Mathematical prerequisites
2.3 Complex-valued functions
EXERCISES
Determine the partial fraction expansion of F(z) given by 2.9
F(z) = z
(z − 1/2)(z − 2).
Determine the partial fraction expansion, into fractions with linear denominators, of 2.10
the function F(z) given by
F(z) = 1
z2+ 4z + 8.
Determine the partial fraction expansion of the function F(z) given by 2.11
a F(z) = z2
(z + 1)2(z + 3), b F(z) = z2+ 1
(z + 1)3.
Determine the partial fraction expansion, into fractions with denominators having 2.12
real coefficients, of the function F(z) given by F(z) = 2z2− 5z + 11
z3− 3z2+ 7z − 5.
2.3 Complex-valued functions
In general, the functions in this book are prescriptions associating a complex number with each real variable t from the domain. We call this a complex-valued function.
Complex-valued function
An important example is the function f(t) = eiωt, whereω is a real constant. This function is called a time-harmonic signal with frequencyω (see chapter 1). Note that according to definition 2.1 one has that eiωt= cos ωt + i sin ωt.
A complex-valued function can be written as f(t) = u(t) + iv(t),
where u and v are real-valued functions of the real argument t. Analogous to com-plex numbers, we use the concepts real and imaginary part of f(t):
u(t) = Re f (t), v(t) = Im f (t).
Re z
A complex-valued function can be represented as a graph by drawing its range in the complex plane. This range is a curve in the complex plane having parametric representation(u(t), v(t)). For the functions eiπt and eiπt/t (a part of) the range is shown in figure 2.4.
For complex-valued functions the notion of a limit can simply be defined by starting from limits of real-valued functions. We utilize the following definition.
Let L be a complex number and f(t) a complex-valued function with real part u(t) DEFINITION 2.2
Limit of a complex-valued function
and imaginary part v(t). Then
tlim→a f(t) = L if and only if lim
t→au(t) = Re L and lim
t→av(t) = Im L.
A consequence of this definition is the following theorem.
Let L be a complex number and f(t) a complex-valued function. Then THEOREM 2.3
From this it follows that if limt→a| f (t) − L | = 0, then limt→au(t) = Re L.
Similarly one proves that limt→av(t) = Im L. This completes the proof. We show that for each realω one has: limt→∞eiωt/t = 0. Since |eiωt/t| = EXAMPLE 2.10
1/t and limt→∞1/t = 0 we have limt→∞|eiωt/t| = 0 and hence, according to
theorem 2.3, limt→∞eiωt/t = 0.
Concepts like continuity and differentiability of a function are defined using lim-Continuity
Differentiability its. For instance, a function is continuous at t = a if limt→a f(t) = f (a) and differentiable at t= a if limt→a( f (t) − f (a))/(t − a) exists. One can show (this is not very hard, but it will be omitted here) that for a complex-valued function f(t) = u(t) + iv(t), the continuity of f (t) is equivalent to the continuity of both the real part u(t) and the imaginary part v(t), and also that the differentiability of f(t) is equivalent to the differentiability of u(t) and v(t). Moreover, one has for the derivative at a point t that
f(t) = u(t) + iv(t). (2.19)
Consequently, for the differentiation of complex-valued functions the same rules ap-ply as for real-valued functions. Complex numbers may be considered as constants here.
If f(t) = eat with a ∈ C, then f(t) = aeat. We can show this as follows: put EXAMPLE 2.11
a= x + iy and write f (t) as f (t) = extei yt = extcos yt+ iextsin yt. The real and imaginary parts are differentiable everywhere with derivatives ext(x cos yt − y sin yt) and ext(x sin yt + y cos yt) respectively. So
f(t) = xext(cos yt + i sin yt) + iyext(cos yt + i sin yt)
= (x + iy)extei yt= aeat.
With the chain rule we can differentiate a composition f(g(t)) of two functions.
The function f(t), however, is defined on (a part of) R. Hence, in the composition g(t) should also be a real-valued function. The chain rule then has the usual form Chain rule
d
dt f(g(t)) = f(g(t))g(t).
A consequence of this is:
d
dt[ f(t)]n= n[ f (t)]n−1f(t) for n= 1, 2, . . ..
Now that the concepts continuity and differentiability of complex-valued func-tions have been introduced, we will proceed with the introduction of two classes of functions that will play an important role in theorems on Fourier series, Fourier integrals and Laplace transforms.
The first class consists of the so-called piecewise continuous functions. To start with, we define in the usual manner the left-hand limit f(t−) and right-hand limit Left-hand limit
Right-hand limit f(t+) of a function at the point t:
f(t−) = lim
h↓0f(t − h) and f(t+) = lim
h↓0f(t + h), provided these limits exist.
A function f(t) is called piecewise continuous on the interval [a, b] if f (t) is con-DEFINITION 2.3
Piecewise continuous function
tinuous at each point of(a, b), except possibly in a finite number of points t1, t2, . . . , tn. Moreover, f(a+), f (b−) and f (ti+), f (ti−) should exist for i = 1, . . . , n.
A function f(t) is called piecewise continuous on R if f (t) is piecewise continuous on each subinterval [a, b] of R.
One can show that a function f(t) which is piecewise continuous on an interval [a, b] is also bounded on [a, b], that is to say: there exists a constant M > 0 such that for all t in [a, b] one has | f (t) | ≤ M. Functions that possess a real or imaginary part with a vertical asymptote in [a, b] are thus not piecewise continuous on [a, b].
Another example is the function f(t) = sin(1/t) for t = 0 and f (0) = 0. This function is continuous everywhere except at t= 0. Since f (0+) does not exist, this function is not piecewise continuous on [0, 1] according to our definition.
Note that the function value f(t) at a point t of discontinuity doesn’t necessarily have to equal f(t+) or f (t−).
A second class of functions to be introduced is the so-called piecewise smooth functions. This property is linked to the derivative of the function. For a piecewise continuous function, we will mean by fthe derivative of f at all points where it exists.
A piecewise continuous function f(t) on the interval [a, b] is called piecewise DEFINITION 2.4
Piecewise smooth function smooth if its derivative f(t) is piecewise continuous.
A function is called piecewise smooth onR if this function is piecewise smooth on each subinterval [a, b] of R.
In figure 2.5 a graph is drawn of a real-valued piecewise smooth function.
0 t1 t2 t3 t4 t
FIGURE 2.5
A piecewise smooth function.
There are now two possible ways of looking at the derivative in the neighbour-hood of a point: on the one hand as the limits f(t+) and f(t−); on the other hand by defining a left-hand derivative f−(t) and a right-hand derivative f+(t) as Left-hand derivative
Right-hand derivative follows:
f−(t) = lim h↑0
f(t + h) − f (t−)
h , f+(t) = lim h↓0
f(t + h) − f (t+)
h , (2.20)
provided these limits exist. Note that in this definition of left-hand and right-hand derivative one does not use the function value at t, since f(t) need not exist at the point t. Often it is the case that f−(t) = f(t−) and f+(t) = f(t+). This holds in particular for piecewise smooth functions, notably at the points of discontinuity of f , as is proven in the next theorem.
Let f(t) be a piecewise smooth function on the interval [a, b]. Then f+(a) = THEOREM 2.4
f(a+), f−(b) = f(b−) and for all a < t < b one has, moreover, that f−(t) = f(t−) and f+(t) = f(t+).
Proof
We present the proof for right-hand limits. The proof for left-hand limits is anal-ogous. So let t ∈ [a, b) be arbitrary. Using the mean value theorem from calcu-lus we will show that the existence of f(t+) implies that f+(t) exists and that f(t+) = f+(t). Since f and f are both piecewise continuous, there exists an h> 0 such that f and fhave no point of discontinuity on(t, t + h]. Possibly f has a discontinuity at t. If we now redefine f at t as f(t+), then f is continuous on [t, t + h]. Moreover, f is differentiable on (t, t + h). According to the mean value theorem there then exists aξ ∈ (t, t + h) such that
f(t + h) − f (t+)
h = f(ξ).
Now let h ↓ 0, then ξ ↓ t. Since f(t+) = limξ↓t f(ξ) exists, it follows from (2.20) that f+(t) exists and that f(t+) = f+(t). When a function is not piecewise smooth, the left- and right-hand derivatives will not always be equal to the left- and right-hand limits of the derivative, as the following example shows.
Let the function f(t) be given by EXAMPLE
f(t) =
t2sin(1/t) for t = 0,
0 for t= 0.
The left- and right-hand derivatives of f(t) at t = 0, calculated according to (2.20), exist and are equal to 0. However, if we first calculate the derivative f(t), then
f(t) =
2t sin(1/t) − cos(1/t) for t = 0,
0 for t= 0.
It then turns out that f(0+) and f(0−) do not exist, since cos(1/t) has no left- or
right-hand limit at t= 0.
The theory of the Riemann integral and the improper Riemann integral for real-valued functions can easily be extended to complex-real-valued functions as well. For complex-valued functions the (improper) Riemann integral exists on an interval if and only if both the (improper) Riemann integral of the real part u(t) and the imag-inary part v(t) exist on that interval. Moreover, one has
Definite integral b does not change by altering the value of the function at the possible jump discontinu-ities. From (2.21) the following properties of definite integrals for complex-valued functions immediately follow:
The fundamental theorem of calculus for complex-valued functions is no different from the one for real-valued functions. Here we will formulate the fundamental theorem for piecewise continuous functions, for which we state without proof that they are Riemann integrable.
Let f(t) be a piecewise continuous function defined on the interval [a, b]. Let THEOREM 2.5
F(x) =x
a f(t) dt. Then F is continuous and piecewise smooth on [a, b].
From the preceding theorem one can then derive that for continuous and piece-wise smooth complex-valued functions, the rule for integration by parts can be ap-plied in the same way as the rule for integration by parts of real-valued functions.
We calculate the integralπ
0 te2i tdt by applying the rule for integration by parts.
EXAMPLE 2.12 The following inequality is often applied to estimate integrals:
For real-valued functions this is a well-known inequality. Here we omit the proof for complex-valued functions. In fact one can consider this inequality as a general-ization of the triangle inequality (theorem 2.1). A direct consequence of inequality (2.22) is the following inequality. If| f (t) | ≤ M on the integration interval [a, b], then
Determine the derivative of the following functions:
2.13
2.4 Sequences and series