CHAPTER 4

Elementary Functions

BY

Dr. Pulak Sahoo

Assistant Professor Department of Mathematics University Of Kalyani West Bengal, India E-mail : [email protected]

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 √ √ have z = reiθ, w = reiθ/2, so that at P, θ = β and w = reiβ/2. After a complete √ √ 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 √ √ 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 = {(z + i)(z − i)}1/2. Therefore, 1 1 arg w = arg(z − i) + arg(z + i). 2 2 This shows that 1 1 change in arg w = {change in arg(z − i)} + {change in arg(z + i)}. 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

√ i(θ1+θ2) 1/2 iθ1/2 iθ2/2 w = {r1r2e } = r1r2e e .

Suppose that we start with a particular value of z corresponding to θ1 = α1 and θ2 = α2. √ 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

√ √ 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

√ √ 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

√ i(θ1+θ2) 1/2 iθ1/2 iθ2/2 w = {r1r2e } = r1r2e e .

Suppose that we start with a particular value of z corresponding to θ1 = α1 and θ2 = α2. √ 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

√ √ 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 √ √ 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, ∞). 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 √ √ −1 + i 3 −1 − i 3 {1, , }, 2 2 i.e. {1, e2πi/3, e4πi/3} = {1, α, β}, 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. We thus define the logarithm of a complex number z, denoted by log z, as the set of all values w = log z for which ew = z. Thus for z 6= 0 we have

w = log z ⇐⇒ ew = z.

Note Since the exponential function never vanishes, there is no logarithm associated with the

7 complex number zero. There are infinitely many values of the logarithm associated with each nonzero complex number as the exponential function assumes every nonzero complex number infinitely often.

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