CHAPTER 4 Elementary Functions Dr. Pulak Sahoo

CHAPTER 4 Elementary Functions Dr. Pulak Sahoo

CHAPTER 4 Elementary Functions BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India E-mail : sahoopulak1@gmail:com 1 Module-3: Multivalued Functions-I 1 Introduction We recall that a function w = f(z) is called multivalued function if for all or some z of the domain, we find more than one value of w. Thus, a function f is said to be single valued if f satisfies f(z) = f[z(r; θ)] = f[z(r; θ + 2π)]: Otherwise, f is called as a multivalued function. We know that a multivalued function can be considered as a collection of single valued functions. For analytical properties of a multivalued function, we just consider those domains in which the functions are single valued. Branch By branch of a multivalued function f(z) defined on a domain D1 we mean a single valued function g(z) which is analytic in some sub-domain D ⊂ D1 at each point of which g(z) is one of the values of f(z). Branch Points and Branch Lines A point z = α is called a branch point of a multivalued function f(z) if the branches of f(z) are interchanged when z describes a closed path about α. We consider the function w = z1=2. Suppose that we allow z to make a complete circuit around the origin in the anticlockwise sense starting from the point P. Thus we p p have z = reiθ; w = reiθ=2; so that at P, θ = β and w = reiβ=2: After a complete p p circuit back to P, θ = β + 2π and w = rei(β+2π)=2 = − reiβ=2: Thus we have not achieved the same value of w with which we started. By making a second complete circuit p p back to P, that is when θ = β + 4π we have w = rei(β+4π)=2 = reiβ=2; which is the original value of w with which we started. 2 Thus we can say that if 0 ≤ θ < 2π we are on one branch of the function w = z1=2, while if 2π ≤ θ < 4π we are on another branch of the function (see Fig. 1). We note Fig. 1: that each branch of the function w = z1=2 is single valued. In order to keep the function single valued, we set up an artificial barrier such as OQ where Q is at infinity (though any other line from O can be used) which we agree not to cross. This barrier is called a branch line or branch cut and the point O is called a branch point. In this case the point z = 0 is the only finite branch point since a circuit around any point other than z = 0 does not give different values of w. The function f(z) = (z − 1)1=2 has a branch point at z = 1: Example 1. Consider the function w = f(z) = (z2 + 1)1=2. (a) Show that the points z = ±i are branch points of f(z). (b) Show also that a complete circuit around both the branch points produces no change in the branches of f(z). Solution. (a) We have w = f(z) = (z2 + 1)1=2 = f(z + i)(z − i)g1=2: Therefore, 1 1 arg w = arg(z − i) + arg(z + i): 2 2 This shows that 1 1 change in arg w = fchange in arg(z − i)g + fchange in arg(z + i)g: 2 2 Let C be a closed curve enclosing the point z = i but not the point z = −i (see Fig. 2). Then as point z goes once counter-clockwise around C, 3 Fig. 2: change in arg(z − i) = 2π; change in arg(z + i) = 0: So, change in arg w = π. Hence w does not return to its original value, i.e. a change in branches has occurred. Since a complete circuit about z = i alters the branches of the function, z = i is a branch point. Similarly, if C is a closed curve enclosing the point z = −i but not the point z = i, it can be shown that z = −i is a branch point. (b) If C encloses both the branch point z = i and z = −i then as point z goes counter- clockwise around C, change in arg(z − i) = 2π; change in arg(z + i) = 2π: So, change in arg w = 2π. Hence w return to its original value, i.e. no change in branches has occurred. Hence a complete circuit around both the branch points produces no change in the branches of f(z). 4 Alternative method iθ1 iθ2 (a) Let z − i = r1e ; z + i = r2e : Then p i(θ1+θ2) 1=2 iθ1=2 iθ2=2 w = fr1r2e g = r1r2e e : Suppose that we start with a particular value of z corresponding to θ1 = α1 and θ2 = α2. p iα1=2 iα2=2 Then w = r1r2e e : As z goes once counterclockwise around z = i, θ1 increases to α1 + 2π; while θ2 remains the same, i.e. θ2 = α2: Hence p p i(α1+2π)=2 iα2=2 iα1=2 iα2=2 w = r1r2e e = − r1r2e e : This shows that we do not obtain the original value of w, i.e. a change of branches has occurred. Therefore z = i is a branch point. Similarly z = −i is also a branch point. (b) In this case, θ1 increases from α1 to α1 + 2π while θ2 increases from α2 to α2 + 2π: Thus p p i(α1+2π)=2 i(α2+2π)=2 iα1=2 iα2=2 w = r1r2e e = r1r2e e : This shows that we obtain the original value of w, i.e. no change of branches has occurred. This completes the solution. Example 2. Discuss the multivaluedness of the function f(z) = (z2 −1)1=2 and introduce cuts to obtain single valued branches. iθ1 iθ2 Solution. Let z − 1 = r1e ; z + 1 = r2e : Then p i(θ1+θ2) 1=2 iθ1=2 iθ2=2 w = fr1r2e g = r1r2e e : Suppose that we start with a particular value of z corresponding to θ1 = α1 and θ2 = α2. p iα1=2 iα2=2 Then w = r1r2e e : As z goes once counterclockwise around z = 1, θ1 increases to α1 + 2π; while θ2 remains the same, i.e. θ2 = α2: Hence p p i(α1+2π)=2 iα2=2 iα1=2 iα2=2 w = r1r2e e = − r1r2e e : Hence we do not obtain the original value of w, i.e. a change of branches has occurred. Therefore z = 1 is a branch point. Similarly z = −1 is another branch point of f(z). In order to obtain single valued branches we introduce two different set of branch cuts (see Fig. 3). 5 Fig. 3: (i) A branch cut between the points −1 and 1 on the real axis. In this case consider the closed contour C enclosing the branch points −1 and 1. Here f(z) returns to the value p p i(α1+2π)=2 i(α2+2π)=2 iα1=2 iα2=2 w = r1r2e e = r1r2e e : So it is a single valued branch. (ii) Two branch cuts on the real axis (−∞; −1) and (1; 1). Here the contour Γ does not enclose any of the branch points, so that f(z) remains single valued as z makes a complete round through Γ initiating from z. Branch Point at Infinity 1 To examine the branch point at infinity, we use the transformation ζ = z and then examine for the point ζ = 0. Example 3. Determine the branch points of the function f(z) = (z3 − 1)1=2: Solution. The roots of the equation z3 − 1 = 0 are p p −1 + i 3 −1 − i 3 f1; ; g; 2 2 i:e: f1; e2πi=3; e4πi=3g = f1; α; βg; say: Therefore, f(z) = (z3 − 1)1=2 = (z − 1)1=2(z − α)1=2(z − β)1=2: So there are branch points at each of cube roots of unity. To examine the branch point at 1 infinity we replace z by ζ and obtain 1 1 f( ) = ( − 1)1=2 = ζ−3=2(1 − ζ3)1=2: ζ ζ3 6 −3=2 3 1=2 1 Now ζ has a branch point at ζ = 0, while (1 − ζ ) is not singular there. Hence ζ has a branch point at ζ = 0. Thus f(z) has a branch point at infinity. Example 4. Determine the branch points of the function f(z) = (z3 − z)1=3: Solution. Here f(z) = (z3 − z)1=3 = z1=3(z − 1)1=3(z + 1)1=3: Therefore, the branch points are z = −1; 0 and 1: To examine the branch point at 1 infinity we replace z by ζ and obtain 1 1 1 1 f( ) = ( )1=3( − 1)1=3( + 1)1=3 ζ ζ ζ ζ 1 = (1 − ζ)1=3(1 + ζ)1=3: ζ 1 Since f( ζ ) does not have a branch point at ζ = 0, f(z) does not have a branch point at infinity. 2 Logarithmic Function We now introduce the logarithmic function of a complex variable z. At first, we recall some properties of the real valued logarithm. For every positive real number x, there exist a unique real number y such that ey = x. We write y = ln x: We also note that for x1; x2 > 0, we have ln(x1x2) = ln x1 + ln x2: The function y = ln x maps the positive reals onto the set of reals, and is the inverse of the function y = ex: Since ex is one-to-one, its inverse is also a one-to-one function.

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