NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 27
Complex Analysis Module: 2: Functions of a Complex Variable Lecture: 4: Analytic functions A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 2 / 27
Derivatives A small note So far, we have seen the mapping of functions that takes values from Z -plane (xy-plane) to the Ω-plane(uv-plane). In the Ω-plane, u and v are functions of x and y. w = u + iv Ω, u = u(x, y), v = v(x, y), x + iy Z (xy plane). A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 3 / 27
Derivatives A small note So far, we have seen the mapping of functions that takes values from Z -plane (xy-plane) to the Ω-plane(uv-plane). In the Ω-plane, u and v are functions of x and y. w = u + iv Ω, u = u(x, y), v = v(x, y), x + iy Z (xy plane). In the one-variable real case, the concept that follows continuity is derivative. Here, before the concept of differentiability, we consider the concept of partial derivatives which is similar to the two-variable real case. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 3 / 27
Partial Derivatives Definition Let u(x, y) be defined at all points in the neighbourhood of (x 0, y 0 ). Then we define the partial derivative of u with respect to x and y, respectively, as u x (x 0, y 0 ) = u y (x 0, y 0 ) = u(x 0 + x, y 0 ) u(x 0, y 0 ) lim x 0 x lim y 0 provided the limit exist, in each case. u(x 0, y 0 + y) u(x 0, y 0 ) y A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 4 / 27
Partial Derivatives Definition Let u(x, y) be defined at all points in the neighbourhood of (x 0, y 0 ). Then we define the partial derivative of u with respect to x and y, respectively, as u x (x 0, y 0 ) = u y (x 0, y 0 ) = u(x 0 + x, y 0 ) u(x 0, y 0 ) lim x 0 x lim y 0 provided the limit exist, in each case. Notation u(x 0, y 0 + y) u(x 0, y 0 ) y Some authors prefer to write the notation u x = u x. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 4 / 27
Partial Derivatives Second order partial derivatives u xx = 2 u x 2, u xy = u yy = 2 u y 2, 2 u x y, u yx = 2 u y x, where u xy and u yx are called second order mixed partial derivatives. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 5 / 27
Partial Derivatives Note Assume that, for the function u(x, y), the first order partial derivatives u x, u y and the second order partial derivatives u xx, u yy, u xy and u yx exist at (x 0, y 0 ). Further if all these partial derivatives are continuous in the neighbourhood (x 0, y 0 ), then the mixed partial derivatives are equal at (x 0, y 0 ); i.e., u xy (x 0, y 0 ) = u yx (x 0, y 0 ). A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 6 / 27
Partial Derivatives Example Let xy(x 2 y 2 ) u(x, y) = x 2 + y 2 for (x, y) (0, 0) 0 for (x, y) = (0, 0). For this function, u xy (0, 0) = 1 and u yx (0, 0) = 1, because the mixed partial derivatives are not continuous at (0, 0). A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 7 / 27
Derivatives Definition Let f : D 1 D 2. Let z 0 D. f is differentiable at z 0 if f (z 0 + z) f (z 0 ) lim 0 z exists. This derivative is denoted as f (z 0 ). A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 8 / 27
Derivatives Note. If f (z 0 ) exists for each z 0 D, then f is said to be differentiable in D. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 9 / 27
Derivatives Example Question.Find the derivative of f (z) = z n. Answer. Using binomial theorem, we write Thus (z + z) n z n f (z) = δz ( lim nz n 1 + z 0 nz n 1 z + = = nz n 1 + n(n 1) z n 2 ( z) 2 + + ( z) n 2 z n(n 1) z n 2 ( z) + + ( z) n 1. 2 ) n(n 1) z n 2 ( z) + + ( z) n 1 = nz n 1. 2 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 10 / 27
Derivatives Note. The above example is similar to the real-variable case. Hence the following results can be adapted from the real variable calculus. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 11 / 27
Derivatives Results Theorem If f and g are differentiable then (f ± g) (z) = f (z) ± g (z), (cf ) (z) = cf (z) for some non-zero real constant c, (fg) (z) = f (z)g (z) + f (z)g(z), ( ) f (z) = g(z)f (z) f (z)g (z) g (g(z)) 2, if g(z) 0. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 12 / 27
Derivatives Chain Rule Theorem Let f (z) be a differentiable function of z and w = f (z). If g(w) is a differentiable function of w, then the composition function g(f (z)) is also differentiable and g (w) = (g(f (z))) = g (f (z))f (z). A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 13 / 27
Derivatives Analytic functions A topological behaviour of a complex valued function, obtained from the concept of its derivative, is called Analytic function. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 14 / 27
Analytic function Definition Let f : D 1 D 2. Let f be differentiable at z 0 and also in the neighborhood of z 0. Then f is said to be analytic at z 0. If this is true for each z 0 D 1 then f is analytic in D 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 15 / 27
Analytic function Example (i). f (z) = z is analytic at everywhere. (ii). All polynomials are analytic in C. (iii). f (z) = z is not analytic at z = 1. 1 z A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 16 / 27
Non-Analytic function Example Question. To check if f (z) = z is analytic at origin. Answer. First we find the derivative of f (z) = z at origin. For this, we need to find f (0 + z) f (0) lim exists. z 0 z This implies f (0 + z) f (0) f ( z) f (0) lim = lim z 0 z z 0 z z = lim z 0 z = Now we consider two different cases. lim ( x, y) 0 z 0 = lim z 0 z x i y x + i y. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 17 / 27
non-analytic function Example case(i): Let x 0 first. Then, ( ) x i y lim lim y 0 x 0 x + i y case(ii): Let y 0 first. Then, ( lim x 0 lim y 0 x i y x + i y = lim ( 1) = 1. y 0 ) = lim (1) = 1. x 0 Since taking the limit in different ways leads to different values, f (z) = z is not differentiable at origin and hence not analytic. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 18 / 27
non-analytic function Example In general f (z) = z is nowhere differentiable and hence is not analytic anywhere in the complex plane C. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 19 / 27
non-analytic function Example In general f (z) = z is nowhere differentiable and hence is not analytic anywhere in the complex plane C. Remark Since f (z) = z is not analytic, any f (z) that can be written in terms of z is not analytic. Since 2Re z = z + z, 2iIm z = z z, f (z) is not analytic if it can be written in the form consisting of z, Re z, Im z. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 19 / 27
Analytic function A consequence One of the important consequences of analytic functions is the following result. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 20 / 27
Analytic function A consequence One of the important consequences of analytic functions is the following result. Theorem Let f be analytic in a open connected set (domain) D and f (z) = 0 everywhere in D. Then f is constant throughout D. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 20 / 27
Analytic function Remark Note that in the previous result, the connectedness of the domain is essential. For example, define f (z) by { α if z < r1 f (z) = β if z > r 1 + 2, for some constant α and r 1 > 0. Then also f (z) = 0. But f (z) is not constant. Here the domain where f (z) is defined is not connected. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 21 / 27
Analytic function Theorem Let f be analytic in a domain D 1 and g be analytic in a domain D 2, then their sum, difference and product are analytic in the domain D 1 D 2. Similarly their quotient f /g is analytic in D 1 D 2, if g(z) 0 for z D 1 D 2. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 22 / 27
Analytic function Choice of domain for resultant function Remark The domain for the resultant function of two analytic function should be chosen very carefully. For the composition of two analytic functions which follows the chain rule, an example is given below to illustrate the choice of the domain. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 23 / 27
Analytic function Choice of domain for resultant function Example Define f (z) = z 2. Clearly this is an entire function (analytic in the whole complex plane). Define g(z) = z 1/2. This is a multi-valued function. Converting this function into polar form and restricting the domain of definition as g(z) = z 1/2 = ρe iφ/2, (ρ > 0, π < φ < π) it can be seen that this function has a derivative at each point of z in the domain of definition and g (z) = 1/2g(z). A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 24 / 27
Analytic function Choice of domain for resultant function Example Hence this function is analytic everywhere in the given domain of definition. For the composition g(f (z)), writing w = f (z) = re iθ, the function is defined as g(f (z)) = g(w) = w 1/2 = re iθ/2, (r > 0, π < θ < π). Even though, the domain of f is C, we should restrict its domain such that the range of f is contained in the domain on which g is defined. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 25 / 27
Analytic function Choice of domain for resultant function Example Hence the largest possible domain in which f can be defined so that the range of f should lie in the domain r > 0, π < θ < π. For this purpose, we write w = ρ 2 e i2φ. This gives π < φ < π when π/2 < φ < π/2. Thus r = ρ 2 and θ = 2φ, where r > 0 and π < θ < π. Hence the plane ρ > 0, π/2 < φ < π/2 is the largest possible domain that can be taken for the definition of f. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 26 / 27
Analytic function Choice of domain for resultant function Example This domain gives the analyticity of f (g(z)) from the analyticity of f and g in the respective domain. Hence g(f (z)) = g(w) = ρ 2 e i2φ/2 = ρe iφ = z, for any point z in the domain ρ > 0, π/2 < φ < π/2. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 27 / 27