CHAPTER 8. COMPLEX NUMBERS

Transcription

1 CHAPTER 8. COMPLEX NUMBERS Why do we need complex numbers? First of all, a simple algebraic equation like x = 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 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 i = 1. In doing so, x = 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 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

2 ward manner, keeping in mind that i = 1. Indeed, given complex numbers c 1 = a 1 +b 1 i and c = a + b i, the sum c 1 + c 1 and the product c 1 c are given by c 1 + c = (a 1 + b 1 i) + (a + b i) = (a 1 + a 1 ) + (b 1 + b )i c 1 c = (a 1 + b 1 i)(a + b i) = a 1 a + b 1 b i + a 1 b i + a b 1 i = (a 1 a b 1 b ) + (a 1 b + a b 1 )i To give a quick example, letting c 1 = 1 i, c = + 3i, we have c 1 + c = 3 + i, c 1 c = 1 5i and c 1 c = 8 + i. Exercise 8.. In each of the following cases, find c 1 + c, c 1 c and c 1 c. Present your answers as simple as possible. (a) c 1 = i, c = 1 i; (b) c 1 = + 3i, c = 3i; (c) c 1 = 1 + i, c 1 i. Exercise 8.3. Find the product c 1 c of c 1 = cos α 1 +i sinα 1 and c = cos α +i sinα. 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 = a + b : a + bi = a + b (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.

3 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.) For examples, + 3i = 3i, i = i, =, 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) = a b i = a + b = z. So z = zz. (8.3) This identity, which resembles v = 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

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 = z w and hence it suffices to verify the latter. By (8.3), we have which is what we want to show. Exercise 8.6. Use (8.6) to deduce zw = zw zw = zw z w = z z w w = z w, (a + b )(c + d ) = (ac + bd) + (ad bc), where a, b, c, d are arbitrary real numbers. Which integers u and v satisfy u +v = ( )( )? Exercise 8.7. Verify the following parallelogram identity for arbitrary complex numbers z and w. z + w + z w = z + w 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 and the numerator of w/z by z, the complex conjugate of the denominator, to obtain w z = wz zz = wz z 4

5 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. 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 i) (1 + i)(1 i) = 1 + i i = i. In division of complex numbers, it is important to recognize the complex conjugate z of z = x + yi and to use zz = z = x + y. Exercise 8.8. Perform each of the following complex division: (a) 8 + i i (b) 5 + i 3 i (c) i i (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 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 = 1, which is the same as ww = 1, (in view of (8.3) above). This leads us to compute w w. Now w w = z a 1 a z z a 1 az in view of zz = z = 1. Done. = zz za az + aa 1 az az + az az = 1 za az + aa 1 az za + aa = 1, Problem 8.9. Verify that, if a is a real number with 0 < a < 1, then for all complex numbers z with z = 1. z a z a 1 = a The real part of a complex number z, denoted by Re z, is given by Re z = z + z (8.7) A complex number z is real if and only if z = z. Exercise Check the last statement as well as identity (8.6). 5

6 Exercise Verify that, if z = 1 and z 1, then is a real number. w = i z + 1 z 1 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 = x + y. 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 cosθ + ir sin θ, or The last expression is called the polar form of z. z = r(cos θ + i sin θ). (8.8) Multiplication of complex numbers is especially revealing when expressed in polar form. Let z 1 = r 1 (cos θ 1 + i sin θ 1 ), z = r (cos θ + i sin θ ) be complex numbers in polar form. Then z 1 z = r 1 r (cos θ 1 + i sin θ 1 )(cosθ + i sin θ ) = r 1 r ((cos θ 1 cos θ sin θ 1 sin θ ) + i(cosθ 1 sin θ + sinθ 1 cosθ )) = r 1 r (cos(θ 1 + θ ) + i sin(θ 1 + θ )). (8.9) Here we have used the addition formula for sine and cosine in trigonometry cos(α + β) = cosαcosβ sin α sin β, sin(α + β) = sinαcos β + cos α sin β. (8.10) We single out the most crucial part of (8.9) (cos θ 1 + i sin θ 1 )(cos θ + i sin θ ) = cos(θ 1 + θ ) + i sin(θ 1 + θ ). (8.11) 6

7 Let us put e iθ = cos θ + i sin θ (8.1) which is called Euler s formula. Then (8.11) becomes e iθ 1 e iθ = e i(θ 1+θ ) and (8.10) becomes (r 1 e iθ 1 )(r e iθ ) = r 1 r e i(θ 1+θ ). By induction, we can prove e iθ 1 e iθ e iθn = e i(θ 1+θ + +θ n). When θ 1 = θ = = θ n = θ, the above identity becomes This is called De Moivre s formula. Using Euler s formula, we have (e iθ ) n = e inθ. (8.13) e πi = 1, e πi = 1, e πi/ = i, e 3πi/ = i, e πi/4 = 1 (1 + i), e 3πi/4 = 1 ( 1 + i), e πi/3 = 1 + i 3, eπi/6 = 3 + i, eπi/3 = i. Exercise 8.1. Check the above identities. Question A logo for T-shirt is I am number e πi. What does that mean? Fix a positive integer n and consider the complex number ω = e πi/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 = (e πi/n ) n = e πi = 1. Furthermore, we can check that the following list of n complex numbers are solutions to the algebraic equation x n = 1: 1, ω, ω,..., ω n 1 (8.14) Indeed, take one of them, say ω k. Then (ω k ) n = ω kn = (ω n ) k = 1 k = 1. It turns out that the above list gives n distinct roots of the polynomial x n 1. 7

8 Question Why are the numbers in the list (8.14) distinct to each other? Example. We are asked to find all solutions to x 3 = 1. First, we express 1 in polar form: 1 = e πi. Then α e πi/3 is a solution to x 3 = 1. Indeed, we can check that α 3 = e πi = 1. Let us write x = αy. Then x 3 = 1 becomes α 3 y 3 = 1, or y 3 = 1. The solutions to y 3 = 1 are y = 1, ω, ω, where ω = e πi/3. So the solution to x 3 = 1 are x = α, αω, αω. Exercise Express the solutions α, αω, αω a + bi. to x 3 = 1 in Cartesian form Exercise Solve each of the following equations: (a) x 4 = 1 (b) x 3 = i. Recall the algebraic identity a n b n = (a b)(a n 1 + a n b + + ab n + b n ). Applying this identity for a = 1 and b = ω, we have 0 = 1 ω n = (1 ω)(1 + ω + ω + + ω n 1 ). Clearly 1 ω 0. Therefore 1 + ω + ω + + 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,..., 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 = e πi/6 = e πi/3 Question 17. Can we get a new identity out of (8) by multiplying (8) by ω to obtain ω + ω + + ω n = 0? 8

9 Finally, we consider the exponential e z for a complex number z = x + iy. Naturally we put e z = e x+iy = e x e iy = e x (cos y + i sin y) For the exponential e x of a real number x, see the Appendix to this chapter. One basic property of e x is that it is always a positive number. So e z = e x e iy = e x e iy = e x. Since x = Re z, the real part of z, we obtain e z = e Re z. (8.16) It is easy to check that, for all complex numbers z 1 and z, e z 1+z = e z 1 e z. (8.17) Exercise 8.18 Check the last identity. Question Are e z and e z 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 cosβ = sin α + β sin α sin β = cos α + β We use a complex trick as follows. sin α β, sin α β. (cos α cos β) + i(sin α i sin b) = e iα e iβ = e i(α+β)/+i(α β)/ e i(α+β)/ i(α β)/ = e i(α+β)/ (e i(α β)/ e i(α β)/ ) = = sin α + β sin α β ( cos α + β 9 + i cos α + β + i sin α + β )( i sin α β ) sin α β.

10 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 π/5. Naturally we consider the complex number ω = e πi/5. Then the fifth roots of unity are 1, ω = e πi/5, ω = e 4πi/5, ω 3 = e 6πi/5, ω 4 = e 8πi/5. geometrically they are vertices of a regular pentagon. Notice that ω 3 conjugate of ω and ω 4 is the complex conjugate of ω. Hence (8) gives is the complex 1 + ω + ω + ω + ω = 0. (8) Let a = ω +ω cos π/5. Then a = (ω +ω) = ω +ω +ω ω = ω +ω +. Hence (8) becomes 1 + a + a = 0, or a + a 1 = 0. Since cos π/5 is positive,we only take the positive solution to a + a 1 = 0, which is a = ( 5 1)/. Thus cos π 5 = a 5 1 =. 4 Question 0. Use a calculator to find numerical values of cos 7 o and ( 5 1)/4. Are they the same? Example. We are asked to give an explicity solution to the equation x = a + ib, where a + ib is a unit modulus number, that is a + b a + ib = 1. Let us write x = u + iv. Then x = a + bi becomes (u + iv) = u v + uvi = a + bi. Comparing with the real and the imaginary parts of both sides, we obtain uv = b and u v = a. Pairing the last identity with a + b = 1, we obtain u = 1 + a and hence 1 + a u = ±. From uv = b we get v = b/u. Let ε = 1 if b 0 and ε = 1 if b < 0. Then b = ε b = ε b = ε 1 a. So v = b u = ± ε 1 a 1 a = ±ε (1 + a). 10

11 Now we use this answer to find the values of cos 15 o cos π/1 and sin 15 o sin π/1. Let a = cos π/1, b = sin π/1 and x = a+bi e πi/1. Then x = e πi/6 = with a = 3/, b = 1/ and ε = 1. Our formula for u and v gives cos π 1 = a = sin π 1 = b = 1 3, which are our answers. [Actually, there is a slick way to simplify our answer. Note that So + 3 = (4 + 3)/ = ( )/ = (1 + 3) /. cos π 1 = 1 (1 + 3) = We can simplify our answer for sin π/1 in the same way.] Exercise 1. Find cos π/8 and sin π/8. Problem. Prove: cos π/7 is a root of the cubic x 3 + x x 1. Example. In the theory of Fourier series, we have to find a closed form of the sum S n 1 + (cosθ + cos θ + + cos nθ) for an arbitrary positive integer n. This requires some good skill to accomplish. Let z = e iθ. Then cos θ = z + z, cos θ = z + z etc. Thus we have Notice that zz = z = 1. So S n = 1 + z + z + z + z + + z n + z n z n + z n z z + z + + z n 1 + z n. zs n = z n 1 + z n + + z n + z n+1. Hence zs n S n = z n+1 z n. Therefore, writing z 1/a for e iθ/a and z 1/a for e iθ/a, we have S n = zn+1 z n z 1 = zn+1 z n z 1 z 1/ z = 1 z n+1/ z n+1/ = sin (n + 1)θ 1/ z 1/ z 1/ sin 1θ. 11

12 This answer is very hard to find without using complex numbers. In Chapter 5, we have studied the inner product for the space R n, consisting of n tuples of complex numbers. We may replace real numbers by complex numbers to obtain a complex space, denoted by C n of n tuples of complex numbers: z = (z 1, z,..., z n ). 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 = (z 1, z,..., z n ) and w = (w 1, w,..., w n ) in C n, their dot product is defined to be Notice that z w = z 1 w 1 + z w + + z n w n. z z = z 1 z 1 + z z + + z n z n = z 1 + z + + z n 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 C n is said to be an orthonormal basis of C n if they are mutually perpendicular unit vectors. Problem 3. As before, write ω = e πi/n. Prove that e j = 1 n (1, ω j 1, ω (j 1), ω 3(j 1),..., ω (n 1)(j 1) ), 1 j n, form an orthonormal basis of C n. (The matrix with these basis vectors as columns defines the finite Fourier transform.) 1

13 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 =, 3, 4 etc., ab is understood in the following way: b = 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: ab = m ( ) 1 n b = m n b. 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/)b is the number satisfying (1/)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, 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) = a + ab + b can be derived as follows: (a + b) = (a + b)(a + b) = a(a + b) + b(a + b) = aa + ab + ba + bb = a + ab + ab + b = a + ab + b. Another important identity is (a b)(a + b) = a b. It is obtained as follows: (a b)(a + b) = a(a + b) + ( b)(a + b) = aa + ab + ( b)a + ( b)b 13

14 More complicated identities such as = a + ab ba b = a b. (a + b) 3 = a 3 + 3a b + 3ab + b 3 and (a b)(a + ab + b ) = a 3 b 3 can be derived in the same manner. The derivation of them is left to you as an exercise.. Exponentiation. How do we assign a meaning to the expression b a? 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, b a is simply b. For a =, 3, 4 etc., we have no difficulty to assign a meaning to b a : b = b b, b 3 = b b b, b 4 = 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 b a in this case, we rewrite it as b a = b n/m = b (1/m)n = ( b 1/m) n. So it is enough to consider b 1/m. What is b 1/m? Well, it is the number, say c, satisfying c m = b (this identity is just (b 1/m ) m = b), right? Wait a minute! There is something wrong! Take the special case m =. The identity c = b tells that b cannot be negative, because the square c of a real number c is never negative. So, from now on, we assume that the base b in b a is positive. Next, we have to worry about that there are more than one c satisfying c m = b. For example, when m = and b = 4, both c = and c = satisfy c = b. Fortunately, for positive b, there is exactly one positive c satisfying c m = b. After taking these precautions, now we can define b 1/m for a given positive number b and positive integer m to be the unique positive number c satisfying c m = b. We remark that b 1/ is usually written as b. More generally, we write m b to stand for b 1/m. What is b a for any positive number a? Again, we approximate a by fractions m/n so that b m/n will give us approximate values of b a. Since the approximation here can be made as accurate as we wish, the meaning of b a is assigned. 14

15 According to the way b a is defined for positive a, the following identities can be verified for positive numbers r, s: b r+s = b r b s, (b r ) s = b rs. (.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) (1/) = (( 1) ) 1/ = ( 1)( 1) = 1 = 1. Horrible! OK, from now on, keep in mind that the base b of an exponential expression b a is always taken to be a positive number, if we allow a to be any real number. What is b a 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 (.1) continue to hold for all real numbers r, s, not necessarily positive. So, assuming (.1) for all real numbers r, s, we determine what b a is for negative a, and what b 0 is. For negative a, we can write a = p ( 1)p. According to the second identity in (.1), we have b a = b ( 1)p = (b 1 ) p. So it is enough to determine b 1. According to the first identity in (.1), we have b 1 b = b 1 b 1 = b ( 1)+1 = b 0. So it boils down to the determination of b 0. According to the first identity in (.1) again, we have b 0 = b 0+0 = b 0 b 0 = (b 0 ). This shows that b 0 satisfies the equation x = x. Now x = x has exactly two solutions, namely x = 0, 1. So b 0 is either 0 or 1. But b 0 cannot be 0. Why? Because b 0 = 0 would give b = b 1+0 = b 1 b 0 = b 0 = 0, contradicting our assumption that b is a positive number. We conclude: b 0 = 1. Let us go back to b 1 b = b 0. Now this can be rewritten as b 1 b = 1. Dividing both sides of the last identity by b, we finally get b 1 = 1 b. Here we make a short conclusion: the expression b a 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 b a into b x. As a function of 15

16 x, the expression b x is called the exponential function with base b. For b =, we get the function x. At some particular points, say x = 1, 1/, /3, 0, 1, 3/, the values of x are given as follows: 1 =, 1/ =, /3 = ( ) 1/3 = 3 4, 0 = 1, 1 = 1, 3/ = 1 3/ = 1. This seems to be fine and natural enough. But x is not called the natural exponential function. The natural exponential function e x 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: e = ! + 1! + 1 3! (3.1) n! Here, n! (read as n factorial) is the product of all integers between (and including) 1 and n. For example 4! = = 4. In general, n! = 1 3 (n 1) n. In fact, (3.1) is the special case x = 1 of the following identity e x = 1 + x + x! + x3 3! + + xn n! + (3.) (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 b x is the neatest when b = e. Indeed, any calculus textbook tells us that d dx ex = e x. (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.) term by term: d dx ex = d dx 1 + d dx x + d x dx! + d x 3 dx 3! + d x 4 dx 4! + = x! + 3x + 4x3 + = 1 + x + x 3! 4!! + x3 3! + = ex. (Here, the procedure of term by term differentiation is taken for granted. But strictly speaking, it needs mathematical justification.) 16

17 There is another reason why the base e is used: amazingly, e x 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.) still makes sense if we allow x to be any complex number. We may define e x for any complex number x to be the right hand side of (3.). 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.) can be rewritten as ) e it = (1 t! + t4 4! t6 6! + ) + i (t t3 3! + t5 5! t7 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 So we have arrived at cos t = 1 t! + t4 4! t6 t3 + sin t = t 6! 3! + t5 5! + t7 7! +. e it = 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 b c, where both b and c are positive numbers? Here, we simply call log b c the logarithm of c. The number b is called the base of the logarithm. The answer is: log b c is the number a satisfying b a = c. In other words, a = log b c is just another way to write b a = c. We give a few numerical examples to illustrate this point. What is log 3 81? Well, if we write log 3 81 = c, then we have 3 c = 81, or 3 c = So c = 4. Thus log 3 81 = 4. Next, what is log? Letting log 4 1 = c, we have 4 c = 1, or 4 c = 1. Hence log = c =. To recapitulate, we put 4 The equivalence of the above identities gives rise to log b c = a b a = c. (4.1) log b b a = a, b log b c = c. (4.) In the expression log b 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 log b c to get the logarithm function 17

18 log b x. Concerning the logarithm function, the following properties are basic log b (cd) = log b c + log b d, log b (c/d) = log b c log b d log b c r = r log b 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 = log b c r. Then b a = c r. Replace c in the last identity by b log b c, we get b a = (b log b c ) r = b r log b c and hence a = log b (b a ) = log b b r log c = r log c. So log b c r = r log b c. What is log b 1? Write log b 1 = a. Then b a = 1. Since we also have b 0 = 1, a = 0. So log b 1 = 0. What is log log 6 18? A calculator can tell you that it is approximately. In fact, it is exactly. Indeed, according the first identity in (4.3), we have log log 6 1 = log 6 (3 1) = log 6 36 = log 6 6 =. This is more convincing than a calculator. When the base b is e, log b c is often rewritten as ln c, that is ln c = log e c. In that case, (4.1) and (4.) can be rewritten as ln c = a e a = c, ln e a = a, e ln c = c. (4.4) As you may know, lnc is called the natural logarithm of c. Question: are the numbers ln 3 and 3 ln 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 = e ln and 3 = e ln 3. So ln 3 = (e ln ) ln3 = e (ln)(ln 3) = e (ln 3)(ln ) = (e ln 3 ) ln = 3 ln. 18

19 Hence ln 3 and 3 ln 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: b a = (e ln b ) a = e a ln b, log b 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) = lnc + ln d, ln(c/d) = lnc ln d ln c r = r ln c (4.5) The derivative of the natural logarithm function is very neat: d dx ln x = 1 x. How do we find the derivative of log 10 x? Answer: use the change of bases formula to rewrite log 10 x as lnx/ ln 10. Thus, d dx log 10 x = d dx ln x ln 10 == 1 ln 10 d dx ln x = 1 ln 10 1 x = 1 (ln 10)x. We give a list of identities mentioned in this article for you to keep in mind (a + b) = a + ab + b (a + b) 3 = a 3 + 3a b + 3ab + b 3 (a b)(a + b) = a b (a b)(a + ab + b ) = a 3 b 3, For b > 0: b 0 = 1 b r b s = b r+s b 1/ = b b 1 = 1/b e 0 = 1 e r+s = e r e s e 1 = 1/e d dx ex = e x For b, c, d > 0: log b (cd) = log b c + log b d log b c r = r log b c log 1 = 0 For c, d > 0: ln(cd) = lnc + lnd ln c r = r ln c ln 1 = 0 b a = (e ln b ) a = e aln b ln 1 = 0 lne = 1 log b a = ln a ln b d dx ln x = 1 x 19

20 EXERCISES (answers posted on the next page) 1. Without using a calculator, find the value of each of the following expressions. (a) 5! (b) 7! ( ) ( ) ( ) 1 1! (c), or (d) (e) (f) 1000! 4! 10 10!! !. Expand each of the following. (a) (x + 3y) (b) (x y) (c) (x 5y) (d) (x + y) 3 (e) (x y) 3 (f) ( 1 x y ) 3 3. Express each of the following as to a power. (a) (b) 1 (c) 8 (d) 8 (e) 3 4 (f) (g) 4 (h) Write each of the following as a single logarithm. (a) log log 3 5 (b) log 15 log 3 (c) log 3 7 log log 3 4 (d) log 5 + log 5 3 (e) 3 ln ln 4 (f) 1 ln Find the value of each expression below. (a) log 16 (b) log (c) log 5 5 (d) e 4ln (e) 3 ln(e 4 ) (f) e 3 ln 8 (g) log 16 (h) log 6 + log 6 3 (i) (log a b)(log b a) (Here a and b are arbitrary positive numbers.) 0

21 Answers to Exercises in Appendix 1. (a) 5! = = 10 (b) 7! 4! = 1 7 = = ( ) 1 (c) = 1! ( ) = = 66 (d) = = !! ( ) (e) = = (f) 1000! = = !. (a) (x + 3y) = x + 6xy + 9y (b) (x y) = x xy + y (c) (x 5y) = 4x 0xy + 5y (d) (x + y) 3 = x 3 + 6x y + 1xy + 8y 3 (e) (x y) 3 = x 3 3x y + 3xy y 3 (f) ( 1 x y ) 3 = 1 8 x3 3 4 x y + 3 xy4 y 6 3. (a) = 1/ (b) 1 = 1 (c) 8 = 3 (d) 8 = 3/ (e) 3 4 = /3 (f) = 3/ 4/3 = 1/6 (g) 4 = 5/ (h) 4 3 = 1/3 = 7/3 4. (a) log log 3 5 = log 3 0 (b) log 15 log 3 = log 5 (c) log 3 7 log log 3 4 = log 3 (d) log 5 + log 5 3 = log 5 18 (e) 3 ln ln 4 = ln (f) 1 ln 1 9 = 1 ln 3 = ln 3 5 (a) log 16 = log 4 = 4 (b) log = log 3 9 = (c) log 5 5 = log5 5 1/ = 1/ (d) e 4ln = (e ln ) 4 = 4 = 16 (e) 3 ln(e 4 ) = 1 (f) e 3 ln 8 = (e ln 8 ) /3 = 8 /3 = 4 (g) log 16 = log /4 = 1/4 Or, log 16 = a 16 a = a = 1/4. (h) log 6 + log 6 3 = log 6 3 = log 6 6 = 1 (i) (log a b)(log b a) = ln b ln a ln a ln b = 1. 1