 z 2 z 1 Exercise 4. Show that, The cross-ratio is invariant under Möbius transformation.

MAT665:ANALYTIC FUNCTION THEORY 13 12.5. Example of Möbius transformation. Find the Möbius transformation f that maps 0 to 1, i to 0 and to 1. f 1 (z) = f(z) = z 0 z.i i 0 = z i f 2 (z) = f(z) = z + 1 z 1.0 1 0 + 1 = z + 1 z + 1 To find f 1 2 solve for z: w = z + 1 z + 1 z = w 1 w + 1 Thus, f(z) = f2 1 f 1 (z) = f 1(z) 1 f 1 (z) + 1 = z i z + i Any alternative approach?? is the desired map. Exercise 5. Find the Möbius transformation discussed in above example using a different approach. 12.6. Composition of Möbius transformation. Theorem 12.5. Every Möbius transformation f is the composition of the maps of the types. z az (rotation and dilation) z z + b (Translation) z 1 z (Inversion). Proof: If f( ) =, then f(z) = az + b is the Möbius transformation which is composition of dilations and rotations first and then translation. If f( ), then c 0 in f(z) = az + b a cz + d = z + b c c z + d c Assume c = 1. So f(z) = az + b a(z + d) + b ad = f(z) = = a + b ad z + d z + d z + d This corresponds to the following composition: z trans. z + d inv. 1 z + d b ad dil. + Rot. z + d trans. a + b ad z + d 13. Conformal Self-maps The simplest open subset of C is C itself. A function f : C C is a conformal map iff there are complex numbers a, b with a 0 such that f(z) = az + b, z C.

14 DR. RITU AGARWAL A generalization of this result about conformal maps is the following: If h : C C is a holomorphic function such that lim h(z) = + z then h is a polynomial. If an entire function has a pole at infinity then it is a polynomial. If f : C C is a conformal map then lim z h(z) = + for both f and f 1 take bounded sets to bounded sets. So f will be a polynomial. But if f has a degree k > 1, then it will not be one-to-one. Thus f is a first degree polynomial. 14. Local geometry of Holomorphic functions Theorem 14.1 (Open mapping theorem). [1, p. 176] If f : U C is a non-constant holomorphic function on a connected open set U, then f(u) is an open set in C. Argument principle tells that if z = P is only zero of f(z) of order n in the disk D(P, r), then the integral on the boundary D(P, r) is given by 1 g (z) dz = n. 2πi g(z) Let g(z) = f(z) Q, Q = f(p ), then 1 2πi D(P,r) D(P,r) f (z) f(z) Q dz 1 as the order of zero is unknown, at least the order is one. Continuity of integral tells that if we perturb Q by a small amount, the value of the integral will not change, so it is still 1. This says that f assumes all values that are near to Q. Which says that the image of f contains a neighbourhood of Q, so it is open. Q.E.D. 15. Schwarz s Lemma Theorem 15.1 (Schwarz s Lemma). [1, Th.6, p.188] Let f be holomorphic on open unit disc D. Assume that (1) f(z) 1 for all z. (2) f(0) = 0. Then f(z) z and f (0) 1. If either f(z) = z for some z 0 or if f (0) = 1, then f is a rotation f(z) = αz for some complex constant α of unit modulus. Proof: Consider the function g(z) = f(z)/z. f being holomorphic, it can be written as f(z) = a 1 z + a 2 z 2 +.... a 0 is missing because f(0) = 0. Therefore, { f(z)/z, z 0 g(z) = f (0), z = 0

MAT665:ANALYTIC FUNCTION THEORY 15 Since g has a removable singularity at origin, we see that g is holomorhic on the entire unit disc. On the circle with center 0 and radius 1 ɛ, we see that g(z) 1 1 ɛ By the maximum modulus principle 1, it follows that g(z) 1 on all of D(0, 1 ɛ). Since 1 ɛ the conclusion is true for all ɛ > 0, we conclude that g 1 on D(0, 1). Moreover, suppose that f(z) = z for some non-zero z in D, or f(0) = 1. Then, g(z) = 1 at some point of D. So by the maximum modulus principle, g(z) is equal to a constant a such that α = 1. Therefore, f(z) = αz, as desired. Q.E.D. Alternative statement:[4, p. 210] Let f : D D be an analytic function of the unit disc into itself such that f(0) = 0. Then (1) f(z) z for all z D. (2) If for some z 0 0, we have f(z 0 ) = z 0 then f is a rotation f(z) = αz for some complex constant α of unit modulus. Corollary 15.2. If f(z) B on a disk z < R and f(0) = 0, then f(z) B z /R and equality occurs at some point only if f(z) = B αz, α = 1. R 15.1. Application of Schwarz Lemma. If ϕ is real, the map z e iϕ z geometrically represents the counterclockwise rotation by an angle ϕ. Let us find a conformal map which maps the unit disk to itself such that for a given complex number α, α < 1; f(α) = 0 and f(0) = α. Also, boundary is mapped to boundary of the unit disk. Such a mapping is given by g α (z) = α z (15.1) 1 αz Then g is analytic on the closed disc z 1. Furthermore, if z = 1, then z = e iθ for some real θ and α e iθ g(z) = (15.2) e iθ (e iθ α) and hence g(z) = 1. g α has an inverse function i.e. g α g α = I By application of second part of Schwarz lemma,we can obtain the following theorem: Theorem 15.3. [4, p. 213] Let f : D D be an analytic automorphism of the unit disc into itself such that f(α) = 0. Then there exists a real number φ such that f(z) = e iφ α z (15.3) 1 αz Exercise 6. Find a conformal map which maps the unit disk to itself such that for a given complex number α, α < 1; f(α) = 0 and f(0) = α. Also, boundary is mapped to boundary of the unit disk. Exercise 7. Prove the Theorem 15.3. 1 Let U C be a domain. Let f be holomorphic function on U. If there is a point P U such that f(p ) f(z) for all z U, then f is constant.

16 DR. RITU AGARWAL 15.2. Schwarz-Pick Lemma. Theorem 15.4 (Schwarz-Pick Lemma ). [1, p.189] Let f be holomorphic on unit disc. Assume that (1) f(z) 1 for all z. (2) f(a) = b for some a, b D(0, 1). Then f (a) 1 b 2 1 a. 2 Moreover, if f(a 1 ) = b 1 and f(a 2 ) = b 2, then b 2 b 1 1 b 1 b a 2 a 1 2 1 a 1 a 2 (15.4) There is uniqueness result in Schwarz-Pick Lemma. If either f (a) = 1 b 2 1 a or b 2 b 1 2 1 b 1 b = 2 a 2 a 1 1 a 1 a 2 with a 1 a 2, then the function f is a conformal self-mapping of D(0, 1) to itself. Exercise 8. 2 Prove the Schwarz-Pick lemma. References [1] Steven G. Krantz, Complex variables, Chapman & Hall/CRC, Indian Reprint 2012. [2] Zeev Nehari, Conformal Mapping, Dover publication, New York, 1952. [3] MOOC Video lectures by P. Bonfert Taylor on Analysis of complex kind, available on youtube. [4] S. Lang, Complex Analysis, Fourth Ed., Springer Verlag, NewYork, 1999. 2 Desirable: Assignments are to be submitted regularly, typed in L A TEX, within one week of discussion in the class. However, as you all are very busy in other more important schedules, assignments are not a compulsion for you. They may be submitted if and when you wish.