Download Report

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 become 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 communication 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 Vall´ee 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
√
−1 for i so that we can reserve the
correct understanding is used, such as writing
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 straightfor1
ward manner, keeping in mind that i2 = −1. Indeed, given complex numbers c1 = a1 +b1 i
and c2 = a2 + b2 i, the sum c1 + c1 and the product c1 c2 are given by
c1 + c2 = (a1 + b1 i) + (a2 + b2 i) = (a1 + a1 ) + (b1 + b2 )i
c1 c2 = (a1 + b1 i)(a2 + b2 i) = a1 a2 + b1 b2 i2 + a1 b2 i + a2 b1 i
= (a1 a2 − b1 b2 ) + (a1 b2 + a2 b1 )i
To give a quick example, letting c1 = 1 − 2i, c2 = 2 + 3i, we have c1 + c2 = 3 + i,
c1 − c2 = −1 − 5i and c1 c2 = 8 + i.
Exercise 8.2. In each of the following cases, find c1 + c2 , c1 − c2 and c1 c2 . Present
√
√
your answers as simple as possible. (a) c1 = i, c2 = 1−i; (b) c1 = 2+ 3i, c2 = 2− 3i;
(c) c1 = 1 + i, c2 − 1 − i.
Exercise 8.3. Find the product c1 c2 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 + b 2
(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 − b2 i2 = 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
3
(distributive Law)
(8.4)
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
|zw|2 = zw zw = zw z w = z z w w = |z|2 |w|2 ,
which is what we want to show.
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 + v 2 =
(92 + 102 )(102 + 112 )?
Exercise 8.7. Verify the following “parallelogram identity”
|z + w|2 + |z − w|2 = 2|z|2 + 2|w|2
for arbitrary complex numbers z and w.
An important use of complex conjugation is to divide complex numbers. Given complex numbers z and w with z 6= 0, to divide w by z, we multiply both the denominator
and the numerator of w/z by z, the complex conjugate of the denominator, to obtain
w
wz
wz
=
= 2
z
zz
|z|
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 conjugate of 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 + y 2 .
Exercise 8.8. Perform each of the following complex division:
(a)
8+i
2−i
(b)
5+i
3 − 2i
(c)
75 + 223i
63 + 17i
(d)
cos θ + i sin θ
.
cos θ − i sin θ
Example. We are asked to check that if |z| = 1 and if
w=
z−a
,
1 − az
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
ww =
z−a z−a
zz − za − az + aa
1 − za − az + aa
=
=
= 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 < a < 1, then
|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
Re z =
z+z
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
(8.7)
Exercise 8.11. Verify that, if |z| = 1 and z 6= 1, then
w=i
is a real number.
z+1
z−1
Take any complex number z = x + iy, with x as its real part and y as its imaginary
p
part. Write r for the modulus of z: r = |z| = x2 + y 2 . Let θ be the angle between
the vector representing z; (see the next figure). Then 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
z1 z2 = r1 r2 (cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 )
= r1 r2 ((cos θ1 cos θ2 − sin θ1 sin θ2 ) + i(cos θ1 sin θ2 + sin θ1 cos θ2 ))
= r1 r2 (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 ).
6
(8.11)
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)
becomes (r1 eiθ1 )(r2 eiθ2 ) = r1 r2 ei(θ1 +θ2 ) . By induction, we can prove
eiθ1 eiθ2 · · · eiθn = ei(θ1 +θ2 +···+θn ) .
When θ1 = θ2 = · · · = θn = θ, the above identity becomes
(eiθ )n = einθ .
(8.13)
This is called De Moivre’s formula.
Using Euler’s formula, we have
e2πi = 1,
eπi/3
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
= +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 ω n = 1. Indeed, by De Moivre’s formula,
ω n = (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:
1, ω, ω 2 , . . . , ω n−1
(8.14)
Indeed, take one of them, say ω k . Then (ω k )n = ω kn = (ω n )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 α3 y 3 = −1, or
y 3 = 1. The solutions to y 3 = 1 are y = 1, ω, ω 2 , where ω = e2πi/3 . So the solution
to x3 = −1 are x = α, αω, αω 2 .
Exercise 8.15. Express the solutions α, αω, αω 2 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
an − bn = (a − b)(an−1 + an−2 b + · · · + abn−2 + bn ).
Applying this identity for a = 1 and b = ω, we have
0 = 1 − ω n = (1 − ω)(1 + ω + ω 2 + · · · + ω n−1 ).
Clearly 1 − ω 6= 0. Therefore
1 + ω + ω 2 + · · · + ω n−1 = 0.
(8.15)
(Aside: This identity is basic to finite Fourier transform, an area applied to signal
transmission and processing.) The terms 1, ω, ω 2 , . . . , ω n−1 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 + ω 2 + · · · + ω n = 0?
8
Finally, we consider the exponential ez for a complex number z = x + iy. Naturally
we put
ez = ex+iy = ex eiy = 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 | = |ex eiy | = |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:
cos α − cos β = −2 sin
α−β
α+β
sin
,
2
2
sin α − sin β = 2 cos
α+β
α−β
sin
.
2
2
We use a “complex trick” as follows.
(cos α − cos β) + i(sin α − i sin b) = eiα − eiβ = ei(α+β)/2+i(α−β)/2 − ei(α+β)/2−i(α−β)/2
¶µ
¶
µ
α−β
α+β
α+β
i(α+β)/2 i(α−β)/2
−i(α−β)/2
2i sin
+ i sin
=e
(e
−e
) = cos
2
2
2
= −2 sin
α+β
α−β
α+β
α−β
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 , ω 2 = e4πi/5 , ω 3 = e6πi/5 ,
ω 4 = e8πi/5 .
geometrically they are vertices of a regular pentagon. Notice that ω 3 is the complex
conjugate of ω 2 and ω 4 is the complex conjugate of ω. Hence (8) gives
1 + ω + ω + ω 2 + ω 2 = 0.
(8)′
Let a = ω + ω ≡ 2 cos 2π/5. Then a2 = (ω + ω)2 = ω 2 + ω 2 + 2ω ω = ω 2 + ω 2 + 2. Hence
(8)′ becomes 1 + a + a2 − 2 = 0, or a2 + 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 − v 2 + 2uvi = a + bi. Comparing with the real
and the imaginary parts of both sides, we obtain 2uv = b and u2 − v 2 = a. Pairing the
last identity with a2 + b2 = 1, we obtain 2u2 = 1 + a and hence
r
1+a
u=±
.
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
r
√
b
ε 1 − a2
1−a
v=
= ±p
= ±ε
.
2u
2
2(1 + a)
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 =
√
a = 3/2, b = 1/2 and ε = 1. Our formula for u and v gives
q
q
√
√
π
1
1
π
2+ 3
sin
2 − 3,
=a=
=b=
cos
12
2
12
2
√
3
2
+ 21 with
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
cos
=
12
2
s
√
√
(1 + 3)2
1+ 3
= √ .
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
Sn ≡ 1 + 2 (cos θ + cos 2θ + · · · + cos nθ)
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 = z 2 + z 2 etc. Thus we have
Sn = 1 + z + z + z 2 + z 2 + · · · + z n + z n
≡ z n + z n−1 + · · · + z + 1 + z + z 2 + · · · + z n−1 + z n .
Notice that zz = |z|2 = 1. So
zSn = z n−1 + z n−2 + · · · + z n + z n+1 .
Hence
zSn − Sn = z n+1 − z n .
Therefore, writing z 1/a for eiθ/a and z 1/a for e−iθ/a , we have
sin (n + 21 )θ
z n+1 − z n z 1/2
z n+1 − z n
1 z n+1/2 − z n+1/2
=
.
Sn =
=
=
z−1
z − 1 z 1/2
2 z 1/2 − z 1/2
sin 12 θ
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 )
and w = (w1 , w2 , . . . , wn ) in Cn , their dot product is defined to be
z • w = z1 w1 + z2 w2 + · · · + zn wn .
Notice that
z • z = z1 z 1 + z2 z 2 + · · · + zn z n
= |z1 |2 + |z2 |2 + · · · + |zn |2 ≥ 0.
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
ej = √ (1, ω j−1 , ω 2(j−1) , ω 3(j−1) , . . . , ω (n−1)(j−1) ),
n
1 ≤ 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:
µ ¶
1
m
b=m
b .
ab =
n
n
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 + 3a2 b + 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
2
c = 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 = br bs ,
(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 =
p
(−1)(−1) =
√
1 = 1. Horrible!
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
ba = b(−1)p = (b−1 )p . So it is enough to determine b−1 . According to the first identity in
(2.1), we have b−1 b = b−1 b1 = b(−1)+1 = b0 . So it boils down to the determination of b0 .
According to the first identity in (2.1) again, we have b0 = b0+0 = b0 b0 = (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 = b1 b0 = b × 0 = 0, contradicting our assumption that b is a positive number.
We conclude:
b0 = 1.
Let us go back to b−1 b = b0 . Now this can be rewritten as b−1 b = 1. Dividing both sides
of the last identity by b, we finally get
1
b−1 = .
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:
21 = 2, 21/2 =
√
2, 22/3 = (22 )1/3 =
√
3
1
1
1
4, 20 = 1, 2−1 = , 2−3/2 = 3/2 = √ .
2
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
+ + + ··· +
+ ···.
(3.1)
1! 2! 3!
n!
Here, n! (read as n factorial) is the product of all integers between (and including) 1 and
e=1+
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
ex = 1 + x +
x2 x3
xn
+
+ ··· +
+ ···
2!
3!
n!
(3.2)
(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 x
e = 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 x
d
d
d x2
d x3
d x4
e =
1+ x+
+
+
+ ···
dx
dx
dx
dx 2!
dx 3!
dx 4!
2x 3x2 4x3
x2 x3
+
+
+ ··· = 1 + x +
+
+ · · · = ex .
2!
3!
4!
2!
3!
(Here, the procedure of term by term differentiation is taken for granted. But strictly
=0+1+
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
¶
µ
¶
µ
t3 t5 t7
t2 t4 t6
it
e = 1 − + − + ··· + i t − + − + ··· .
2! 4! 6!
3! 5! 7!
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
cos t = 1 −
t2 t4 t6
+ − + ···
2! 4! 6!
sin t = t −
t3 t5 t7
+ + + ···.
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 logb c, where both b and c
are positive numbers? Here, we simply call logb c the logarithm of c. The number b is
called the base of the logarithm.
The answer is: logb c is the number a satisfying ba = c. In other words, a = logb c
is just another way to write ba = c. We give a few numerical examples to illustrate this
point. What is log3 81? Well, if we write log3 81 = c, then we have 3c = 81, or 3c = 34 .
So c = 4. Thus log3 81 = 4. Next, what is log2 14 ? Letting log2
2c = 2−2 . Hence log2
1
4
1
4
= c, we have 2c = 41 , or
= c = −2. To recapitulate, we put
logb c = a
⇐⇒
ba = c.
(4.1)
The equivalence of the above identities gives rise to
logb ba = a,
blogb c = 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
logb (cd) = logb c + logb d,
logb (c/d) = logb c − logb d
logb cr = r logb c
(4.3)
Here, b, c, d are positive numbers and r is any real number. These identities are not too
hard to verify. For example, to verify the last identity, we write a = logb cr . Then ba = cr .
Replace c in the last identity by blogb c , we get
ba = (blogb c )r = br logb c
and hence a = logb (ba ) = logb br log c = r log c. So logb cr = r logb c.
What is logb 1? Write logb 1 = a. Then ba = 1. Since we also have b0 = 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
log6 3 + log6 12 = log6 (3 × 12) = log6 36 = log6 62 = 2.
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:
ba = (eln b )a = ea ln b ,
logb a =
ln a
.
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 ln x
1
d
1 1
1
d
log10 x =
==
ln x =
=
.
dx
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 + 3a2 b + 3ab2 + b3
(a − b)(a + b) = a2 − b2
(a − b)(a2 + ab + b2 ) = a3 − b3 ,
For b > 0: b0 = 1
e0 = 1
br bs = br+s
er+s = er es
b1/2 =
√
For b, c, d > 0: logb (cd) = logb c + logb d
ba = (eln b )a = ea ln b
ln 1 = 0
logb a =
b−1 = 1/b
d x
e = ex
dx
e−1 = 1/e
For c, d > 0: ln(cd) = ln c + ln d
b
logb cr = r logb c
ln cr = r ln c ln 1 = 0
ln a
ln b
ln e = 1
19
log 1 = 0
d
1
ln x =
dx
x
EXERCISES (answers posted on the next page)
1. Without using a calculator, find the value of each of the following expressions.
µ ¶
µ ¶
µ ¶
12!
1000!
12
16
75
7!
(c)
, or
(d)
(e)
(f)
(a) 5! (b)
4!
10! 2!
998!
10
3
4
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) ( 12 x − y 2 )3
3. Express each of the following as 2 to a power.
(a)
√
2
(b)
1
2
(c) 8
(d)
√
8
(e)
√
3
4
√
8
(f) √
3
16
√
(g) 4 2
√
(h) 4 3 2
4. Write each of the following as a single logarithm.
(a) log3 4 + log3 5
(d) log5 2 + 2 log5 3
(b) log2 15 − log2 3
(e) 3 ln 2 − ln 4
(c) log3 7 − log3 14 + log3 4
(f) −
1
1
ln
2
9
5 Find the value of each expression below.
√
(a) log2 16 (b) log3 19 (c) log5 5 (d) e4 ln 2
(g) log16 2
(h) log6 2 + log6 3
(i) (loga b)(logb a)
(Here a and b are arbitrary positive numbers.)
20
(e) 3 ln(e4 )
2
(f) e 3 ln 8
Answers to Exercises in Appendix
1 × 2 × ··· × 7
7!
=
= 5 × 6 × 7 = 210
1. (a) 5! = 1 × 2 × 3 × 4 × 5 = 120 (b)
4!
1 × ··· × 4
µ ¶
µ ¶
12
12!
12 × 11
16
16 × 15 × 14
(c)
=
=
= 66 (d)
= 560
=
10
10! 2!
2
1×2×3
3
µ ¶
1000!
75
75 × 74 × 73 × 72
= 1215450 (f)
= 1000 × 999 = 999000.
=
(e)
1×2×3×4
998!
4
2. (a) (x + 3y)2 = x2 + 6xy + 9y 2
(b) (x − y)2 = x2 − 2xy + y 2
(c) (2x − 5y)2 = 4x2 − 20xy + 25y 2
(d) (x + 2y)3 = x3 + 6x2 y + 12xy 2 + 8y 3
(f) ( 21 x − y 2 )3 = 81 x3 − 43 x2 y 2 + 32 xy 4 − y 6
(e) (x − y)3 = x3 − 3x2 y + 3xy 2 − y 3
3. (a)
√
2 = 21/2
(b)
√
23/2
8
=
= 21/6
(f) √
3
4/3
2
16
1
2
= 2−1
(c) 8 = 23
√
(g) 4 2 = 25/2
4. (a) log3 4 + log3 5 = log3 20
(d)
√
8 = 23/2
(e)
√
3
4 = 22/3
√
(h) 4 3 2 = 22 21/3 = 27/3
(b) log2 15 − log2 3 = log2 5
(c) log3 7 − log3 14 + log3 4 = log3 2 (d) log5 2 + 2 log5 3 = log5 18
1
1
1
(e) 3 ln 2 − ln 4 = ln 2 (f) − ln = − ln 3−2 = ln 3
2
9
2
5 (a) log2 16 = log2 24 = 4 (b) log3
(d) e4 ln 2 = (eln 2 )4 = 24 = 16
(g) log16 2 = log16 161/4 = 1/4
1
9
= − log3 9 = −2 (c) log5
(e) 3 ln(e4 ) = 12
5 = log5 51/2 = 1/2
2
(f) e 3 ln 8 = (eln 8 )2/3 = 82/3 = 4
Or, log16 2 = a ⇒ 16a = 2 ⇒ a = 1/4.
(h) log6 2 + log6 3 = log6 2 × 3 = log6 6 = 1
(i) (loga b)(logb a) =
√
ln b ln a
= 1.
ln a ln b
21