17.4 Functions of a Complex Variable
INTRODUCTION
One of the most important concepts in mathematics is that of a function. You may recall from previous courses that a function is a certain kind of correspondence between two sets; more specifically: A function f from a set A to a set B is a rule of correspondence that assigns to each element in A one and only one element in B. If b is the element in the set B assigned to the element a in the set A by f, we say that b is the image of a and write b = f(a). The set A is called the domain of the function f(but is not necessarily a domain in the sense defined in Section 17.3). The set of all images in B is called the range of the function. For example, suppose the set A is a set of real numbers defined by 3 ≤ x < ∞ and the function is given by f(x) = ; then f(3) = 0, f(4) = 1, f(8) =
, and so on. In other words, the range of f is the set given by 0 ≤ y < ∞. Since A is a set of real numbers, we say f is a function of a real variable x.
Functions of a Complex Variable
When the domain A in the foregoing definition of a function is a set of complex numbers z, we naturally say that f is a function of a complex variable z or a complex function for short. The image w of a complex number z will be some complex number u + iv; that is,
(1)
where u and v are the real and imaginary parts of w and are real-valued functions. Inherent in the mathematical statement (1) is the fact that we cannot draw a graph of a complex function w = f(z) since a graph would require four axes in a four-dimensional coordinate system.
Some examples of functions of a complex variable are
Each of these functions can be expressed in form (1). For example,
f(z) = z2 − 4z = (x + iy)2 − 4(x + iy) = (x2 − y2 − 4x) + i(2xy − 4y).
Thus, u(x, y) = x2 − y2 − 4x, and v(x, y) = 2xy − 4y.
Although we cannot draw a graph, a complex function w = f(z) can be interpreted as a mapping or transformation from the z-plane to the w-plane. See FIGURE 17.4.1.
FIGURE 17.4.1 Mapping from z-plane to w-plane
EXAMPLE 1 Image of a Vertical Line
Find the image of the line Re(z) = 1 under the mapping f(z) = z2.
SOLUTION
For the function f(z) = z2 we have u(x, y) = x2 − y2 and v(x, y) = 2xy. Now, Re(z) = x and so by substituting x = 1 into the functions u and v, we obtain u = 1 − y2 and v = 2y. These are parametric equations of a curve in the w-plane. Substituting y = v/2 into the first equation eliminates the parameter y to give u = 1 − v2/4. In other words, the image of the line in FIGURE 17.4.2(a) is the parabola shown in Figure 17.4.2(b). ≡
FIGURE 17.4.2 Image of x = 1 is a parabola
We shall pursue the idea of f(z) as a mapping in greater detail in Chapter 20.
It should be noted that a complex function is completely determined by the real-valued functions u and v. This means a complex function w = f(z) can be defined by arbitrarily specifying u(x, y) and v(x, y), even though u + iv may not be obtainable through the familiar operations on the symbol z alone. For example, if u(x, y) = xy2 and v(x, y) = x2 − 4y3, then f(z) = xy2 + i(x2 − 4y3) is a function of a complex variable. To compute, say, f(3 + 2i), we substitute x = 3 and y = 2 into u and v to obtain f(3 + 2i) = 12 − 23i.
Complex Functions as Flows
We also may interpret a complex function w = f(z) as a two-dimensional fluid flow by considering the complex number f(z) as a vector based at the point z. The vector f(z) specifies the speed and direction of the flow at a given point z. FIGUREs 17.4.3 and 17.4.4 show the flows corresponding to the complex functions f1(z) = and f2(z) = z2, respectively.
FIGURE 17.4.3 f1(z) = (normalized)
FIGURE 17.4.4 f2(z) = z2 (normalized)
If x(t) + iy(t) is a parametric representation for the path of a particle in the flow, the tangent vector T = x′(t) + iy′(t) must coincide with f(x(t) + iy(t)). When f(z) = u(x, y) + iv(x, y), it follows that the path of the particle must satisfy the system of differential equations
We call the family of solutions of this system the streamlines of the flow associated with f(z).
EXAMPLE 2 Streamlines
Find the streamlines of the flows associated with the complex functions (a) f1(z) = and (b) f2(z) = z2.
SOLUTION
(a) The streamlines corresponding to f1(z) = x − iy satisfy the system
and so x(t) = c1et and y(t) = c2e−t. By multiplying these two parametric equations, we see that the point x(t) + iy(t) lies on the hyperbola xy = c1c2.
(b) To find the streamlines corresponding to f2(z) = (x2 − y2) + i 2xy, note that dx/dt = x2 − y2, dy/dt = 2xy, and so
This homogeneous differential equation has the solution x2 + y2 = c2y, which represents a family of circles that have centers on the y-axis and pass through the origin. ≡
Limits and Continuity
The definition of a limit of a complex function f(z) as z → z0 has the same outward appearance as the limit in real variables.
DEFINITION 17.4.1 Limit of a Function
Suppose the function f is defined in some neighborhood of z0, except possibly at z0 itself. Then f is said to possess a limit at z0, written
if, for each ϵ > 0, there exists a δ > 0 such that |f(z) − L| < ϵ whenever 0 <|z − z0| < δ.
In words, f(z) = L means that the points f(z) can be made arbitrarily close to the point L if we choose the point z sufficiently close to, but not equal to, the point z0. As shown in FIGURE 17.4.5, for each ϵ-neighborhood of L (defined by |f(z) − L| < ϵ) there is a δ-neighborhood of z0 (defined by |z − z0| < δ) so that the images of all points z ≠ z0 in this neighborhood lie in the ϵ-neighborhood of L.
FIGURE 17.4.5 Geometric meaning of a complex limit
The fundamental difference between this definition and the limit concept in real variables lies in the understanding of z → z0. For a function f of a single real variable x, f(x) = L means f(x) approaches L as x approaches x0 either from the right of x0 or from the left of x0 on the real number line. But since z and z0 are points in the complex plane, when we say that
f(z) exists, we mean that f(z) approaches L as the point z approaches z0 from any direction.
The following theorem summarizes some properties of limits:
THEOREM 17.4.1 Limit of Sum, Product, Quotient
Suppose f(z) = L1 and
g(z) = L2. Then
(i) [f(z) + g(z)] = L1 + L2
(ii) f(z)g(z) = L1L2
(iii) L2 ≠ 0.
DEFINITION 17.4.2 Continuity at a Point
A function f is continuous at a point z0 if
As a consequence of Theorem 17.4.1, it follows that if two functions f and g are continuous at a point z0, then their sum and product are continuous at z0. The quotient of the two functions is continuous at z0 provided g(z0) ≠ 0.
A function f defined by
(2)
where n is a nonnegative integer and the coefficients ai, i = 0, 1, . . ., n, are complex constants, is called a polynomial of degree n. Although we shall not prove it, the limit result z = z0 indicates that the simple polynomial function f(z) = z is continuous everywhere—that is, on the entire z-plane. With this result in mind and with repeated applications of Theorem 17.4.1 (i) and (ii), it follows that a polynomial function (2) is continuous everywhere. A rational function
where g and h are polynomial functions, is continuous except at those points at which h(z) is zero.
Derivative
The derivative of a complex function is defined in terms of a limit. The symbol Δz used in the following definition is the complex number Δx + iΔy.
DEFINITION 17.4.3 Derivative
Suppose the complex function f is defined in a neighborhood of a point z0. The derivative of f at z0 is
(3)
provided this limit exists.
If the limit in (3) exists, the function f is said to be differentiable at z0. The derivative of a function w = f(z) is also written dw/dz.
As in real variables, differentiability implies continuity:
If f is differentiable at z0, then f is continuous at z0.
Moreover, the rules of differentiation are the same as in the calculus of real variables. If f and g are differentiable at a point z, and c is a complex constant, then
(4)
(5)
(6)
(7)
(8)
The usual Power Rule for differentiation of powers of z is also valid:
(9)
Combining (9) and (8) gives the Power Rule for Functions:
(10)
EXAMPLE 3 Using the Rules of Differentiation
Differentiate
(a) f(z) = 3z4 − 5z3 + 2z
(b) f(z) =
(c)
SOLUTION
(a) Using the Power Rule (9) along with the Sum Rule (5), we obtain
f′(z) = 3 · 4z3 − 5 · 3z2 + 2 = 12z3 − 15z2 + 2.
(b) From the Quotient Rule (7),
(c) In the Power Rule for Functions (10), we identify and
so that
≡
In order for a complex function f to be differentiable at a point z0, must approach the same complex number from any direction. Thus in the study of complex variables, to require the differentiability of a function is a greater demand than in real variables. If a complex function is made up, such as f(z) = x + 4iy, there is a good chance that it is not differentiable.
EXAMPLE 4 A Function That Is Nowhere Differentiable
Show that the function f(z) = x + 4iy is nowhere differentiable.
SOLUTION
With Δz = Δx + iΔy, we have
f (z + Δz) − f(z) = (x + Δx) + 4i(y + Δy) − x − 4iy = Δx + 4iΔy
and so (11)
Now, if we let Δz → 0 along a line parallel to the x-axis, then Δy = 0 and the value of (11) is 1. On the other hand, if we let Δz → 0 along a line parallel to the y-axis, then Δx = 0 and the value of (11) is seen to be 4. Therefore, f(z) = x + 4iy is not differentiable at any point z. ≡
Analytic Functions
While the requirement of differentiability is a stringent demand, there is a class of functions that is of great importance whose members satisfy even more severe requirements. These functions are called analytic functions.
DEFINITION 17.4.4 Analyticity at a Point
A complex function w = f(z) is said to be analytic at a point z0 if f is differentiable at z0 and at every point in some neighborhood of z0.
A function f is analytic in a domain D if it is analytic at every point in D.
The student should read Definition 17.4.4 carefully. Analyticity at a point is a neighborhood property. Analyticity at a point is, therefore, not the same as differentiability at a point. It is left as an exercise to show that the function f(z) = |z |2 is differentiable at z = 0 but is differentiable nowhere else. Hence, f(z) = |z |2 is nowhere analytic. In contrast, the simple polynomial f(z) = z2 is differentiable at every point z in the complex plane. Hence, f(z) = z2 is analytic everywhere. A function that is analytic at every point z is said to be an entire function. Polynomial functions are differentiable at every point z and so are entire functions.
REMARKS
Recall from algebra that a number c is a zero of a polynomial function if and only if x − c is a factor of f(x). The same result holds in complex analysis. For example, since f(z) = z4 + 5z2 + 4 = (z2 + 1)(z2 + 4), the zeros of f are −i, i, −2i, and 2i. Hence, f(z) = (z + i)(z − i)(z + 2i)(z − 2i). Moreover, the quadratic formula is also valid. For example, using this formula, we can write
See Problems 21 and 22 in Exercises 17.4.
17.4 Exercises Answers to selected odd-numbered problems begin on page ANS-45.
In Problems 1–6, find the image of the given line under the mapping f(z) = z2.
- y = 2
- x = −3
- x = 0
- y = 0
- y = x
- y = −x
In Problems 7–14, express the given function in the form f(z) = u + iv.
- f(z) = 6z − 5 + 9i
- f(z) = 7z − 9i
− 3 + 2i
- f(z) = z2 − 3z + 4i
- f(z) = 3
+ 2z
- f(z) = z3 − 4z
- f(z) = z4
- f(z) = z + 1/z
In Problems 15–18, evaluate the given function at the indicated points.
- f(z) = 2x − y2 + i(xy3 − 2x2 + 1)
(a) 2i
(b) 2 − i
(c) 5 + 3i
- f(z) = (x + 1 + 1/x) + i(4x2 − 2y2 − 4)
(a) 1 + i
(b) 2 − i
(c) 1 + 4i
- f(z) = 4z + i
+ Re(z)
(a) 4 − 6i
(b) −5 + 12i
(c) 2 − 7i
- f(z) = ex cos y + iex sin y
(a) πi/4
(b) −1 − πi
(c) 3 + πi/3
In Problems 19–22, the given limit exists. Find its value.
(4z3 − 5z2 + 4z + 1 − 5i)
In Problems 23 and 24, show that the given limit does not exist.
In Problems 25 and 26, use (3) to obtain the indicated derivative of the given function.
- f(z) = z2, f′(z) = 2z
- f(z) = 1/z, f′(z) = −1/z2
In Problems 27–34, use (4)–(10) to find the derivative f′(z) for the given function.
- f(z) = 4z3 − (3 + i)z2 − 5z + 4
- f(z) = 5z4 − iz3 + (8 − i)z2 − 6i
- f(z) = (2z + 1)(z2 − 4z + 8i)
- f(z) = (z5 + 3iz3)(z4 + iz3 + 2z2 − 6iz)
- f(z) = (z2 − 4i)3
- f(z) = (2z − 1/z)6
- f(z) =
- f(z) =
In Problems 35–38, give the points at which the given function will not be analytic.
- Show that the function f(z) =
is nowhere differentiable.
- The function f(z) = |z|2 is continuous throughout the entire complex plane. Show, however, that f is differentiable only at the point z = 0. [Hint: Use (3) and consider two cases: z = 0 and z ≠ 0. In the second case let Δz approach zero along a line parallel to the x-axis and then let Δz approach zero along a line parallel to the y-axis.]
In Problems 41–44, find the streamlines of the flow associated with the given complex function.
- f(z) = 2z
- f(z) = iz
- f(z) = 1/
- f(z) = x2 − iy2
In Problems 45 and 46, use a graphics calculator or computer to obtain the image of the given parabola under the mapping f(z) = z2.
- y =
x2
- y = (x − 1)2