CHAPTER 8. COMPLEX NUMBERS Why Do We Need Complex Numbers? First of All, a Simple Algebraic Equation Like X2 = −1 May Not Have

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CHAPTER 8. COMPLEX NUMBERS

Why do we need complex numbers?

First of all, a simple algebraic equation like x2 = 1 may not have a real solution. − Introducing complex numbers validates the so called fundamental theorem of algebra: every polynomial with a positive degree has a root. However, the usefulness of complex numbers is much beyond such simple applications. Nowadays, complex numbers and complex functions have been developed into a rich theory called complex analysis and be- come a power tool for answering many extremely difficult questions in mathematics and theoretical physics, and also finds its usefulness in many areas of engineering and com- munication technology. For example, a famous result called the prime number theorem, which was conjectured by Gauss in 1849, and defied efforts of many great mathematicians, was finally proven by Hadamard and de la Vallée Poussin in 1896 by using the complex theory developed at that time. A widely quoted statement by Jacques Hadamard says: “The shortest path between two truths in the real domain passes through the complex domain”.

The basic idea for complex numbers is to introduce a symbol i, called the imaginary unit, which satisﬁes i2 = 1. − In doing so, x2 = 1 turns out to have a solution, namely x = i; (actually, there − is another solution, namely x = i). We remark that, sometimes in the mathematical − literature, for convenience or merely following tradition, an incorrect expression with correct understanding is used, such as writing √ 1 for i so that we can reserve the − letter i for other purposes. But we try to avoid incorrect usage as much as possible. So here we use that letter “i” to stand nothing other than the imaginary unit.

A general complex number is of the form c = a + bi (or c = a + ib), where a, b are real numbers. We call a the real part of c and b the imaginary part of c. Sometimes they are denoted by Re c and Im c respectively.

Question 8.1. What is wrong by saying that “√2 is a real number and hence it is not a complex number”?

Addition and multiplication of complex numbers can be carried out in a straightfor-

1 ward manner, keeping in mind that i2 = 1. Indeed, given complex numbers c = a +b i − 1 1 1 and c2 = a2 + b2i, the sum c1 + c1 and the product c1c2 are given by

c1 + c2 =(a1 + b1i)+(a2 + b2i)=(a1 + a1)+(b1 + b2)i

2 c1c2 =(a1 + b1i)(a2 + b2i)= a1a2 + b1b2i + a1b2i + a2b1i

=(a a b b )+(a b + a b )i 1 2 − 1 2 1 2 2 1 To give a quick example, letting c = 1 2i, c = 2+3i, we have c + c = 3+ i, 1 − 2 1 2 c c = 1 5i and c c =8+ i. 1 − 2 − − 1 2 Exercise 8.2. In each of the following cases, ﬁnd c + c , c c and c c . Present 1 2 1 − 2 1 2 your answers as simple as possible. (a) c = i, c = 1 i; (b) c = 2+√3i, c = 2 √3i; 1 2 − 1 2 − (c) c =1+ i, c 1 i. 1 2 − −

Exercise 8.3. Find the product c1c2 of c1 = cos α1 +i sin α1 and c2 = cos α2 +i sin α2. Present your answers as simple as possible.

The expression a + bi is called the Cartesian representation of c, (vs the polar representation described in the future), since geometrically it is represented by the point or the vector in the Cartesian xy–plane with coordinates (a, b). (See the following ﬁgure.)

The magnitude of the vector representing c is called the absolute value or the modulus of c, and is denoted by c . Thus, if c = a + bi, then c = √a2 + b2: | | | | a + bi = √a2 + b2 (8.1) | | The xy–plane for representing complex numbers is called the complex plane or Argand plane and is denoted by C. An alternative way to write “c is a complex number” is “c C”, read as “c belongs to C”. ∈ 2 Every complex number z has a “twin sister” z, called the complex conjugate of z. The twins z and z do not quite have exactly the same look. They are more like mirror images to each other. The complex conjugate of c = a + bi, denoted by c, is deﬁned to be a bi: − a + bi = a bi. (8.2) − For examples, 2 + 3i = 2 3i, i = i, 2 = 2, 1 i =1+ i, etc. − − −

Question 8.4. In each of the following cases, what is the complex conjugate z and the modulus z of the given complex number z? (a) z = i, (b) z = 3, (c) z =4+3i | | − − (d) z = cos θ + i sin θ.

Notice that, for a complex number z = a + bi, we have

zz =(a + bi)(a bi)= a2 b2i2 = a2 + b2 = z 2. − − | | So

z 2 = zz. (8.3) | |

This identity, which resembles v 2 = v v, is one of the most useful facts about complex | | · numbers, despite of its simplicity.

Usual algebraic identities such as

uv = vu (commutative law), (u + v)w = uw + vw (distributive Law) (8.4)

3 hold for complex numbers u, v and w. Furthermore, taking complex conjugates “goes along” with algebraic operations well:

z + w = z + w, zw = z w, z/w = z/w. (8.5)

Exercise 8.5. Check all identities in (8.4) and (8.5).

Example. We are asked to verify the identity

zw = z w (8.6) | | | || | for two complex numbers z and w. Notice that zw = z w is equivalent to | | | || | zw 2 = z 2 w 2 | | | | | | and hence it suﬃces to verify the latter. By (8.3), we have

Exercise 8.6. Use (8.6) to deduce

(a2 + b2)(c2 + d2)=(ac + bd)2 +(ad bc)2, − where a, b, c, d are arbitrary real numbers. Which integers u and v satisfy u2 + v2 = (92 + 102)(102 + 112)?

Exercise 8.7. Verify the following “parallelogram identity”

An important use of complex conjugation is to divide complex numbers. Given complex numbers z and w with z = 0, to divide w by z, we multiply both the denominator 6 and the numerator of w/z by z, the complex conjugate of the denominator, to obtain w wz wz = = z zz z 2 | | 4 in order to convert the denominator of w/z into a real number. Then we can divide the real part and the imaginary part of wz by the real number z 2. For example, to divide | | 1 i by 1+ i, we proceed as follows, by noticing that the complex conjugateof 1+ i − is 1 i: − 1 i (1 i)(1 i) 1+ i2 2i − = − − = − = i. 1+ i (1 + i)(1 i) 11 + 12 − − In division of complex numbers, it is important to recognize the complex conjugate z of z = x + yi and to use zz = z 2 = x2 + y2. | | Exercise 8.8. Perform each of the following complex division: 8+ i 5+ i 75 + 223i cos θ + i sin θ (a) (b) (c) (d) . 2 i 3 2i 63 + 17i cos θ i sin θ − − −

Example. We are asked to check that if z = 1 and if | | z a w = − , 1 a z − where a is any ﬁxed complex number with a < 1, then w = 1. First, let us notice | | | | that w = 1 is the same as w 2 = 1, which is the same as ww = 1, (in view of (8.3) | | | | above). This leads us to compute w w. Now z a z a zz za az + aa 1 za az + aa w w = − − = − − = − − = 1, 1 a z 1 az 1 az az + az az 1 az za + aa − − − − − − in view of zz = z 2 = 1. Done. | | Problem 8.9. Verify that, if a is a real number with 0

z a | − | = a z a 1 | − − | for all complex numbers z with z = 1. | | The real part of a complex number z, denoted by Re z, is given by z + z Re z = (8.7) 2 A complex number z is real if and only if z = z.

Exercise 8.10. Check the last statement as well as identity (8.6).

5 Exercise 8.11. Verify that, if z =1 and z = 1, then | | 6 z + 1 w = i z 1 − is a real number.

Take any complex number z = x+iy, with x as its real part and y as its imaginary part. Write r for the modulus of z: r = z = x2 + y2. Let θ be the angle between | | the vector representing z; (see the next ﬁgure). Thenp x = r cos θ and y = r sin θ. Hence z = r cos θ + ir sin θ, or z = r(cos θ + i sin θ). (8.8)

The last expression is called the polar form of z.

Multiplication of complex numbers is especially revealing when expressed in polar form. Let

z1 = r1(cos θ1 + i sin θ1), z2 = r2(cos θ2 + i sin θ2) be complex numbers in polar form. Then

z1z2 = r1r2(cos θ1 + i sin θ1)(cos θ2 + i sin θ2)

= r r ((cos θ cos θ sin θ sin θ )+ i(cos θ sin θ + sin θ cos θ )) 1 2 1 2 − 1 2 1 2 1 2

= r1r2(cos(θ1 + θ2)+ i sin(θ1 + θ2)). (8.9) Here we have used the addition formula for sine and cosine in trigonometry

cos(α + β) = cos α cos β sin α sin β, sin(α + β) = sin α cos β + cos α sin β. (8.10) − We single out the most crucial part of (8.9)

(cos θ1 + i sin θ1)(cos θ2 + i sin θ2) = cos(θ1 + θ2)+ i sin(θ1 + θ2). (8.11)

6 Let us put eiθ = cos θ + i sin θ (8.12) which is called Euler’s formula. Then (8.11) becomes eiθ1 eiθ2 = ei(θ1+θ2) and (8.10) iθ1 iθ2 i(θ1+θ2) becomes (r1e )(r2e )= r1r2e . By induction, we can prove

iθ1 iθ2 iθn i(θ1+θ2+ +θn) e e e = e ··· . ··· When θ = θ = = θ = θ, the above identity becomes 1 2 ··· n (eiθ)n = einθ. (8.13)

This is called De Moivre’s formula.

Using Euler’s formula, we have

e2πi = 1, eπi = 1, eπi/2 = i, e3πi/2 = i, − − 1 1 eπi/4 = (1 + i), e3πi/4 = ( 1+ i), √2 √2 − 1 √3 √3 i 1 √3 eπi/3 = + i , eπi/6 = + , e2πi/3 = + i . 2 2 2 2 −2 2

Exercise 8.12. Check the above identities.

Question 8.13. A logo for T-shirt is “I am number eπi ”. What does that mean? − Fix a positive integer n and consider the complex number ω = e2πi/n, called a primitive nth root of unity. (Remark: the Greek letter “ω”, with “Ω” as its upper case, is read as “omega”. Since it is the last letter of Greek’s alphabet, it is sometimes used for eternity.) Then we have ωn = 1. Indeed, by De Moivre’s formula,

ωn =(e2πi/n)n = e2πi = 1.

Furthermore, we can check that the following list of n complex numbers are solutions to the algebraic equation xn = 1:

2 n 1 1, ω, ω , ..., ω − (8.14)

Indeed, take one of them, say ωk. Then (ωk)n = ωkn = (ωn)k = 1k = 1. It turns out that the above list gives n distinct roots of the polynomial xn 1. − 7 Question 8.14. Why are the numbers in the list (8.14) distinct to each other?

Example. We are asked to ﬁnd all solutions to x3 = 1. First, we express 1 in − − polar form: 1 = eπi. Then α eπi/3 is a solution to x3 = 1. Indeed, we can check − ≡ − that α3 = eπi = 1. Let us write x = αy. Then x3 = 1 becomes α3y3 = 1, or − − − y3 = 1. The solutions to y3 = 1 are y = 1, ω, ω2, where ω = e2πi/3. So the solution to x3 = 1 are x = α, αω, αω2. − Exercise 8.15. Express the solutions α, αω, αω2 to x3 = 1 in Cartesian form − a + bi. Exercise 8.16. Solve each of the following equations: (a) x4 = 1 (b) x3 = i. − Recall the algebraic identity

n n n 1 n 2 n 2 n a b =(a b)(a − + a − b + + ab − + b ). − − ··· Applying this identity for a = 1 and b = ω, we have

n 2 n 1 0 = 1 ω = (1 ω)(1 + ω + ω + + ω − ). − − ··· Clearly 1 ω = 0. Therefore − 6 2 n 1 1+ ω + ω + + ω − = 0. (8.15) ··· (Aside: This identity is basic to ﬁnite Fourier transform, an area applied to signal 2 n 1 transmission and processing.) The terms 1, ω, ω , ..., ω − form a list of all nth roots of unity. They are located at the vertices of a regular n-gon in the complex plane.

where n = 6 and ω = e2πi/6 = eπi/3 Question 17. Can we get a new identity out of (8) by multiplying (8) by ω to obtain ω + ω2 + + ωn = 0? ··· 8 Finally, we consider the exponential ez for a complex number z = x + iy. Naturally we put ez = ex+iy = exeiy = ex(cos y + i sin y)

For the exponential ex of a real number x, see the Appendix to this chapter. One basic property of ex is that it is always a positive number. So

ez = exeiy = ex eiy = ex. | | | | | || | Since x = Re z, the real part of z, we obtain

ez = eRe z. (8.16) | |

It is easy to check that, for all complex numbers z1 and z2,

ez1+z2 = ez1 ez2 . (8.17)

Exercise 8.18 Check the last identity.

Question 8.19. Are ez and ez¯ equal? Why?

***The rest of the present chapter is optional. It contains more interesting but harder materials about complex numbers.

Example. We are asked to verify the following identities: α + β α β cos α cos β = 2 sin sin − , − − 2 2 α + β α β sin α sin β = 2 cos sin − . − 2 2 We use a “complex trick” as follows.

iα iβ i(α+β)/2+i(α β)/2 i(α+β)/2 i(α β)/2 (cos α cos β)+ i(sin α i sin b)= e e = e − e − − − − − −

i(α+β)/2 i(α β)/2 i(α β)/2 α + β α + β α β = e (e − e− − )= cos + i sin 2i sin − − 2 2 2 µ ¶ µ ¶ α + β α β α + β α β = 2 sin sin − + 2i cos sin − . − 2 2 2 2 9 Comparing the real and the imaginary parts of both sides, we obtain the required identity.

Example. We are asked to ﬁnd the value of cos 2π/5. Naturally we consider the complex number ω = e2πi/5. Then the ﬁfth roots of unity are

1, ω = e2πi/5, ω2 = e4πi/5, ω3 = e6πi/5, ω4 = e8πi/5. geometrically they are vertices of a regular pentagon. Notice that ω3 is the complex conjugate of ω2 and ω4 is the complex conjugate of ω. Hence (8) gives

2 2 1+ ω + ω + ω + ω = 0. (8)′

Let a = ω + ω 2 cos 2π/5. Then a2 =(ω + ω)2 = ω2 + ω2 + 2ω ω = ω2 + ω2 + 2. Hence ≡ 2 2 (8)′ becomes 1 + a + a 2 = 0, or a + a 1 = 0. Since cos 2π/5 is positive,we only − − take the positive solution to a2 + a 1 = 0, which is a =(√5 1)/2. Thus − − 2π a √5 1 cos = = − . 5 2 4

Question 20. Use a calculator to ﬁnd numerical values of cos 72o and (√5 1)/4. Are − they the same?

Example. We are asked to give an explicity solution to the equation x2 = a + ib, where a + ib is a unit modulus number, that is a2 + b2 a + ib 2 = 1. Let us write x = u + iv. ≡ | | Then x2 = a + bi becomes (u + iv)= u2 v2 + 2uvi = a + bi. Comparing with the real − and the imaginary parts of both sides, we obtain 2uv = b and u2 v2 = a. Pairing the − last identity with a2 + b2 = 1, we obtain 2u2 =1+ a and hence

1+ a u = . ±r 2 From 2uv = b we get v = b/2u. Let ε = 1 if b 0 and ε = 1 if b < 0. Then ≥ − b = ε b = ε√b2 = ε√1 a2. So | | − b ε√1 a2 1 a v = = − = ε − . 2u ± 2(1 + a) ± r 2 p

10 Now we use this answer to ﬁnd the values of cos 15o cos π/12 and sin 15o sin π/12. ≡ ≡ Let a = cos π/12, b = sin π/12 and x = a + bi eπi/12. Then x2 = eπi/6 = √3 + 1 with ≡ 2 2 a = √3/2, b = 1/2 and ε = 1. Our formula for u and v gives

π 1 π 1 cos = a = 2+ √3 sin = b = 2 √3, 12 2 12 2 − q q which are our answers. [Actually, there is a slick way to simplify our answer. Note that

2 2+ √3=(4+2√3)/2=(1+2√3+ √3 )/2=(1+2√3)2/2.

So π 1 (1 + √3)2 1+ √3 cos = = . 12 2 s 2 2√2 We can simplify our answer for sin π/12 in the same way.]

Exercise 21. Find cos π/8 and sin π/8.

Problem 22. Prove: 2 cos 2π/7 is a root of the cubic x3 + x2 2x 1. − − Example. In the theory of Fourier series, we have to ﬁnd a closed form of the sum

S 1 + 2 (cos θ + cos 2θ + + cos nθ) n ≡ ··· for an arbitrary positive integer n. This requires some good skill to accomplish. Let z = eiθ. Then 2 cos θ = z + z, 2 cos 2θ = z2 + z2 etc. Thus we have

S =1+ z + z + z2 + z2 + + zn + zn n ···

n n 1 2 n 1 n z + z − + + z +1+ z + z + + z − + z . ≡ ··· ··· Notice that zz = z 2 = 1. So | |

n 1 n 2 n n+1 zS = z − + z − + + z + z . n ··· Hence zS S = zn+1 zn. n − n − 1/a iθ/a 1/a iθ/a Therefore, writing z for e and z for e− , we have

n+1 n n+1 n 1/2 n+1/2 n+1/2 1 z z z z z 1 z z sin (n + 2 )θ Sn = − = − = − = . z 1 z 1 z1/2 2 z1/2 z1/2 sin 1 θ − − − 2

11 This answer is very hard to ﬁnd without using complex numbers.

In Chapter 5, we have studied the inner product for the space Rn, consisting of n– tuples of complex numbers. We may replace real numbers by complex numbers to obtain a complex space, denoted by Cn of n tuples of complex numbers:

z =(z1, z2,...,zn).

Addition and scalar multiplication can be defined in the same way. To define the dot product, it is essential to use complex conjugation: for complex vectors z = (z1, z2,...,zn) n and w =(w1, w2,...,wn) in C , their dot product is defined to be

z w = z w + z w + + z w . • 1 1 2 2 ··· n n Notice that z z = z z + z z + + z z • 1 1 2 2 ··· n n = z 2 + z 2 + + z 2 0. | 1| | 2| ··· | n| ≥ This allow us to deﬁne the magnitude of a complex vector z by putting z = √z z. | | • As for real vectors, we say that a vector z in C is a unit vector if z = 1, and we | | say that two vectors are perpendicular if their dot product is zero: z w z w = 0. ⊥ ⇔ • A set of n vectors in Cn is said to be an orthonormal basis of Cn if they are mutually perpendicular unit vectors.

Problem 23. As before, write ω = e2πi/n. Prove that

1 j 1 2(j 1) 3(j 1) (n 1)(j 1) e = (1, ω − , ω − , ω − , ..., ω − − ), 1 j n, j √n ≤ ≤ form an orthonormal basis of Cn. (The matrix with these basis vectors as columns deﬁnes the ﬁnite Fourier transform.)

12 Appendix. Exponentials and Logarithms

§1. Multiplication. How do we normally understand the product ab a b of two ≡ × (real) numbers a and b? Well, it takes several steps. We begin with the case that a is a positive integer. When a = 1, ab is just b. When a = 2, 3, 4 etc., ab is understood in the following way: 2b = b + b, 3b = b + b + b, 4b = b + b + b + b. Next, we consider the case when a is a fraction (or more precisely, a rational number), say a = m/n, where m, n are positive integers. In that case, we can rewrite ab as follows: m 1 ab = b = m b . n n µ ¶ So it is enough to consider (1/n)b. What is (1/n)b? Answer: it is the number, say c, satisfying nc = b. Thus (1/2)b is the number satisfying 2 (1/2)b = b, (1/3)b is the number satisfying 3 (1/3)b = b, etc. Now we can describe ab for any positive number ab. Although a here is not necessarily a fraction (for example, √2 is known to be irrational), we can approximate a by fractions as accurate as we wish. So, when we approximate a by m/n, as we know, (m/n)b is a number close to ab. This is good enough for assigning a meaning to ab because the approximation here can be as good as we wish. Finally, what is ab when a is a negative number? In that case, we can write a = p for some positive number p, − giving us ab =( p)b = pb = p( b). So it is enough to know what b is. What is b? − − − − − Well, it is the number c satisfying b + c = 0, that is, b +( b) = 0. After taking these − steps, now we can claim that we ‘understand’ what multiplication means. Concerning multiplication, the following rules are basic: Associative Law: (ab)c = a(bc). (It tells us the expression abc is unambiguous.) Commutative Law: ab = ba. Distributive Law: (a + b)c = ac + bc, c(a + b)= ca + cb. All other elementary identities concerning multiplication are more or less from these three. For example, the well-known identity (a + b)2 = a2 + 2ab + b2 can be derived as follows:

(a + b)2 =(a + b)(a + b)= a(a + b)+ b(a + b)= aa + ab + ba + bb

= a2 + ab + ab + b2 = a2 + 2ab + b2. Another important identity is (a b)(a + b)= a2 b2. It is obtained as follows: − − (a b)(a + b)= a(a + b)+( b)(a + b)= aa + ab +( b)a +( b)b − − − − 13 = a2 + ab ba b2 = a2 b2. − − − More complicated identities such as

(a + b)3 = a3 + 3a2b + 3ab2 + b3 and (a b)(a2 + ab + b2)= a3 b3 − − can be derived in the same manner. The derivation of them is left to you as an exercise.

§2. Exponentiation. How do we assign a meaning to the expression ba? You can call this expression a power function if a is ﬁxed and b is allowed to vary. You can call it an exponential function if b is ﬁxed and a is allowed to vary. Anyway, in this expression, normally b is referred to as the base and a is referred to as the exponent.

Again, we proceed in small steps. First we consider the case when a is a positive integer. When a = 1, ba is simply b. For a = 2, 3, 4 etc., we have no diﬃculty to assign a meaning to ba: b2 = b b, b3 = b b b, b4 = b b b b b etc. Next, We consider the × × × × × × × case when a is a fraction, say a = n/m, where m and n are positive integers. In order to assign a meaning to ba in this case, we rewrite it as

n ba = bn/m = b(1/m)n = b1/m . ¡ ¢ So it is enough to consider b1/m. What is b1/m? Well, it is the number, say c, satisfying cm = b (this identity is just (b1/m)m = b), right?

Wait a minute! There is something wrong! Take the special case m = 2. The identity c2 = b tells that b cannot be negative, because the square c2 of a real number c is never negative. So, from now on, we assume that the base b in ba is positive. Next, we have to worry about that there are more than one c satisfying cm = b. For example, when m = 2 and b = 4, both c = 2 and c = 2 satisfy c2 = b. Fortunately, for − positive b, there is exactly one positive c satisfying cm = b. After taking these precautions, now we can deﬁne b1/m for a given positive number b and positive integer m to be the unique positive number c satisfying cm = b. We remark that b1/2 is usually written as √b. More generally, we write √m b to stand for b1/m.

What is ba for any positive number a? Again, we approximate a by fractions m/n so that bm/n will give us approximate values of ba. Since the approximation here can be made as accurate as we wish, the meaning of ba is assigned.

14 According to the way ba is deﬁned for positive a, the following identities can be veriﬁed for positive numbers r, s:

br+s = brbs, (br)s = brs. (2.1)

(We skip the proofs here because they are rather boring.) Here, let me remind you once again our crucial assumption that b is positive. Why is this crucial? Well, otherwise, something really terrible may happen: these identities would turn into what I call “weapons of math destruction”. Indeed, allowing b to be 1, we would get − 1=( 1)1 =( 1)2(1/2) = (( 1)2)1/2 = ( 1)( 1) = √1 = 1. Horrible! − − − − − − p OK, from now on, keep in mind that the base b of an exponential expression ba is always taken to be a positive number, if we allow a to be any real number.

What is ba for a equal to 0, or for negative a? The answer is not obvious. Actually it depends on what we want. What we want is, identities (2.1) continue to hold for all real numbers r, s, not necessarily positive. So, assuming (2.1) for all real numbers r, s, we determine what ba is for negative a, and what b0 is. For negative a, we can write a = p ( 1)p. According to the second identity in (2.1), we have − ≡ − a ( 1)p 1 p 1 b = b − = (b− ) . So it is enough to determine b− . According to the first identity in 1 1 1 ( 1)+1 0 0 (2.1), we have b− b = b− b = b − = b . So it boils down to the determination of b . According to the first identity in (2.1) again, we have b0 = b0+0 = b0b0 = (b0)2. This shows that b0 satisfies the equation x = x2. Now x = x2 has exactly two solutions, namely x = 0, 1. So b0 is either 0 or 1. But b0 cannot be 0. Why? Because b0 = 0 would give b = b1+0 = b1b0 = b 0 = 0, contradicting our assumption that b is a positive number. × We conclude: b0 = 1.

1 0 1 Let us go back to b− b = b . Now this can be rewritten as b− b = 1. Dividing both sides of the last identity by b, we ﬁnally get

1 1 b− = . b Here we make a short conclusion: the expression ba is well–deﬁned for all positive numbers b and all real numbers a (including negative numbers).

3. The exponential function. Now we keep the base b ﬁxed and let the exponent a vary. It is more appropriate to use the letter x for a to turn ba into bx. As a function of

15 x, the expression bx is called the exponential function with base b. For b = 2, we get the function 2x. At some particular points, say x = 1, 1/2, 2/3, 0, 1, 3/2, the values of 2x − − are given as follows:

1 1/2 2/3 2 1/3 3 0 1 1 3/2 1 1 2 = 2, 2 = √2, 2 = (2 ) = √4, 2 = 1, 2− = , 2− = = . 2 23/2 2√2 This seems to be fine and natural enough. But 2x is not called the natural exponential function. The natural exponential function ex is the one with a strange number denoted by e as its base, which can be defined to be the sum of an infinite series as follows: 1 1 1 1 e =1+ + + + + + . (3.1) 1! 2! 3! ··· n! ··· Here, n! (read as n factorial) is the product of all integers between (and including) 1 and n. For example 4! = 1 2 3 4 = 24. In general, × × × n! = 1 2 3 (n 1) n. × × × · · · × − × In fact, (3.1) is the special case x = 1 of the following identity

x2 x3 xn ex =1+ x + + + + + (3.2) 2! 3! ··· n! ··· (Here we have no intention of explaining where this identity comes from.) Why do we use this strange number e and call it the natural choice of the base for an exponential function? The answer is given in another math subject called calculus: the derivative of the function bx is the neatest when b = e. Indeed, any calculus textbook tells us that d ex = ex. dx (This identity is one of the most beautiful things in math.) It can be seen as follows. (To understand this line of argument, we assume that you have some experience with diﬀerentiation. If not, skip this line.) Diﬀerentiate the series in (3.2) term by term:

d d d d x2 d x3 d x4 ex = 1+ x + + + + dx dx dx dx 2! dx 3! dx 4! ··· 2x 3x2 4x3 x2 x3 =0+1+ + + + =1+ x + + + = ex. 2! 3! 4! ··· 2! 3! ··· (Here, the procedure of term by term diﬀerentiation is taken for granted. But strictly speaking, it needs mathematical justiﬁcation.)

16 There is another reason why the base e is used: amazingly, ex can be linked up to the trigonometric functions cos x and sin x, if we use complex numbers! The ﬁrst step to see this is by observing that the series on the right hand side of (3.2) still makes sense if we allow x to be any complex number. We may deﬁne ex for any complex number x to be the right hand side of (3.2). In particular, we may take x = it, where t is any real number. A straightforward manipulation of complex numbers shows that, when x = it, (3.2) can be rewritten as

t2 t4 t6 t3 t5 t7 eit = 1 + + + i t + + . − 2! 4! − 6! ··· − 3! 5! − 7! ··· µ ¶ µ ¶ The Taylor series for the cosine and sine functions (you will learn this if you are going to take a second year calculus course) give

t2 t4 t6 t3 t5 t7 cos t = 1 + + sin t = t + + + . − 2! 4! − 6! ··· − 3! 5! 7! ··· So we have arrived at eit = cos t + i sin t.

This is called the Euler formula. This is another one of the most beautiful things in mathematics! (Euler is pronounced as ‘oiler’.)

4. Logarithms. What is the meaning of the expression log c, where both b and c § b are positive numbers? Here, we simply call logb c the logarithm of c. The number b is called the base of the logarithm.

a The answer is: logb c is the number a satisfying b = c. In other words, a = logb c is just another way to write ba = c. We give a few numerical examples to illustrate this c c 4 point. What is log3 81? Well, if we write log3 81 = c, then we have 3 = 81, or 3 = 3 . 1 1 c 1 So c = 4. Thus log3 81 = 4. Next, what is log2 4 ? Letting log2 4 = c, we have 2 = 4 , or c 2 1 2 = 2− . Hence log = c = 2. To recapitulate, we put 2 4 − log c = a ba = c. (4.1) b ⇐⇒ The equivalence of the above identities gives rise to

a logb c logb b = a, b = c. (4.2)

In the expression logb c, we often keep the base b ﬁxed and allow c to vary. In that situation we prefer use the letter x instead of c in logb c to get the logarithm function

17 logb x. Concerning the logarithm function, the following properties are basic

log (cd) = log c + log d, log (c/d) = log c log d log cr = r log c (4.3) b b b b b − b b b Here, b, c, d are positive numbers and r is any real number. These identities are not too r a r hard to verify. For example, to verify the last identity, we write a = logb c . Then b = c . Replace c in the last identity by blogb c, we get

ba =(blogb c)r = br logb c

a r log c r and hence a = logb(b ) = logb b = r log c. So logb c = r logb c.

a 0 What is logb 1? Write logb 1= a. Then b = 1. Since we also have b = 1, a = 0. So

logb 1 = 0.

What is log6 3 + log6 18? A calculator can tell you that it is approximately 2. In fact, it is exactly 2. Indeed, according the ﬁrst identity in (4.3), we have

log 3 + log 12 = log (3 12) = log 36 = log 62 = 2. 6 6 6 × 6 6 This is more convincing than a calculator.

When the base b is e, logb c is often rewritten as ln c, that is

ln c = loge c.

In that case, (4.1) and (4.2) can be rewritten as

ln c = a ea = c, ln ea = a, eln c = c. (4.4) ⇔ As you may know, ln c is called the natural logarithm of c.

Question: are the numbers 2ln 3 and 3ln 2 the same? They look diﬀerent. But if you use a calculator to check their numerical values, you would believe that they are the same, or at least they are approximately the same. How do you know they are actually the same? Well, we use the following simple trick: convert them into exponentials with the same base e. According to the last identity of (4.3), we have 2 = eln 2 and 3 = eln 3. So

2ln 3 =(eln 2)ln 3 = e(ln 2)(ln 3) = e(ln 3)(ln 2) =(eln 3)ln 2 = 3ln 2.

18 Hence 2ln 3 and 3ln 2 are indeed equal. The neat answer here tells us that sometimes we need to change bases in order to answer a question concerning exponential or logarithm expressions. The following “change of bases” formula is handy for such a purpose: ln a ba =(eln b)a = ea ln b, log a = . b ln b We leave the veriﬁcation of the second identity to you as an exercise.

For natural logarithms, (4.3) becomes

ln(cd)=ln c + ln d, ln(c/d)=ln c ln d ln cr = r ln c (4.5) − The derivative of the natural logarithm function is very neat: d 1 ln x = . dx x

How do we ﬁnd the derivative of log10 x? Answer: use the change of bases formula to rewrite log10 x as ln x/ ln 10. Thus, d d ln x 1 d 1 1 1 log x = == ln x = = . dx 10 dx ln 10 ln 10 dx ln 10 x (ln 10)x

We give a list of identities mentioned in this article for you to keep in mind

(a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2b + 3ab2 + b3

(a b)(a + b)= a2 b2 (a b)(a2 + ab + b2)= a3 b3, − − − −

0 r s r+s 1/2 1 For b> 0: b = 1 b b = b b = √b b− = 1/b d e0 = 1 er+s = eres e 1 = 1/e ex = ex − dx

r For b, c, d > 0: logb(cd) = logb c + logb d logb c = r logb c log 1 = 0 d 1 For c, d> 0: ln(cd)=ln c + ln d ln cr = r ln c ln 1 = 0 ln x = dx x ln a ba =(eln b)a = ea ln b log a = b ln b ln1=0 ln e = 1

19 EXERCISES (answers posted on the next page)

1. Without using a calculator, ﬁnd the value of each of the following expressions. 7! 12 12! 16 75 1000! (a) 5! (b) (c) , or (d) (e) (f) 4! 10 10! 2! 3 4 998! µ ¶ µ ¶ µ ¶ 2. Expand each of the following.

(a) (x + 3y)2 (b) (x y)2 (c) (2x 5y)2 − − (d) (x + 2y)3 (e) (x y)3 (f) ( 1 x y2)3 − 2 −

3. Express each of the following as 2 to a power.

3 √8 3 (a) √2 (b) 1 (c) 8 (d) √8 (e) √4 (f) (g) 4√2 (h) 4√2 2 √3 16

4. Write each of the following as a single logarithm.

(a) log 4 + log 5 (b) log 15 log 3 (c) log 7 log 14 + log 4 3 3 2 − 2 3 − 3 3 1 1 (d) log 2 + 2 log 3 (e) 3 ln 2 ln4 (f) ln 5 5 − −2 9

5 Find the value of each expression below.

2 1 √ 4 ln 2 4 3 ln 8 (a) log2 16 (b) log3 9 (c) log5 5 (d) e (e) 3ln(e ) (f) e

(g) log16 2 (h) log6 2 + log6 3 (i) (loga b)(logb a)

(Here a and b are arbitrary positive numbers.)

20 Answers to Exercises in Appendix

7! 1 2 7 1. (a) 5! = 1 2 3 4 5=120 (b) = × × · · · × = 5 6 7 = 210 × × × × 4! 1 4 × × × · · · × 12 12! 12 11 16 16 15 14 (c) = = × =66 (d) = × × = 560 10 10! 2! 2 3 1 2 3 µ ¶ µ ¶ × × 75 75 74 73 72 1000! (e) = × × × = 1215450 (f) = 1000 999 = 999000. 4 1 2 3 4 998! × µ ¶ × × × 2. (a) (x + 3y)2 = x2 + 6xy + 9y2 (b) (x y)2 = x2 2xy + y2 − − (c) (2x 5y)2 = 4x2 20xy + 25y2 (d) (x + 2y)3 = x3 + 6x2y + 12xy2 + 8y3 − − (e) (x y)3 = x3 3x2y + 3xy2 y3 (f) ( 1 x y2)3 = 1 x3 3 x2y2 + 3 xy4 y6 − − − 2 − 8 − 4 2 −

√ 1/2 1 1 3 √ 3/2 √3 2/3 3. (a) 2 = 2 (b) 2 = 2− (c) 8 = 2 (d) 8 = 2 (e) 4 = 2 3/2 √8 2 3 (f) = = 21/6 (g) 4√2 = 25/2 (h) 4√2 = 2221/3 = 27/3 √3 16 24/3

4. (a) log 4 + log 5 = log 20 (b) log 15 log 3 = log 5 3 3 3 2 − 2 2 (c) log 7 log 14 + log 4 = log 2 (d) log 2 + 2 log 3 = log 18 3 − 3 3 3 5 5 5 1 1 1 2 (e) 3 ln 2 ln4=ln2 (f) ln = ln 3− = ln 3 − −2 9 −2

5 (a) log 16 = log 24 = 4 (b) log 1 = log 9= 2 (c) log √5 = log 51/2 = 1/2 2 2 3 9 − 3 − 5 5

2 (d) e4 ln 2 =(eln 2)4 = 24 =16 (e) 3ln(e4)=12 (f) e 3 ln 8 =(eln 8)2/3 = 82/3 = 4

(g) log 2 = log 161/4 = 1/4 Or, log 2= a 16a = 2 a = 1/4. 16 16 16 ⇒ ⇒ (h) log 2 + log 3 = log 2 3 = log 6 = 1 6 6 6 × 6 ln b ln a (i) (log b)(log a)= = 1. a b ln a ln b