The Language of Sets
The language of sets pervades all of mathematics. It provides a convenient shorthand for expressing mathematical statements. Loosely speaking, a set can be defined as a collection of objects, called themembers of the set. This definition will suffice for us. We use some shorthand to indicate certain rela- tionships between sets and elements. Usually, sets will be designated by upper- case letters such as A, B, etc., and elements will be designated by lowercase letters such asa,b, etc. As usual, setAis asubsetof setB if every element of Ais an element ofB, and aproper subset if it is a subset but not equal toB.
Two setsAandBare said to beequal if they have exactly the same elements.
Set Symbols
Some shorthand:
∅denotes the empty set, i.e., the set with no members.
a∈Ameans “ais a member of the setA.”
A=B means “the setAis equal to the setB.”
A⊆B means “Ais a subset ofB.”
A⊂B means “Ais a proper subset ofB.”
There are two ways in which we may define a set: we maylist its elements, such as in the definition A ={0,1,2,3}, or specify them by rule such as in the definition A={x| xis an integer and 0 ≤x≤3}. (Read this as “A is the set ofxsuch thatxis an integer and0≤x≤3.”) With this notation we can give formal definitions of set intersections and unions:
Definition 1.3.Set Union, Intersection, Difference Let Aand B be sets.
Then the intersection of A and B is defined to be the set A ∩ B = {x| x∈Aandx∈B}. The union of A and B is the set A ∪ B = {x| x∈Aor x∈B} (inclusive or, which means that x ∈ A or x ∈ B or both). Thedifference ofAandB is the setA−B={x| x∈Aandx ∈B}.
Example 1.7.LetA={0,1,3}andB={0,1,2,4}. Then
A∪ ∅=A, A∩ ∅=∅,
A∪B={0,1,2,3,4}, A∩B={0,1}, A−B={3}.
About Numbers
One could spend a whole course fully developing the properties of number sys- tems. We won’t do that, but we will review some of the basic sets of numbers, and assume that the reader is familiar with properties of numbers we have not mentioned here. At the start of it all is the kind of numbers that everyone knows something about: the natural or counting numbers. This is the set
Natural Numbers N={1,2, . . .}.
One could view most subsequent expansions of the concept of number as a matter of rising to the challenge of solving new equations. For example, we cannot solve the equation
x+m=n, m, n∈N,
for the unknownxwithout introducing subtraction and extending the notion of natural number that of integer. The set of integers is denoted by Integers Z={0,±1,±2, . . .}.
Next, we cannot solve the equation
ax=b, 0 =a, b∈Z,
for the unknown x without introducing division and extending the notion of integer to that of rational number. The set of rationals is denoted by Rational Numbers Q={a/b | a, b∈Z and b = 0}.
Rational-number arithmetic has some characteristics that distinguish it from integer arithmetic. The main difference is that nonzero rational numbers have multiplicative inverses: the multiplicative inverse of a/b is b/a. Such a Field of Numbers number system is called afield of numbers. In a nut-
shell, afield of numbers is a system of objects, called numbers, together with operations of addition, subtraction, multiplication, and division that satisfy the usual arithmetic laws; in particular, it must be possible to subtract any number from any other and divide any number by a nonzero number to obtain another such number. The associative, commutative, identity, and inverse
laws must hold for each of addition and multiplication; and the distributive law must hold for multiplication over addition. The rationals form a field of numbers; the integers don’t since division by nonzero integers does not always yield an integer.
The jump from rational to real numbers cannot be entirely explained by algebra, although algebra offers some insight as to why the number system still needs to be extended. There is no rational number whose square is 2. Thus the equation
x2= 2
cannot be solved using rational numbers alone. (Story has it that this is lethal knowledge, in that followers of a Pythagorean cult claim that the gods threw overboard from a ship one of their followers, Hippasus of Metapontum, who was unfortunate enough to discover that fact.) There is also the problem of numbers like π and the mathematical constant e which do not satisfy any polynomial equation. The heart of the problem is that if we consider only rationals on a number line, there are many “holes” that are filled by numbers like π and √
2. Filling in these holes leads us to the set R of real numbers, which are in one-to-one correspondence with the points on a number line.
We won’t give an exact definition of the set of real numbers. Recall that every real number admits a (possibly infinite) decimal representation, such as 1/3 = 0.333. . . or π= 3.14159. . . .This provides us with a loose definition:
Real Numbers Real numbers are numbers that can be expressed by a decimal representation, i.e., limits of finite decimal expansions. Equivalently, real numbers can be thought of as points on the real number line. As usual, the set of all real numbers is denoted byR. In addition, we employ the usual interval notations for real numbersa,b such thata≤b:
[a, b] ={x∈R|a≤x≤b}, [a, b) ={x∈R|a≤x < b}, (a, b) ={x∈R|a < x < b}.
There is one more problem to overcome. How do we solve a system like x2+ 1 = 0
over the reals? The answer is we can’t: ifxis real, thenx2≥0, sox2+ 1>0. Complex Numbers We need to extend our number system one more time,
and this leads to the set Cof complex numbers. We definei to be a quantity such thati2=−1and
C={a+bi | a, b∈R}.
Standard Form We say that the formz=a+biis the standard form of z.In this case the real part ofzis(z) =aand the imaginary part is defined
Fig. 1.5: Standard and polar coordinates in the complex plane.
as (z) = b.(Notice that the imaginary part ofz is a real number: it is the real coefficient ofi.) Two complex numbers areequal precisely when they have the same real part and the same imaginary part. All of this could be put on a more formal basis by initially defining complex numbers to be ordered pairs of real numbers. We will not do so, but the fact that complex numbers behave like ordered pairs of real numbers leads to an important geometrical insight:
complex numbers can be identified with points in the plane.
Instead of an x- and y-axis, one lays out a real and an imaginary axis Real and Imaginary Parts (which are still usually labeled withxand y)
and plots complex numbers a + bi as in
Figure 1.5. This results in thecomplex plane.Arithmetic in Cis carried out using the usual laws of arithmetic forRand the algebraic identityi2=−1to reduce the result to standard form. In addition, there are several more useful ideas about complex numbers that we will need.
The length, or absolute value, of a complex number in standard Absolute Value form, z=a+bi, is defined as the nonnegative real num-
ber |z|=√
a2+b2, which is the distance from the origin toz. Thecomplex conjugate ofz is defined asz=a−bi(see Figure1.5). Thus we have:
Laws of Complex Arithmetic
(a+bi) + (c+di) = (a+c) + (b+d)i (a+bi)ã(c+di) = (ac−bd) + (ad+bc)i
a+bi = a−bi
|a+bi| = √ a2+b2
If meaning is clear, the productz1ãz2 is often abbreviated toz1z2.
Example 1.8.Letz1= 2 + 4iandz2= 1−3i.Computez1−3z2 andz1z2.
Solution.We have that
z1−3z2= (2 + 4i)−3(1−3i) = 2 + 4i−3 + 9i =−1 + 13i and
z1z2= (2+4i)(1−3i) = 2+4i−2ã3i−4ã3i2= (2+12)+(4−6)i) = 14−2i.
Here are some easily checked and very useful facts about absolute value and complex conjugation:
Laws of Conjugation and Absolute Value
|z1z2| = |z1| |z2|
|z1+z2| ≤ |z1|+|z2|
|z|2 = zz z1+z2 = z1+z2
z1z2 = z1z2 z1/z2 =z1z2/|z2|2
Example 1.9.Letz1= 2 + 4iandz2= 1−3i.Verify that |z1z2|=|z1| |z2|.
Solution.From Example1.8,z1z2=14−2i, so that|z1z2|=
142+(−2)2=
√200, while |z1|=√
22+ 42 =√
20 and |z2|=
12+ (−3)2 =√
10. Hence
|z1z2|=√ 10√
20 =|z1| |z2|.
Example 1.10.Verify that the conjugate of the product is the product of conjugates.
Solution.This is just the fifth fact in the preceding list. Letz1=x1+ iy1 andz2=x2+iy2be in standard form, so thatz1=x1−iy1andz2=x2−iy2. We calculate
z1z2= (x1x2−y1y2) + i(x1y2+x2y1), so that
z1z2= (x1x2−y1y2)−i(x1y2+x2y1).
Also,
z1z2= (x1−iy1)(x2−iy2) = (x1x2−y1y2)−i(x1y2+x2y1) =z1z2. The complex number z = i solves the equation z2+ 1 = 0 (no surprise here: it was invented expressly for that purpose). The big surprise is that once we have the complex numbers in hand, we have a number system so complete that we can solve any polynomial equation in it. We won’t offer a proof of this fact; it’s very nontrivial. Suffice it to say that nineteenth-century mathematicians considered this fact so fundamental that they dubbed it the
“Fundamental Theorem of Algebra,” a terminology we adopt.
Theorem 1.1.Fundamental Theorem of Algebra Let p(z) = anzn + an−1zn−1+ã ã ã+a1z +a0 be a nonconstant polynomial in the variable z with complex coefficientsa0, . . . , an.Then the polynomial equationp(z) = 0 has a solution in the fieldCof complex numbers.
Note that the fundamental theorem doesn’t tell us how to find a root of a polynomial, only that it exists. There are numerical techniques for approx- imating such roots. But for polynomials of degree greater than four, there are no general algebraic expressions in terms of radicals (like the quadratic formula) for their roots.
In vector space theory the numbers in use are called scalars, and Scalars we will use this term. Unless otherwise stated or suggested by the
presence ofi, the field of scalars in which we do arithmetic is assumed to be the field of real numbers. However, we shall see later, when we study eigensystems, that even if we are interested only in real scalars, complex numbers have a way of turning up quite naturally.
The following example shows how to “rationalize” a complex denominator.
Example 1.11.Solve the linear equation(1−2i)z= (2 + 4i)for the complex variable z. Also compute the complex conjugate and absolute value of the solution.
Solution. The solution requires that we put the complex number z = (2+4i)/(1−2i)in standard form. Proceed as follows: multiply both numerator and denominator by(1−2i) = 1 + 2ito obtain that
z= 2 + 4i
1−2i =(2 + 4i)(1 + 2i)
(1−2i)(1 + 2i) = 2−8 + (4 + 4)i 1 + 4 = −6
5 +8 5i. Next we see that
z= −6 5 +8
5i =−6 5 −8
5i and
|z|= 1
5(−6 + 8i) = 1
5|(−6 + 8i)|= 1 5
(−6)2+ 82= 10
5 = 2.
Practical Complex Arithmetic
We conclude this section with a discussion of the more advanced aspects of complex arithmetic. This material will not be needed until Chapter4. Recall from basic algebra the roots theorem: the linear polynomialz−ais a factor of a polynomial f(z) = a0+a1z+ã ã ã+anzn if and only if a is a root of the polynomial, i.e.,f(a) = 0. If we team this fact up with the Fundamental Theorem of Algebra, we see an interesting fact about factoring polynomials overC: every polynomial can be completely factored into a constant times a product of linear polynomials of the form z−a. The numbers a that occur are exactly the roots off(z). Of course, these roots could be repeated roots, as in the case of f(z) = 3z2−6z+ 3 = 3(z−1)2. But how can we use the Fundamental Theorem of Algebra in a practical way to find the roots of a polynomial? Unfortunately, the usual proofs of the Fundamental Theorem of Algebra don’t offer a clue, because they are nonconstructive, i.e., they prove that solutions must exist, but do not show how to explicitly construct such a solution. Usually, we have to resort to numerical methods to get approximate solutions, such as the Newton’s method used in calculus. For now, we will settle on a few ad hoc methods for solving some important special cases.
First-degree equations offer little difficulty: the solution to az = b is z = b/a, as usual. There is one detail to attend to: what complex number is represented by the expressionb/a? We saw how to handle this by the trick of “rationalizing” the denominator in Example1.11.
Quadratic equations are also simple enough: use the quadratic formula, which says that the solutions to az2+bz+c= 0, wherea = 0, are given by
Quadratic Formula
z= −b±√
b2−4ac
2a .
One little catch: what does the square root of a complex number mean? For nonnegative real numbers r the expression √
r is called the principal square root of r and its meaning is unambiguous. For complex numbers it is not.
What we are really asking is this: How do we solve the equation z2 =dfor z, wheredis a complex number? Let’s try for a little more: How do we solve zn=dfor all possible solutionsz, wheredis a nonzero complex number? In a few cases, such an equation is quite easy to solve. We know, for example, thatz=±iare solutions toz2=−1, so these are all the solutions. Similarly, one can check by hand that ±1,±iare all solutions toz4= 1.Consequently, z4−1 = (z−1)(z+1)(z−i)(z+i).Roots of the equationzn= 1are sometimes called thenth roots of unity. Thus the4th roots of unity are±1and±i.But what about something likez3= 1 + i?
The key to answering this question is another form of a nonzero complex numberz=a+bi.In reference to Figure1.5we can writez=r(cosθ+i sinθ) = reiθ, where θ is a real number, ris a positive real, and eiθ is defined by the following expression, which is called Euler’s formula:
Definition 1.4.Polar form eiθ= cosθ+ i sinθ.
Notice that |eiθ| =
cos2θ+ sin2θ = 1, so that |reiθ| = |r||eiθ| = r, provided r is nonnegative. The expression reiθ with r = |z| and the angle θ measured counterclockwise in radians is called the polar form of z. The numberθis sometimes called anargument ofz.It is important to notice that θis not unique. If the angleθ0works for the nonzero complex numberz, then so does θ=θ0+ 2πk, for any integer k, since sinθ andcosθ are periodic of period2π.It follows that a complex number may have more than one polar form. For example, i = eiπ/2= ei5π/2 (here r= 1). In fact, the most general polar expression foriisi = ei(π/2+2kπ),wherekis an arbitrary integer.
Example 1.12.Find the possible polar forms of 1 + i.
Solution.Sketch a picture of1 +iin the complex plane and we see that the angle θ0 = π/4 works fine as a measure of the angle from the positive x-axis to the radial line from the origin toz. Moreover, the absolute value of zis√
1 + 1 =√
2. However, we can adjust the angleθ0by any multiple of2π, a full rotation and get a polar form for z. So the most general form forz is z=√
2ei(π/4+2kπ), wherekis any integer.
As the notation suggests, polar forms obey the laws of exponents. A simple application of the laws for the sine and cosine of a sum of angles shows that for anglesθandψ we have the identity
ei(θ+ψ)= eiθeiψ.
By using this formula n times, we obtain thateinθ = eiθn
, which can also be expressed asde Moivre’s Formula:
(cosθ+ i sinθ)n= cosnθ+ i sinnθ.
Now for solvingzn =d: First, find the general polar form ofd, sayd = aei(θ0+2kπ),whereθ0is theprincipal angleford, i.e.,0≤θ0<2π,anda=|d|. Next, writez=reiθ, so that the equation to be solved becomes
rneinθ=aei(θ0+2kπ).
Taking absolute values of both sides yields thatrn =a, whence we obtain the unique value of r= √n
a= n
|d|.What about θ? The most general form for nθ is
nθ=θ0+ 2kπ.
Hence, we obtain that
θ= θ0 n +2kπ
n .
Notice that the values of ei2kπ/n start repeating themselves as k passes a multiple of n, since ei2π = e0 = 1. Therefore, one gets exactly n distinct values for eiθ, namely
θ=θ0 n +2kπ
n , k= 0, . . . , n−1.
These points are equally spaced around the unit circle in the complex plane, starting with the pointeiθ0.Thus we have obtainedndistinct solutions to the equationzn=d,whered = 0, namely
General solution tozn=d
z=a1/nei(θ0/n+2kπ/n), k= 0, . . . , n−1, where0 =d=aeiθ0
Example 1.13.Solve the equationz3= 1 + ifor the unknownz.
Solution.The solution goes as follows: We have seen that1+ihas a polar form
1 + i =√ 2eiπ/4.
Then according to the previous formula, the three solutions to our cubic are z = (√
2)1/3ei(π/4+2kπ)/3= 21/6ei(1+8k)π/12, k= 0,1,2.
See Figure1.6for a graph of these complex roots.
–1
–1 1
21/6
21/6eiπ/12 y
x 21/6ei9π/12 1
21/6ei17π/12
21/2ei/4= 1 + i
Fig. 1.6: Roots ofz3= 1 + i.
We conclude with a little practice with square roots and the quadratic formula. In regard to square roots, as we have noted, the expressionw=√
dis ambiguous when dealing with complex numbers. In view of this difference, we will generally avoid using the radical sign with complex numbers (exceptions:
the quadratic formula and the case of√
−dwith positived, in which case the interpretation is√
−d= i√ d).
Example 1.14.Compute a square root of the numbers −4 andi.
Solution.Observe that−4 = 4ã(−1).It is reasonable to expect the laws of exponents to continue to hold, so we should have (−4)1/2= 41/2ã(−1)1/2.
Now we know that i2 = −1, so we can take i = (−1)1/2 and obtain that
√−4 = (−4)1/2= 2i.Let’s check it:(2i)2= 4i2=−4.
We have to be a bit more careful withi.We’ll just borrow the idea of the formula for solvingzn=d.First, putiin polar form asi = 1ãeiπ/2.Now raise each side to the1/2power to obtain
i1/2= 11/2ã(eiπ/2)1/2
= 1ãeiπ/4= cos(π/4) + i sin(π/4)
= √1
2(1 + i).
A quick check confirms that((1 + i)/√
2)2= 2i/2 = i.
Example 1.15.Solve the equationz2+z+ 1 = 0.
Solution.According to the quadratic formula, the answer is z= −1±√
12−4
2 =−1
2±i
√3
2 .
Example 1.16.Solvez2+z+1+i = 0and factor the polynomialz2+z+1+i.
Solution.This time we obtain from the quadratic formula that z=−1±
1−4(1 + i)
2 = −1±
−(3 + 4i)
2 .
What is interesting about this problem is that we don’t know the polar angle θ for z = −(3 + 4i). Fortunately, we don’t have to. We know that sinθ =
−4/5and cosθ=−3/5. We also have the standard half angle formulas from trigonometry to help us:
cos2θ/2 = 1 + cosθ
2 = 1
5 and sin2θ/2 = 1−cosθ
2 = 4
5.
Sinceθis in the third quadrant of the complex plane,θ/2is in the second, so cosθ/2 = −1√
5 and sinθ/2 = √2 5.
Notice that | −(3 + 4i)|= 5. Hence, a square root of−(3 + 4i)is given by w=√
5 −1√
5 +√2 5i
=−1 + 2i.
Check that w2 =−(3 + 4i), so the two roots to our quadratic equation are given by
z= −1±(−1 + 2i)
2 =−1 + i,−i.
In particular, we see that z2+z+ 1 + i = (z+ 1−i)(z+ i).