Complex Algebra

Chapter 18

Complex Arithmetic Complex numbers were developed because there was a need to expand the notion of numbers to include solutions of algebraic equations whose prototype is x2 + 1 = 0. Such developments are not atypical in the history of mathematics. The invention of irrational numbers occurred because of a need for a number that could solve an equation of the form x2 − 2 = 0. Similarly, rational numbers were the offspring of the operations of multiplication and division and the quest for a number that gives, for example, 4 when multiplied by 3, or, equivalently, a number that solves the equation 3x − 4 = 0. There is a crucial difference between complex numbers and all the numbers mentioned above: All rational, irrational, and, in general, real numbers correspond to measurable physical quantities. However, there is no single measurable physical quantity that can be described by a complex number. A natural question then is this: What need is there for complex numbers if no physical quantity can be measured in terms of them? The answer is that although no single physical quantity can be expressed in terms of complex numbers, a pair of physical quantities can be neatly described by a single complex number. For example, a wave with a given amplitude and phase can be concisely described by a complex number. Another, more fundamental, reason is that equations that describe the behavior of subatomic particles are inherently complex.

18.1

Cartesian Form of Complex Numbers

We demand a number system broad enough to include solutions to the equation or x2 = −1. x2 + 1 = 0 Clearly the solution(s) cannot be real because a real number raised to the second power gives a positive real number, and we want x2 to be negative.

478 Cartesian form of a complex number

Complex Arithmetic So we broaden the concept of numbers by considering complex numbers. Such numbers are of the form √ z = x + iy with i ≡ −1 and i2 = −1. (18.1) It turns out that we don’t need to introduce any other numbers to solve all algebraic equations—equations of the form p(x) = 0 with p(x) a polynomial. In fact, the fundamental theorem of algebra, to which we shall return, states that all roots of any algebraic equation an xn + an1− xn−1 + · · · + a1 x + a0 = 0

complex plane, real and imaginary parts

with arbitrary real or complex coefficients a0 , a1 , . . . , an , are in the complex number system. In this sense, then, the complex number system is the most complete system. A complex number can be conveniently represented as a point (or equivalently, as a vector) in the xy-plane, called the complex plane, as shown in Figure 18.1. In Equation (18.1), x is called the real part of z, written Re(z), and y is called the imaginary part of z, written Im(z). Similarly, the horizontal axis in Figure 18.1 is named the real axis, and the vertical axis is named the imaginary axis. The set of all complex numbers—or the set of points in the complex plane—is denoted by C. We can define various operations on C that are extensions of similar operations on the real number system, R. The only proviso is that i2 = −1, and that the final form of an equation must be written as Equation (18.1)—with real and imaginary parts. For instance, the sum of two complex numbers, z1 = x1 + iy1 and z2 = x2 + iy2 , is z1 + z2 = (x1 + x2 ) + i(y1 + y2 ). This sum can be represented in the complex plane as the vector sum of z1 and z2 , as shown in Figure 18.2. The product of z1 and z2 can also be obtained: z1 z2 = (x1 + iy1 )(x2 + iy2 ) = x1 x2 + x1 (iy2 ) + iy1 x2 + iy1 (iy2 ) = x1 x2 + i(x1 y2 + y1 x2 ) − y1 y2 = x1 x2 − y1 y2 + i(x1 y2 + y1 x2 ). Thus, Re(z1 z2 ) = x1 x2 − y1 y2 , Im(z1 z2 ) = x1 y2 + x2 y1 .

(18.2)

Im z y Re x

Figure 18.1: Complex numbers as points or vectors in a plane.

18.1 Cartesian Form of Complex Numbers Im

479

z1 + z2 z1 z2 Re

Figure 18.2: Addition of complex numbers as addition of vectors. To obtain this equation, we have implicitly used the fact that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. The factor i in z allows new operations for complex numbers that do not exist for real numbers. One such operation is complex conjugation. The complex conjugate, z ∗ or z¯, of z is defined as z ∗ ≡ z¯ = (x + iy)∗ = x − iy

complex conjugation

(18.3)

which is obtained from z by replacing i with −i. We note immediately that zz ∗ = (x + iy)(x − iy) = x2 + y 2 = z ∗ z which is a positive real number. The positive square root of zz ∗ is called the absolute value of z and denoted by |z|. It is simply the length of the vector representing z in the xy-plane. Thus, we have " ! √ √ 2 2 |z| = zz ∗ = z ∗ z = x2 + y 2 = (Re(z)) + (Im(z)) . (18.4) We can also define the division of two complex numbers using complex conjugation. Box 18.1.1. To find the real and imaginary parts of a quotient, multiply the numerator and denominator by the complex conjugate of the denominator. So, for the ratio of z1 /z2 , we get z1 z1 z2∗ (x1 + iy1 )(x2 − iy2 ) x1 x2 + y1 y2 + i(y1 x2 − x1 y2 ) = = = ∗ 2 z2 z2 z2 |z2 | |z2 |2 x1 x2 + y1 y2 y1 x2 − x1 y2 = +i . |z2 |2 |z2 |2 Thus, Re

#

z1 z2

$

x1 x2 + y1 y2 = x22 + y22

and

Im

#

z1 z2

$

=

y1 x2 − x1 y2 . x22 + y22

(18.5)

absolute value

480

Complex Arithmetic In particular,

properties of absolute value of complex numbers

z∗ 1 x − iy = 2 = 2 z |z| x + y2

and

1 = −i. i

Some useful properties of absolute values are as follows: ! ! ! z1 ! |z1 | , |z1 z2 | = |z1 | |z2 |, !! !! = z2 |z2 | ! ! ! ! ! |z1 | − |z2 | ! ≤ |z1 + z2 | ≤ |z1 | + |z2 |.

(18.6)

This last inequality is called the triangle inequality and it comes directly from the vector property of complex numbers. The right half of it can be generalized to more than two complex numbers: ! n ! n !" ! " ! ! ≤ z |zk | . (18.7) ! k! ! ! k=1

k=1

Example 18.1.1. Here we present some sample manipulations with complex num-

bers:

(1 + i)2 = (1)2 + (i)2 + 2i = 1 − 1 + 2i = 2i,

1 1 + i − (1 − i) 2i 2i 1 − = = = i, = 1−i 1+i (1 − i)(1 + i) |1 + i|2 2 1 1 1 1 = =− , (1 + i)−4 = = (1 + i)2 (1 + i)2 (2i)(2i) −4 4 (2 + i)(3 + i) 1 5 + i5 1 2+i = = 2 = +i , 3−i |3 − i|2 3 + (−1)2 2 2 # ! ! 2 + 22 ! 2i − 1 ! (−1) | − 1 + i2| ! ! ! i − 2 ! = | − 2 + i| = # (−2)2 + 12 = 1.

The equation |z − a| = b, where a is a fixed complex number and b is real and positive, describes a circle of radius b with center at a ≡ ax + iay . This is easily seen because b2 = |z − a|2 = |(x + iy) − (ax + iay )|2

= |(x − ax ) + i(y − ay )|2 = (x − ax )2 + (y − ay )2 .

We note that |z − a| is the distance between the two complex numbers z and a. Therefore, |z − a| = b—with a a constant and z a variable—is the collection of all points z that are at a distance b from a. ! properties of complex conjugation of complex numbers

Complex conjugation satisfies some nice properties that we list below: $ %∗ z1 z∗ ∗ ∗ ∗ ∗ ∗ ∗ (z1 + z2 ) = z1 + z2 , (z1 z2 ) = z1 z2 , = 1∗ , z2 z2 1 Re(z) = 12 (z + z ∗ ), Im(z) = (z − z ∗ ), (18.8) 2i ∗ ∗ n ∗ ∗ n (z ) = z, (z ) = (z ) .

18.1 Cartesian Form of Complex Numbers

481

The complex conjugate of a function of z is easily obtained by substituting z ∗ for z in that function.1 This can be summarized as (f (z))∗ = f (z ∗ )

(18.9)

which is equivalent to replacing every i with −i in the expression for f (z). Historical Notes In the first half of the sixteenth century there was hardly any change from the attitude or spirit of Arabs, whose work had put practical arithmetical calculations in the forefront of mathematics, but merely an increase in the kind of activity Europeans had learned from Arabs. Moreover, the technological advances spurred by the Renaissance demanded further refinement in magnitudes such as trigonometric tables and astronomical observations. By 1500 or so, zero was accepted as a number and irrational numbers were used more freely in calculations. However, the problem of whether irrationals were really numbers still troubled people. Michael Stifel (1486?–1567), the German mathematician, argued that Since, in proving geometrical figures, when rational numbers fail us irrational numbers . . . prove exactly those things which rational numbers could not prove . . . we are compelled to assert that they truly are numbers . . . . On the other hand, . . . that cannot be called a true number which is of such a nature that it lacks precision [decimal representation].

He then argues that only whole numbers or fractions can be called true numbers, and since irrationals are neither, they are not real numbers. Even a century later, Pascal, Barrow, and Newton thought of irrational numbers as being understood in terms of geometric magnitude; they were mere symbols that had no existence independent of continuous geometrical magnitude. Negative numbers were treated with equal suspicion by the sixteenth- and seventeenth-century mathematicians. They were considered “absurd.” Jerome Cardan (1501–1576), the great Italian mathematician of the Renaissance, was willing to accept the negative numbers as roots of equations, but considered them as “fictitious,” while he called the positive roots real. Fran¸cois Vieta (1540–1603), a lawyer by profession but recognized far more as the foremost mathematician of the sixteenth century, discarded negative numbers entirely. Descartes accepted them in part, but called negative roots of equations false, on the grounds that they represented numbers less than nothing. An interesting argument against negative numbers was given by Antoine Arnauld (1612–1694), a theologian and mathematician who was a close friend of Pascal. Arnauld questioned the equality −1 : 1 = 1 : (−1) because, he said, −1 is less than +1; hence, How could a smaller number be to a greater as a greater is to a smaller? Without having fully overcome their difficulties with irrational and negative numbers, the Europeans were hit by another problem: the complex numbers! They obtained these new numbers by extending the arithmetic operation of square root 1 This statement is not strictly true for all functions. However, only a mild restriction is to be imposed on them for the statement to be true. We shall not go into details of such restrictions because they require certain complex analytic tools which go beyond the scope of this book. See Hassani, S. Mathematical Physics: A Modern Introduction to Its Foundations, Springer-Verlag, 1999, Chapter 11 for details.

to find the complex conjugate of a function, change all its i’s to −i.

482

Complex Arithmetic to whatever numbers appeared in solving quadratic equations. Thus Cardan sets up and solves the problem of dividing 10 into two parts whose √ product is 40. The equation is x(10−x) = 40, for which he obtains the roots 5± −15 and then he says “Putting aside the mental torture involved,” multiply these two roots and note that the product is 25 − (−15) or 40. He then states, “So progresses arithmetic subtlety the end of which, as is said, is as refined as it is useless.” Descartes also rejected complex roots and coined them “imaginary.” Even Newton did not regard complex roots as significant, most likely because in his day they lacked physical meaning. The confusion surrounding complex numbers is illustrated by the oft-quoted statement by Leibniz, “The Divine Spirit found a sublime outlet in that wonder of analysis, that portent of the ideal world, that amphibian between being and not being, which we call the imaginary root of negative unity.”

18.2

a very important relation

The introduction of polar coordinates in the complex plane makes available a powerful tool with which to facilitate complex manipulations. Figure 18.3 shows a complex number and its polar coordinates. In terms of these polar coordinates, z can be written as z = x + iy = r cos θ + ir sin θ = r(cos θ + i sin θ).

(18.10)

Assuming that series of complex numbers can be manipulated as those of real numbers,2 we obtain the useful relation between imaginary exponentials and trigonometric functions. In Chapter 10 we presented the Maclaurin series for the exponential and trigonometric functions. Let us assume that those functions are valid for complex numbers as well. Then, we have eiθ =

∞ ∞ ∞ ∞ ∞ ! ! ! ! (iθ)n (iθ)n (iθ)n (iθ)2k ! (iθ)2k+1 = + = + n! n! n! (2k)! (2k + 1)! n=even n=0 n=odd

=

∞ !

k=0

k=0

∞

! θ2k θ2k+1 +i = cos θ + i sin θ (−1)k (−1)k (2k)! (2k + 1)!

k=0

(18.11)

k=0

Im z r sin θ

polar representation of a complex number

Polar Form of Complex Numbers

r

z| =| θ r cos θ

Re

Figure 18.3: Complex numbers in polar coordinates. 2 This assumption turns out to be correct. In particular, the power series expansion used in the following example plays a central role in complex analysis.

18.2 Polar Form of Complex Numbers

483

because i2k = (i2 )k = (−1)k . This is probably the most important relation in complex number theory.

Box 18.2.1. The trigonometric and imaginary exponential functions are related by the Euler equation: eiθ = cos θ + i sin θ. The use of Equation (18.11) in (18.10) leads to another way of representing complex numbers: r=

z = reiθ , Note that

! x2 + y 2 ,

θ = tan−1

"y # x

.

(18.12)

Box 18.2.2. The angle θ is not uniquely determined: Any multiple of 2π can be added to it without affecting z. We can use Equation (18.12) together with x = r cos θ, y = r sin θ to convert from Cartesian coordinates to polar coordinates, and vice versa. The coordinate θ is called the argument of z and written θ = arg(z). Example 18.2.1. Let us look at some numerical examples of polar-Cartesian conversion. In many cases, a diagram can be very helpful. For instance, take i whose real part is obviously zero and whose imaginary part is 1. If we were to use the formula, we would have tan θ = 1/0 which is not defined. However, Figure 18.4 shows that z = i lies on the positive imaginary axis, and, thus, θ = π/2. Since we can always add a multiple of 2π to the angle, we have i = eiπ/2+i2nπ ,

n = 0, ±1, ±2, . . . .

Similarly, the same figure makes it clear that −i = e−iπ/2+i2nπ = ei3π/2+i2nπ , Im

n = 0, ±1, ±2, . . . .

Im

θ = π/2

Re Re

r=1

r=1

i

θ = − π/2

−i

Figure 18.4: Cartesian and polar coordinates for i and −i.

argument of a complex number

484

Complex Arithmetic Im Im

Im θ=π

−1

1+ i

Re

θ = π/4

r=1

−1 + i2 Im

2 + i3

Re

θ

Re

θ

θ = − π/4

Re

1− i

Figure 18.5: Cartesian and polar coordinates for some other complex numbers. Referring to Figure 18.5, the reader may verify the following polar representations of complex numbers: −1 = eiπ+i2nπ , √ 1 + i = 2 eiπ/4+i2nπ , √ √ 1 − i = 2 e−iπ/4+i2nπ = 2 ei7π/4+i2nπ , √ √ −1 2 + i3 = 13 ei tan (3/2)+i2nπ = 13 ei0.983+i2nπ , √ √ −1 −1 + i2 = 5 ei tan (−2)+i2nπ = 5 ei2.03+i2nπ . In all cases, n is an integer and angles are in radians.

!

The complex conjugate of z in polar coordinates is z ∗ = x − iy = r cos θ − ir sin θ = r cos(−θ) + ir sin(−θ) = re−iθ . This equation confirms the earlier statement that complex conjugation is equivalent to replacing i with −i. Generally speaking, polar coordinates are useful for operations of multiplication, division, and exponentiation, and Cartesian coordinates for addition and subtraction. Example 18.2.2. We can use the polar representation of complex numbers to find some trigonometric identities. In all of the following, we set r = 1: 1 = eiθ e−iθ = (cos θ + i sin θ)(cos θ − i sin θ) = cos2 θ + sin2 θ. Now consider the identity ei(θ1 +θ2 ) = cos(θ1 + θ2 ) + i sin(θ1 + θ2 ) which can also be written as ei(θ1 +θ2 ) = eiθ1 eiθ2 = (cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 + i(sin θ1 cos θ2 + sin θ2 cos θ1 ). Equating the real and imaginary parts of the last two equations, we obtain cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 , sin(θ1 + θ2 ) = sin θ1 cos θ2 + sin θ2 cos θ1 .

18.2 Polar Form of Complex Numbers

485

Similarly, equating the real and imaginary parts of ei3θ = cos 3θ + i sin 3θ and ! "3 ei3θ = eiθ = (cos θ + i sin θ)3 = cos3 θ + 3i cos2 θ sin θ − 3 sin2 θ cos θ − i sin3 θ

gives the following trigonometric identity:

cos 3θ = 4 cos3 θ − 3 cos θ, sin 3θ = 3 sin θ − 4 sin3 θ.

!

From einθ = cos nθ + i sin nθ

and

einθ = (eiθ )n = (cos θ + i sin θ)n

we obtain the so-called de Moivre theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ.

de Moivre theorem

(18.13)

Equation (18.11) and its complex conjugate lead to the following useful results: # $ cos θ = 12 eiθ + e−iθ , $ 1 # iθ e − e−iθ . (18.14) sin θ = 2i

As mentioned earlier, the exponential nature of polar coordinates makes them especially useful in multiplication, division, and exponentiation. For instance, z1 r1 eiθ1 r1 = = ei(θ1 −θ2 ) , iθ 2 z2 r2 e r2 $# $ # z1 z2 = r1 eiθ1 r2 eiθ2 = r1 r2 ei(θ1 +θ2 ) , √ $1/2 # $1/2 √ iθ/2 # √ z = reiθ = reiθ = r1/2 eiθ = re ,

(18.15)

and so forth. All of these relations have interesting geometric interpretations. For example, the second equation says that when you multiply a complex number z1 by another complex number z2 , you dilate the magnitude of z1 by a factor r2 and increase its angle by θ2 . That is, multiplication involves both a dilation and a rotation. In particular, if we multiply a complex number by eiωt where t is time, we get a vector of constant length in the xy-plane that is rotating with angular velocity ω. Example 18.2.3. A plane wave is represented by a periodic function such as A cos(kx − ωt)

or

B sin(kx − ωt).

two important relations

486

Complex Arithmetic On the other hand, sine and cosine are related by ! π" . sin(kx − ωt) = − cos kx − ωt + 2

complex amplitude

roots of complex numbers

Therefore, one can concentrate solely on the cosine function with a phase angle added to its argument. Thus a typical periodic plane wave is represented as A cos (kx − ωt + α). To make connection with the material of this section, we note that " ! " ! A cos (kx − ωt + α) = A Re ei(kx−ωt+α) = Re Aei(kx−ωt+α) ! " ! " = Re Aeiα ei(kx−ωt) = Re Zei(kx−ωt) ,

where Z is a complex number—called complex amplitude—of magnitude A and argument α. It is therefore convenient to represent plane waves by the complex function Zei(kx−ωt) which includes the phase of the wave as the argument of Z. !

Another interesting application of these ideas is finding roots of complex numbers. Suppose we are interested in all the nth roots of Z; i.e., all z’s satisfying z n = Z. To find the roots of a complex number Z, write it in polar form in the most general way: Z = ReiΘ+i2πk , Thus,

z n = ReiΘ+i2πk

k = 0, ±1, ±2, . . . , with

k = 0, ±1, ±2, . . . .

Taking the nth root of both sides, we obtain z = Z 1/n = R1/n eiΘ/n+i2πk/n ,

k = 0, ±1, ±2, . . . ,

and Box 18.2.3. The distinct nth roots {zk } of Z = ReiΘ are zk = R1/n eiΘ/n+i2πk/n ,

k = 0, 1, 2, . . . , n − 1.

(18.16)

We see that the number of nth roots of a complex number is exactly n. It is clear that zk of Equation (18.16) repeats itself for k ≥ n.

Example 18.2.4. Let us find the three cube roots of unity. With n = 3 and

Z = ei2πk , we have

zk = ei2πk/3 ,

k = 0, 1, 2,

or z0 = e0 = 1, 2π + i sin 3 4π = cos + i sin 3

z1 = ei2π/3 = cos z2 = ei4π/3

√ 2π 1 3 =− +i , 3 2 2 √ 3 4π 1 =− −i . 3 2 2

18.2 Polar Form of Complex Numbers It is instructive to show directly that √ "3 ! 1 3 − +i =1 and 2 2

487

!

√ "3 1 3 − −i = 1. 2 2

Here are some more examples of finding roots: $1/2 #√ √ 1+i= 2 eiπ/4+i2nπ = 21/4 eiπ/8+inπ n = 0, 1, # π $& % #π$ + i sin = 1.1 + i0.456, z0 = 21/4 eiπ/8 = 21/4 cos 8 8 1/4 iπ/8+iπ 1/4 iπ/8 z1 = 2 e = −2 e = −1.1 − i0.456.

or

The equation z 3 = i has the roots $1/3 # √ 3 i = eiπ/2+i2nπ = eiπ/6+i2nπ/3 , z0 = eiπ/6 = cos z1 = eiπ/6+i2π/3 z2 = eiπ/6+i4π/3

n = 0, 1, 2,

√ #π$ 3 1 + i sin = +i , 6 6 2 2 √ ! " ! " 5π 5π 3 1 + i sin =− = cos +i , 6 6 2 2 ! " ! " 3π 3π + i sin = −i. = cos 2 2

#π$

The reader is urged to show that zk3 = i for k = 0, 1, 2. Note how careful we were to include the factor of ei2nπ when taking roots of complex numbers. !

All nth roots of Z = ReiΘ are equally spaced on a circle of radius R1/n in the complex plane. Figure 18.6 shows two circles on which the sixth and the eighth roots of unity are located. Im

Im

π/3

(a)

π/4

Re

Re

(b)

Figure 18.6: The (a) sixth and (b) eighth roots of unity. Example 18.2.5. In certain applications of electromagnetic wave propagation (as in conductors) it becomes necessary to find an analytic expression for the Cartesian representation of the square root of a complex number. In this example, we derive such an expression.

Cartesian form of the square root of a complex number

488

Complex Arithmetic We are trying to calculate the Cartesian representation of the square root of z = x + iy. First we express z in polar form; next we take its square root, and finally reexpress the result in Cartesian form. Thus, ! y where r = x2 + y 2 , tan θ = , n = 0, ±1, ±2, . . . . z = rei(θ+2nπ) x Taking the square root of both sides yields √ z = z 1/2 = r 1/2 ei(θ+2nπ)/2 = (x2 + y 2 )1/4 eiθ/2+inπ " # θ θ = ±(x2 + y 2 )1/4 eiθ/2 = ±(x2 + y 2 )1/4 cos + i sin 2 2

because einπ = 1 if n is even and einπ = −1 if n is odd. All that is left now is to express the trigonometric functions in terms of x and y: " #1/2 %1/2 $ 1 1 θ = √ 1+ √ cos = 12 (1 + cos θ) 2 2 1 + tan2 θ & & '1/2 '1/2 1 1 1 |x| = √ 1+ ! = √ 1+ ! . 2 2 1 + (y/x)2 x2 + y 2 Similarly,

& '1/2 |x| θ 1 sin = √ 1− ! . 2 2 x2 + y 2 Collecting all these formulas together and simplifying, we obtain ( *1/2 )! *1/2 + ! 1 )! 2 x + iy = ± √ x + y 2 + |x| +i x2 + y 2 − |x| . (18.17) 2 The complexity of the expression for the square root rests on our insistence on an analytic form. The process of converting the Cartesian form of a complex number to polar, taking the square root, and converting the result back to Cartesian form ! is a far easier process than the one leading to Equation (18.17).

18.3

Fourier Series Revisited

The connection between the trigonometric and exponential functions can be utilized to write the Fourier series expansion of periodic functions more succinctly. If we substitute e2inπx/L + e−2inπx/L 2nπx = , L 2 2nπx e2inπx/L − e−2inπx/L sin = , L 2i in Equation (10.38) and collect the similar exponential terms, we obtain ∞ . , 1 (an − ibn ) e2inπx/L + (an + ibn ) e−2inπx/L f (x) = a0 + 2 cos

= a0 +

1 2

n=1 ∞ , n=1

(an − ibn ) e2inπx/L +

1 2

∞ ,

n=1

(an + ibn ) e−2inπx/L . (18.18)

18.3 Fourier Series Revisited

489

In the second sum, let n = −m to obtain 2nd sum =

−∞ !

1 2

m=−1

(a−m + ib−m ) e

2imπx/L

=

−∞ !

1 2

n=−1

(a−n + ib−n ) e2inπx/L ,

(18.19) where in the last step, we switched the dummy index back to n. If we now introduce new coefficients An defined as ⎧ 1 ⎪ if 1 ≤ n ≤ ∞, ⎨ 2 (an − ibn ) An = 12 (a−n + ib−n ) if − ∞ ≤ n ≤ −1, ⎪ ⎩ a0 if n = 0,

and use Equation (18.19) in (18.18), we obtain f (x) =

+∞ !

where

An e2inπx/L

n=−∞

L = b − a,

(18.20)

which is the equation we are after. To find An directly from this equation, multiply both sides by e−2ikπx/L , integrate from a to b, and use the readily obtainable relation ' & b 0 if n ̸= k e2i(n−k)πx/L = (18.21) = Lδnk , L if n = k a where δnk is the Kronecker delta. It follows that & & 1 b 1 b Ak = f (x)e−2ikπx/L dx or An = f (x)e−2inπx/L dx. L a L a

(18.22)

It is customary to redefine the coefficients in the summation of Equation (18.20) in such a way that the summation giving f (x) and the integral giving i.e., have the same constant in front of them. To this An are more symmetric, √ end, define fn ≡ LAn . Then (18.20) and (18.22) become +∞ 1 ! √ fn e2inπx/L , f (x) = L n=−∞

1 fn = √ L

&

b

f (x)e−2inπx/L dx

(18.23)

a

Note that the coefficients fn are complex; however, when f (x) is a real function, the exponentials and their complex coefficients add up in such a way that the final result can be expressed as an infinite sum of trigonometric functions with real coefficients. In fact, we can show this generally using Equations (18.23). First, we note that, for real f (x), fn∗

1 = √ L

&

a

b

f (x)e

+2inπx/L

1 dx = √ L

&

a

b

f (x)e−2i(−n)πx/L dx = f−n . (18.24)

Fourier series in terms of complex exponentials

490

Complex Arithmetic Next, we split the sum in (18.23) into positive integers, negative integers, and zero: −1 ∞ 1 ! 1 ! f0 f (x) = √ fn e2inπx/L + √ + √ fn e2inπx/L . L n=−∞ L L n=1

(18.25)

Changing the dummy index n to −m, the first sum can be rewritten as −1 ∞ ! 1 1 ! 1st sum = √ f−m e−2imπx/L = √ f−m e−2imπx/L L −m=−∞ L m=1 ∞ ∞ 1 ! ∗ −2imπx/L 1 ! ∗ −2inπx/L = √ fm e = √ fn e , L m=1 L n=1

where we used Equation (18.24) and changed m back to n at the end. Substituting the last equation in (18.25) yields ∞ # 1 ! " ∗ −2inπx/L f0 fn e f (x) = √ + √ + fn e2inπx/L L L n=1 ∞ " # f0 2 ! Re fn e2inπx/L =√ +√ L L n=1

showing that f (x) is indeed real. Equation (18.23) implies that f0 is also real when f (x) is. It is not hard to show that the expression in the parentheses of the first line is the sum of a sine and a cosine with real coefficients. Example 18.3.1. Let us redo the square potential—whose Fourier series was calculated in Example 10.6.1—using exponentials. From Equation (18.23), for n ̸= 0, we obtain $ 2T $ T 1 1 V (t)e−2inπt/(2T ) dt = √ V0 e−inπt/T dt fn = √ 2T 0 2T 0 & %T % T V0 T V0 −inπt/T % e [1 − (−1)n ] = = √ % −inπ 2 inπ 2T 0

because einπ = (eiπ )n = (−1)n . Similarly, f0 = V0 in the Fourier series expansion 1 V (t) = √ 2T

+∞ !

'

T /2. We now substitute these

fn e2inπt/2T

n=−∞

to get V (t) =

−1 ∞ ! ! V0 V0 V0 + [1 − (−1)n ]e2inπt/2T + [1 − (−1)n ]e2inπt/2T . 2 2inπ 2inπ n=−∞ n=1

18.4 A Representation of Delta Function

491

If we change the dummy index of the first sum from n to −m, and back to n again, and put the two sums together, we obtain V (t) = =

# " V0 ! V0 + [1 − (−1)n ] einπt/T − e−inπt/T 2 2inπ n=1 ∞

∞ # V0 2V0 ! 1 " inπt/T e − e−inπt/T + 2 π n=odd 2in $ %& ' =2i sin(nπt/T )

∞ 2V0 ! sin[(2k + 1)πt/T ] V0 + , = 2 π 2k + 1 k=0

which is the expansion we obtained in Example 10.6.1 using trigonometric functions. !

18.4

A Representation of Delta Function

Consider the function DT (x − x0 ) defined as ( T 1 DT (x − x0 ) ≡ ei(x−x0 )t dt. 2π −T

(18.26)

The integral is easily evaluated, with the result )T 1 ei(x−x0 )t )) 1 sin T (x − x0 ) DT (x − x0 ) = = . 2π i(x − x0 ) )−T π x − x0

The graph of DT (x) as a function of x for various values of T is shown in Figure 18.7. Note that the width of the curve decreases as T increases. The area under the curve can be calculated: ( ( ( ∞ 1 ∞ sin T (x − x0 ) 1 ∞ sin y dy = 1. DT (x − x0 ) dx = dx = π −∞ x − x0 π −∞ y −∞ $ %& ' =π

Figure 18.7 shows that DT (x − x0 ) becomes more and more like the Dirac delta function as T gets larger and larger. In fact, we have δ(x − x0 ) = lim DT (x − x0 ) = lim T →∞

T →∞

1 sin T (x − x0 ) . π x − x0

To see this, we note that for any finite T we can write DT (x − x0 ) =

T sin T (x − x0 ) . π T (x − x0 )

Furthermore, for values of x that are very close to x0 , T (x − x0 ) → 0

and

sin T (x − x0 ) → 1. T (x − x0 )

(18.27)

492

Complex Arithmetic

–5

0

5

Figure 18.7: The function sin T x/x also approaches the Dirac delta function as the width of the curve approaches zero. The value of T is 0.5 for the dashed curve, 2 for the heavy curve, and 15 for the light curve.

delta function as integral of imaginary exponential

Thus, for such values of x and x0 , we have DT (x − x0 ) ≈ (T /π), which is large when T is large. This is as expected of a delta function: δ(0) = ∞. On the other hand, the width of DT (x − x0 ) around x0 is given, roughly, by the distance between the points at which DT (x − x0 ) drops to zero: T (x − x0 ) = ±π, or x − x0 = ±π/T . This width is roughly ∆x = 2π/T , which goes to zero as T grows. Again, this is as expected of the delta function. Therefore, from (18.26) and (18.27), we have the following important representation of the Dirac delta function: δ(x − x0 ) =

1 2π

!

∞

(18.28)

ei(x−x0 )t dt.

−∞

Equation (18.28) can be generalized to higher dimensions, because (at least in Cartesian coordinates) the multi-dimensional Dirac delta function is just the product of the one-dimenstional delta functions. Using the more common k instead of t as the variable of integration, the two-dimensional Dirac delta function can be represented as δ(r − r0 ) =

1 (2π)2

!

∞

−∞

!

∞

−∞

eik·(r−r0 ) dkx dky ≡

1 (2π)2

!!

eik·(r−r0 ) d2 k,

Ω∞

(18.29) where Ω∞ means over all kx ky -plane and in the last integral we substituted d2 k for dkx dky . Similarly, the three dimensional Dirac delta function has the following representation: !! 1 δ(r − r0 ) = eik·(r−r0 ) d3 k, (18.30) (2π)3 Ω∞ where d3 k means a triple integral over k and Ω∞ means over all k-space.

18.5 Problems

18.5

493

Problems

18.1. Find the real and imaginary parts of the following complex numbers: (a) (2 − i)(3 + 2i). i (d) . 1+i 1 + 2i . (g) 2 − 3i 5 . (j) (1 − i)(2 − i)(3 − i)

(b) (2 − 3i)(1 + i). 1+i (e) . 2−i 2 (h) . 1 − 3i 1 + 2i 2 − i (k) + . 3 − 4i 5i

(c) (a − ib)(2a + 2ib). 1 + 3i (f) . 1 − 2i 1−i (i) . 1+i

18.2. Convert the following complex numbers to polar form and find all cube roots of each: (a) 2 − i. (e) − i.

√ (i) 1 + i 3. (m) 2 − 5i.

(b) 2 − 3i. i (f) . 1+i 2 + 3i . (j) 3 − 4i (n) 1 + i.

(c) 3 − 2i. 1+i (g) . 2−i

(d) i. 1 + 3i (h) . 1 − 2i

(o) 1 − i.

(p) 5 + 2i.

(k) 27i.

(l) − 64.

18.3. Using polar coordinates, show that: √ √ (b) (1 + i 3)−10 = 2−11 (−1 + i 3).

(a) (−1 + i)7 = −8(1 + i).

18.4. Find the real and imaginary parts of the following: √ (a) (1 + i 3)3 . √ (e) (1 + i 3)63 . (i)

$

√ %217 1+i 3 √ . 3+i

√ 4

(b) (2 + i) . (c) i. " #81 √ 1−i (f) . (g) 6 −i. 1+i 53

(j) (1 + i)22 .

(k)

√ 6 1 − i.

! √ 3 (d) 1 + i 3.

√ (h) 4 −1.

(l) (1 − i)4 .

18.5. Find the Cartesian form of all complex numbers z which satisfy (a) z 3 + 1 = 0, and (b) z 4 − 16i = 0. 18.6. Find the absolute value of

a + ib 3 + 4i and . 3 − 4i a − ib

18.7. Derive the following trigonometric identities: cos 3θ = 4 cos3 θ − 3 cos θ, sin 3θ = 3 sin θ − 4 sin3 θ.

18.8. Show that Equation (18.11) leads to Equation (18.14).

494

Complex Arithmetic 18.9. Show that z is real if and only if z = z ∗ .

√ 18.10. Show that | Re(z)| + | Im(z)| ≥ |z| ≥ (| Re(z)| + | Im(z)|)/ 2.

18.11. Let z1 = x1 + iy1 and z2 = x2 + iy2 represent two planar vectors z1 and z2 . Show that z1 z2∗ = z1 · z2 − iˆ ez · z1 × z2 .

18.12. Sketch the set of points determined by each of the following conditions: (a) |z − 2 + i| = 2.

(d) Im(z + i) = 2. ∗

(b) |z + 2i| ≤ 4.

(e) 2z + 3z = 1. ∗

(c) |z + i| = |z − i|. (f) z 2 + (z ∗ )2 = 2.

Hint: Find a relation between x and y. 18.13. Show that the equation of a circle of radius r centered at z0 can be written as |z|2 − 2 Re(zz0∗ ) = r2 − |z0 |2 .

18.14. Given that z1 z2 ̸= 0, show that (a) Re(z1 z2∗ ) = |z1 ||z2 |, and |z1 + z2 | = |z1 | + |z2 |, if and only if arg(z1 ) − arg(z2 ) = 2nπ, for n = 0, ±1, ±2, . . . . (b) What does the second equality mean geometrically? 18.15. Assume that z ̸= 1 and z n = 1. Show that 1 + z + z 2 + · · · + z n−1 = 0. 18.16. Substitute x + iy for z in z 2 + z + 1 = 0 and solve the resulting equations for x and y. Compare these with the roots obtained by solving the equation in z directly.

18.17. Find the roots of z 4 + 4 = 0 and use them to factor z 4 + 4 into a product of quadratic polynomials with real coefficients. Hint: First factor z 4 + 4 into linear terms. 18.18. Evaluate the following roots and plot them on the complex plane: √ √ √ √ 8 (b) 4 −1. (c) 1. (d) 5 −32. (a) 5 1 + i. √ √ √ √ (e) 3 + 4i. (f) 3 −1. (g) 4 −16i. (h) 6 −1.

18.19. Use binomial expansion to show directly that ! ! √ "3 √ "3 1 3 3 1 =1 and − −i = 1. − +i 2 2 2 2 # 18.20. Use eax = eax /a to find the indefinite integral of sin2 x. Verify that the derivative of your answer is indeed sin2 x. # 18.21. Use eax = eax /a and eibx = cos(bx)+ i sin(bx) to verify the following relations by integrating a certain complex exponential: $ eax eax cos(bx) dx = 2 [a cos(bx) + b sin(bx)], a + b2 $ eax eax sin(bx) dx = 2 [a sin(bx) − b cos(bx)], a + b2 where a and b are assumed to be real constants.

18.5 Problems

495 !N

18.22. (a) Using k=1 rk = (rN +1 −r)/(r−1), evaluate the sum In particular, show that N "

ei(α−βk) = ei(α−β)

k=1

!N

k=1

e−iβk .

e−iβN − 1 . e−iβ − 1

(b) Now show that if β = 2π/N , then N "

k=1

cos(α − βk) = 0 =

N "

k=1

sin(α − βk).

18.23. Express cos 4θ and sin 4θ in terms of powers of cos θ and sin θ. 18.24. Use mathematical induction to show the de Moivre theorem. 18.25. Using binomial expansion and the de Moivre theorem, show that [n/2]

cos nθ =

"

(−1)

m

m=0 [n/2]

sin nθ =

"

(−1)

m=0

m

#

$ n sin2m θ cosn−2m θ, 2m

#

$ n sin2m+1 θ cosn−2m−1 θ, 2m + 1

where [x] stands for the greatest integer less than or equal to x. 18.26. Derive Equation (18.17) from the equations preceding it. 18.27. Find the following sums, where α and β are real: (a) cos α + cos(α + β) + cos(α + 2β) + · · · + cos(α + nβ). (b) sin α + sin(α + β) + sin(α + 2β) + · · · + sin(α + nβ).

Hint: Use the result of Problem 18.22. 18.28. Show that %

a

b

e

2i(n−k)πx/L

=

&

0 if L if

n ̸= k, n = k,

where b = a + L. 18.29. Use Equations (18.20) and (18.21) to obtain Equation (18.22) 18.30. Find the Fourier series expansion of Problem 10.22 using complex exponentials. 18.31. An electric voltage V (t) is given by # $ πt V (t) = V0 sin , 0≤t≤T 2T

and repeats itself with period T . Find the Fourier series expansion of V (t) using complex exponential functions.

496

Complex Arithmetic 18.32. A periodic voltage is given by the formula ! V0 sin(πt/2T ) if 0 ≤ t ≤ T, V (t) = 0 if T ≤ t ≤ 2T, in the interval (0, 2T ). Find the Fourier series representation of this voltage using complex exponential functions. 18.33. A periodic voltage with period 4T is given by & ⎧ % 2 ⎨V 1 − t if − T ≤ t ≤ T, 0 T2 V (t) = ⎩ 0 if T ≤ |t| ≤ 2T.

Write the Fourier series of V (t) using complex exponential functions.

18.34. The function f (x) is given by the integral ' ∞ f (x) = g(y)eixy dy. −∞

Find g(y) as an integral over f (x). Hint: Multiply both sides of the equation by e−ixz and integrate over x, changing the order of integration on the righthand side and using (18.28).