3.1 Basic properties of real numbers - continuation Inmum and supremum of a set of real numbers

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1 Chapter 3 Real numbers The notion of real number was introduced in section 1.3 where the axiomatic denition of the set of all real numbers was done and some basic properties of the set of all real numbers were shown. This chapter is dedicated to the properties of the set R and its subsets. These properties follows from the Dedekind cut D(R) and from the axioms of ordering. 3.1 Basic properties of real numbers - continuation Inmum and supremum of a set of real numbers Having in mind we will work with the sets of real numbers, we will give a specic characterization of the inmum and supremum, respectively, in R. THEOREM Let B R be nonempty. Then: 1. inf B = b R i (a) x B : b x; (b) ɛ > 0 x 0 B : x 0 < b + ɛ. 2. sup B = b R i (α) x B : x b; (β) ɛ > 0 x 0 B : b ɛ < x 0. Proof. We will prove the rst part of the theorem. The second part can be proven in fully analogical way. Let inf B = b R. The (a) property coincides with the rs property of an inmum mentioned in Theorem The same theorem implies from b < b + ɛ ( R) that x 0 < b + ɛ and this veries the (b) property. Now, let the properties (a) and (b) hold true. For y R such that b < y we put ɛ = y b > 0. 5

2 3.1 Basic properties of real numbers - continuation 6 Then the property (b) implies there exists x 0 in B such that x 0 < b + (y b) = b. This completes the proof of the second property of inmum from Theorem We have shown in the example that the set B = {x Q; x 2 3, x 0} is bounded from above (in Q) but has no supremum in Q. The Dedekind cut for real numbers implies that the subset in R behave essentially dierently from the subsets in Q with respect to supremum. THEOREM Any nonempty set M R that is bounded from above has a supremum in R. Proof. Let us dene two sets A and B as follows: The following relations hold true: B := {y R; x M : x < y}, A := R \ B. = A R, = B R, A B = R. We will prove that for any element a A and any element b B we have: a < b. Let us suppose there are two numbers a 0 A, b 0 B such that a 0 b 0, then for all x M we have x < b 0 a 0. Hence a 0 must be in B and this is the contradiction with the fact that A B =. We have shown that the pair of sets A, B is a Dedekind cut (see section 1.3) and therefore there is a real number c such that a c b for all a A and b B. By using Theorem we will show that c = sup M. The inclusion M A veries the (α) property for c. We have to prove that the (β) property also holds true. Let ɛ > 0. Then c ɛ < c b, b B, and therefore c ɛ / B. If for any x M would be x < c ɛ then it would be also c ɛ B and this would be in contradiction with the statement that c ɛ / B. Thus, the (β) property holds true and this means: ɛ > 0 x 0 M : c ɛ < x 0. There is, naturally, an analogical theorem about inmum: THEOREM Any nonempty set M R that is bounded from below has an inmum in R. Proof. Let k R is a lower bound of a set M R, then k is a upper bound of the set M 1 = {y R; y = x, x M}. Theorem guarantees existence of sup M 1 = c 1 R. It is obvious that inf M = c 1. 6

3 3.1 Basic properties of real numbers - continuation 7 Example Let = M 1 M R and let M be bounded from above. The sup M 1 sup M in R. Solution: The existence of the following two real numbers sup M =: b, sup M 1 =: b 1 is guaranteed by Theorem Using the denition of supremum (Denition 1.5.4) we obtain x M 1 M : x b and b 1 b. Example Show that the set of all irrational numbers R \ Q is nonempty. Solution: Example tells us that the set B = {x Q; x 2 < 3, x 0} R is bounded from above in Q, and therefore it is bounded from above in R also. However, sup B / Q. Thus, with respect to Theorem we obtain sup B R and subsequently sup B R \ Q Archimedean property By using Theorem we will show that the Archimedean property holds true for the real numbers. (Compare this with the Archimedean axiom IV Q from Denition 1.3.1). THEOREM It holds true: Proof. By contradiction. Let x R n N : x < n. y R n N : n y. This means that the set of all natural numbers N is bounded from above and therefore (Theorem 3.1.2) it exists sup N = b R. With respect to the (β) property from Theorem implies for ɛ = 1 that there exists n 0 in N such that b 1 < n 0. Subsequently, we have b < n N and this is in contradiction with that b = sup N On density of the rational numbers and the irrational numbers in R The set R, as shown in Example 3.1.2, is the extension of the set of all rational numbers Q and these two sets diers by nonempty set of all irrational numbers. We will show in the following theorem that both sets Q and R \ Q are "dense" in R. We will start with the supply lemma: 7

4 3.1 Basic properties of real numbers - continuation 8 Lemma The statement: x R n Z : n x < n + 1. The integer n is dened by x uniquely. Proof. Let x R. By Archimedean property we have n 1 N, n 2 N : x < n 1, x < n 2. Therefore n 1 < x < n 2. Let us construct the integers n 1, n 1 + 1,..., n 2 1, n 2 and denote by n the greatest of them such that n x. Such an integer is unique and it obeys n x < n + 1. Denition Let x R. The integer n such that n x < n + 1 is called the integral part of the real number x and it is denoted by [x]. 3 5 For example 2 = 0, 8 = 2. THEOREM Let a < b be a pair of real numbers. Then there exist such a rational number x and such an irrational number y that a < x < b, and a < y < b. Proof. By the Archimedean property we know there exists such a natural number n that 1 n < b a, b a > 0. Let k = [na]. Then we have k na < k + 1 and also The inequality k/n a implies k n a < k + 1 n. k + 1 n a + 1 n So we have shown that the rational number 8 < a + (b 1) = b. x = k + 1 n

5 3.1 Basic properties of real numbers - continuation 9 obeys a < x < b. We know that the number sup B from Example is irrational. Let us choose x in such a way that a sup B < x < b sup B. (The existence of such a rational x is guaranteed). Then y = x + sup B is an irrational number for which the requested inequality holds true. Example Show that for any x R we have: x = sup{u Q; u < x}. Solution: The (α) property of supremum from Theorem is obvious. Let ɛ > 0. Theorem implies that r 0 Q : x ɛ < r 0 < x. Thus, r 0 {u Q; u < x} and the (β) property of Theorem holds true Root of a real number The following theorem guarantees the existence and the uniqueness of the nth root of a positive real number. THEOREM It holds true that a > 0 n N y > 0 : y n = a. The number y is dened by the numbers a and n uniquely. Proof. Let B := {x > 0; x n a}. The set B is nonempty since b := a a + 1 B, in fact: 0 < b < 1 and bn < b < a. Furthermore, the set B is bounded from above by the number c := a + 1. In fact, if some y 0 B would obey y 0 > a + 1 then it would be yn n > (a + 1) n, and this would be in contradiction with our assumption that y 0 belongs to B. At the moment Theorem guarantees existence of R + y := sup B. This number y must obey just one of the followings: y n < a, y n > a, y n = a. The rst two cases can be excluded: 9

6 y n + h n Basic properties of real numbers - continuation 10 (1) let us suppose y n < a. We can choose a positive real number h < 1 such that a y n h < (y + 1) n y, (a n yn > 0). + Œ + Œ By making use of binomial theorem we obtain n n (y + h) n = y n y n 1 h + h n 1 n Œ y n n n Œ = y n + h [(y + 1) n y n ] < y n + (a y n ) = a. Thus we have shown that y + h B and this is the contradiction since y = sup B. (2) One can prove that y n > a cannot hold true by analogy. (3) Finally, the relation must hold true. y n = a The uniqueness can be shown indirectly. Let y 1, y 2 R + be two numbers such that Then we obtain the following contradiction This completes the proof. y n 1 = a, y n 2 = a, y 1 < y 2. a = y n 1 < y n 2 = a. Theorem allows us to write the following denition Denition Let a > 0 and n N. The number y > 0 for which y n = n is called the nth root of the number a. It us denoted by the symbol n a or by a 1/n. 2. For a = 0 we put n 0 := For a < 0 and an odd n N we dene n a = n a. It follows from the proof of Theorem and from Examples and that 3 = sup{x R + ; x 2 < 3} = sup{x Q; x 2 < 3, x 0} is an irrational number. 10

7 3.1 Basic properties of real numbers - continuation Extended set of real numbers For the further purposes in Chapter 6 we will need some extension R of the set of all real numbers R. The set R will be closed under operation of taking a limit, i.e. a mapping with values in R R will have a limit in R. Such a set can be dened with help of relation of natural ordering in R: Denition where for all x R: < x < +. R := R {, + }, For our purposes we need not to dene any operation with improper elements ±, it will be sucient to know the topological structure of the set R only (see Theorem 5.2.5). By denition and Theorems and we obtain: THEOREM Any nonempty set M R has a supremum as well as an inmum in R. For example, sup Z = +, inf Z = (in R), where sup Z and inf Z do not exist in real numbers. Problems 1. Prove in details the second part of Theorem Prove that Theorem as well as Theorem are equivalent with the Dedekind cut mentioned in the third section of the rst Chapter. 3. Let = M 1 M R and let M 1 be bounded from below. Prove that inf M 1 inf M in R. 4. Show that for any real number x we have 5. Show that x = inf{u Q : x < u} = sup{u Q : u < x}. a R, a > 1 b R + 0 n Z : a n 1 b < a n. (Hint: you can use the Archimedean property.) 11

8 3.2 Notion of interval, intervals over the real axis Let = M R and M 1 = {y R; y = x, x M}. Show that sup M 1 = inf M, inf M 1 = sup M. 7. Find supremum as well as inmum in R and in R of the following mappings, a) f : x x n x, x R \ {1}, b) =: (a n ) n N n ( 1) nšn N n, n N c1) f 1 : x [x], x {t R; 0 < t < 2}, c2) f 2 : x [x], x {t R; 0 t 2}, c3) f 3 : x x [x], x {t R; 0 t 1}. 8. Prove Theorem Answers 7 a) There are no min f, max f, inf f, sup f in R and min f = inf f = and max f = sup f = + in R. b) In the set R: do not exist and In the set R: min a n, min b n, max b n, sup n N n N n N b n n N inf a n = 1, max a n = sup a n = 2, inf b n = 0. max b n = sup b n = +. Further relations coincides with those valid in R. c) In the set R min f 1, and max f 1 do not exist, inf f 1 = 0, sup f 1 = 1; min f 2 = inf f 2 = 0, max f 2 = sup f 2 = 2; min f 3 = inf f 3 = 0, max f 3 does not exist and sup f 3 = Notion of interval, intervals over the real axis Studying continuity, dierentiability and integrability of a mapping (function) we will see that these properties do depend not only on a relation that denes the mapping but they do depend also on denition domain of the mapping. This is the reason why we will be interested in more details in most frequently used subsets of the real axis - intervals. 12

9 3.2 Notion of interval, intervals over the real axis Intervals Denition A set I R is called an interval (from R) i it has the following properties: a) The set I contains at least two distinct points, b) If x 1 I, x 2 I then ( x R : x 1 < x < x 2 ) x I. 2 A cartesian product of m intervals (from R) is called an m-dimensional interval (from R). It is easy to verify that if a R, b R and if a < b then any of the following sets is an interval from R: I 1 = {x R : a x b} =: [a, b] (closed interval) I 2 = {x R : a x < b} =: [a, b) from the right etc.) I 3 = {x R : a < x b} =: (a, b] from the left etc.) (half-closed interval or half-open interval, open (half-closed interval or half-open interval, open I 1 = {x R : a < x < b} =: (a, b) (open interval) Let us remark that any of the sets I i, i = 1, 2, 3, 4 represents four intervals from R with respect to whether a, b R or a = or b = +. For example, for a, b R the interval I 1 can behave as [a, b], [a, + ], [, b], [, + ] = R. Analogically for other intervals. A natural question arises whether any interval can be identied with one of the types I i, i = 1, 2, 3, 4. The answer to this question together with full characterization of the set of all intervals is given in the following theorem. THEOREM A set I R is an interval from R if and only if there exists an index i in {1, 2, 3, 4} such that I = I i. Proof. 1. It is obvious from the text above this theorem that if I = I i, i = 1, 2, 3, 4 then I is an interval (in R). 2. Let know I be an interval. Then (Theorem 3.1.7) there exist a := inf I and b := sup I in R. Since I contains at least two distinct points, we have a < b. The denitions of inmum and supremum of the set I imply: x I : a x b. ( ) 13

10 3.2 Notion of interval, intervals over the real axis 14 Four distinct cases can appear: a) a I, b I, b) a I, b / I, c) a / I, b I, a) a / I, b / I. Let us consider case a). (In other cases the proof is similar.) Eq. ( ). implies that I [a, b]. Let x [a, b]. If x = a or x = b then x I. Let now x (a, b). Since a = inf I it exists an x 1 I such that x 1 < x (see Theorem 3.1.5). Similarly, b = sup I implies that it exists an x 2 I such that x < x 2. So we have that x 1 < x < x 2. The (b) property of the interval we have x I, thus [a, b] I. We have shown that I = [a, b]. Example Write down some examples of two dimensional intervals from R 2 and nd out what is their number. Solution: By Theorem and the second part of Denition the following sets are intervals in R 2 (a, b, c, d R, a < b, c < d): [a, b] [c, d], [a, b] [, d], [a, + ] [c, d], [a, + ] [, d], [, b] [c, d], [, b] [, d], [, + ] [c, d], [, + ] [, d], [a, b] [c, + ], [a, b] [, ], [a, + ] [c, + ], [a, + ] [, + ], [, b] [c, + ], [, b] [, + ], [, + ] [c, + ], [, + ] [, + ]. Other two dimensional intervals can be derived from I i I j, i, j = 1, 2, 3, 4 such that i+j 2. This means we have 4 4 = 256 distinct two dimensional intervals in R Embedded intervals Denition Let the sequences of real numbers (a n ) n N and (b n ) n N obey: Then the sequence of closed intervals in R a 1 a 2 a n b n b 2 b 1. ([a n, b n ] n N ) is called the system of embedded intervals in R. For any system of embedded intervals we have: [a n+1, b n+1 ] [a n, b n ], (n N). 14

11 3.2 Notion of interval, intervals over the real axis 15 THEOREM Let ([a n, b n ] n N ) be \ a system of embedded intervals in R. Then i.e. c R n N : c [a n, b n ]. [a n, b n ], n N Proof. The sequence (a n ) n N is bounded from above by any of the numbers b n and therefore (see Theorem 3.1.2) we know that it exists the number sup(a n ) n N =: a R and a b n. Boundedness from below of the sequence (b n ) n N by a implies that there exists inf(b n ) n N =: b R (see Theorem 3.1.3) and furthermore a n a b b n, for all n N. Thus we have shown that at least a belongs to any interval of the system ([a n, b n ] n N ). The intersection \ n N 1 n, 1 n equals to one-point set {0}. The intersection \ of the embedded system of open intervals n N0, 1 n is empty (but this fact is not in contradiction with statement of Theorem 3.2.2) An example of uncountable set THEOREM The interval [0, 1] is uncountable set. Proof. The proof will be done indirectly. Let us suppose the interval in question [0, 1] is an countable set. Then it follows from Theorem that this interval is the set of values of an injective sequence of the elements of R. Thus, there exists a sequence (c n ) n N, c n R 3 such that its values coincides with [0, 1]. Let us divide the interval [0, 1] into three parts: 0, 1 1, 3 3, 2 2, 3, 1. Then c 1 [0, 1] does not belong to at least one of these parts of the interval [0, 1]. Let us denote it by [a 1, b 1 ], so c 1 / [a 1, b 1 ]. This interval can be also divided into three parts analogically as we have done with [0, 1] one step before, i.e.: a 1, b 1 a 1 3, b1 a 1 3, 2(b 1 a 1 ) 3 1 2(b1 a 1 ),, b. 3 15

12 3.2 Notion of interval, intervals over the real axis 16 The one of the above written intervals which is such that c 2 does not belong to it will be denoted as [a 2, b 2 ], obviously [a 2, b 2 ] [a 1, b 1 ]. By induction we can construct the sequence of embedded closed intervals in R with property: n N : c n / [a n, b n ]. By making use of Theorem we know that Since c [0, 1] we have: and this contradicts the previous statement Real line c R n N : c [a n, b n ]. n 0 N : c = c n0 / [a n0, b n0 ], In order to be able to interpret geometrically various fact and properties of mappings we will give a description of relation between the set of all real numbers R and the set of points of a line. This relation will be described using the real line axiom. We have already used this axiom in an intuitive way. Let us denote by ρ(a, B) the length of the line segment with end-points A and B. (We suppose we can measure the lengths of line segments.) Let us have two distinct points O and J on the line o. Let us suppose ρ(o, J) = 1. Then we can dene a mapping: in the following way: f : R o (α) For x 1 R + we dene f(x 1 ) =: P x o is such a point on a half-line OJ that ρ(o, P x ) = x 1 (β) For x 2 R we dene f(x 1 ) =: Q x o is such a point on a half-line OJ that ρ(o, Q x ) = x 2 (γ) For x = 0 we dene f(x) = O. This situation is shown on Figure 3.1. Real line axiom The mapping f : R o with properties (α), (β), (γ) is bijective. 16

13 3.2 Notion of interval, intervals over the real axis 17 x x1 Q x O J P x Figure 3.1: o The line o which is with respect to Real line axiom the image of the set R under the mapping f is called real line, its point O is called the beginning and the line segment OJ is called the unity. One can dene on the real line an inner binary relation of addition of two points A, B o (we denote it by the symbol ) and an external operation of multiplication of a point A o by a number s R (we denote it by the symbol ) as follows: A B =: E o, s A =: F o, ( ) where the point E is obtained as the end point of the addition of oriented segments OA and OB (see Figure 3.2). The point O is the beginning of the real line o and ρ(o, B) = ρ(o, B ). o O J A = O B B Figure 3.2: The point F can be constructed by homothety as shown in Figure 3.3. The point S o is an image f(s) (f is the mapping from the real line axiom) and p o is a line crossing through the beginning O of the real line o. F p S O J S A F o Figure 3.3: The following theorem holds true. THEOREM The vector space of the points of real line over R (o, R;,, =) is isomorphic with the vector space of the real numbers over R (R, R; +,, =). 17

14 3.2 Notion of interval, intervals over the real axis 18 Proof. It is easy to nd out that (o, R;,, =) is in fact the vector space. Let us consider the bijective mapping f dened in the real line axiom. For a, b R we put f(a) =: A, f(b) =: B. Then we obtain by the properties (α), (β), (γ) and ( ) that f(a + b) = A B = f(a) f(b), f(sa) = s (A) = s f(a). This theorem allows for considering the set of real numbers R and the real line o as equivalent from the algebraic point of view. Furthermore, we can write a = A(= f(a)) for any a R and we can use the real line o as a geometrical tool for representation of the real numbers. Problems 1. Let I 1, I 2 R be intervals. Decide when the sets are intervals. I 1 I 2, I 1 I 2 2. Write down few patterns of two dimensional intervals from R 2 and nd out their number. 3. Prove the following: A set I R is an interval if and only if it can be written as one of the next nine possibilities (a, b, c, d R, a < b): [a, b], [a, b), (a, b], (a, b), [c, + ), (c, + ), (, d], (, d), (, + ). (Hint: see the proof of Theorem 3.2.1) 4. Prove the following: for any a, b R, a < b the intervals [a, b], (a, b) and R are uncountable sets. (Hint: the function dened by f : [0, 1] [a, b] f : x (b a)x + a is bijective. To prove that the set R is uncountable you can use the function ) 5. Prove Theorem in details. g : (0, 1) R, g(x) = 1 x + 1 x 1. 18

15 3.3 Further extensions of the real numbers 19 Answers 1 The set I 1 I 2 is an interval i I 1 I 2. The set I 1 I 2 is an interval i it contains at least two distinct points Further extensions of the real numbers The extension of the set of all rational numbers to the set of real numbers was motivated by practical purposes, e.g. to give sense to the solution to the algebraic equation: x 2 = 3. Another question is related to the fact that we have not dened even root of a negative number (see Denition 3.1.2). The equation x 2 = 3 has no solution in R. However, high school knowledge tells us that this equation can be solved in the set of complex numbers C. And this set C is, in some sense, also an extension of the set of all real numbers. In this section we will remind basic properties of the set of all complex numbers Field of complex numbers Since the set R is isomorphic with the set of all pairs R {0} (an isomorphism is given by the mapping: f : x (x, 0), x R), the following denition will be a reasonable generalization of the set of all real numbers. Denition The set of all complex numbers C is the cartesian product R R on which we have dened an inner binary operation of addition and multiplication (the equivalence of pairs is dened in the fourth section of the rst Chapter) in this way: 1 Let z 1 = (x 1, y 1 ) C, z 2 = (x 2, y 2 ) C. Then the addition of the complex numbers z 1 and z 2 is the complex number: z 1 + z 2 = (x 1 + x 2, y 1 + y 2 ). 2 The product of z 1 and z 2 is the complex number: z 1 z 2 = (x 1 x 2 y 1 y 2, x 1 y 2 + x 2 y 1 ). 19

16 3.3 Further extensions of the real numbers 20 The following theorem is a simple consequence of previous denition: THEOREM The set of all complex numbers taken as (C, +,, =) is a eld and taken as (C, R, +,, =) is a vector space over R. Proof. It suces to verify the properties of eld and vector space, respectively (they are introduced in Denitions and 1.6.5). The external operation - product " " of a vector z = (x, y) C with a scalar s R (the set {(x, 0); x R} is isomorphic with R) in the vector space (C, R, +,, =) is dened by: sz = (s, 0) (x, y) = (sx, sy) C. The neutral element with respect to addition is 0 := (0, 0) C. The neutral element with respect to multiplication is (1, 0) C. The inverse element (with respect to addition) to an element z = (x, y) C is z = ( x, y). The inverse element (with respect to multiplication) to an element z = (x, y) is dened if z 0 and is denoted by z 1 (x 0, y 0 ) and is given by the condition z 1 z = (1, ) (xx 0 yy 0, xy 0 + yx 0 ) = (1, 0). This system of equation has unique solution that reads Thus, x 0 = z 0 = x x 2 + y, y 2 0 = y x 2 2Œ + y. 2 x x 2 + y, y. 2 x 2 + y If z = (x, y) C then we dene: x =: Rez, y =: Imz and we call them the real part and the imaginary part of the complex number z. If we denote complex number (0, 1) as i ((0, 1) =: i, the imaginary unit) then and Complex number i 2 = ( 1, 0) = 1, z = (x, y) = (x, 0) + (0, y) = (x, 0) + (0, 1)(y, 0) =: x + iy. z = (x, y) = x iy 20

17 3.3 Further extensions of the real numbers 21 is called the complex conjugate to the complex number z = (x, y) = x + iy. It is obvious that z z = x 2 + y 2, z + z = 2x, x = 1 (z + z), 2 y = 1 (z z). 2i Complex plane The geometrical interpretation of complex numbers is given in the following theorem THEOREM Let o 1, o 2 be two real axes with common beginning. Then the vector space of complex numbers (C, R; +,, =) is isomorphic with the vector space (o 1 o 2, R,,, =). An inner binary operation means the vector addition and the external binary operation represents the multiplication of a vector by a real scalar. Proof. With help of bijective mapping f introduced in Axiom we can dene an bijection from R 2 onto o 1 o 2 : by Figure 3.4. g(x, y) = Z o 1 o 2, (x, y) R R, y =: B o 2 Z O x =: A o 1 Figure 3.4: Denitions of the operations and can be also easily formulated by Figures 3.5 and 3.6. Let (x 1, y 1 ), (x 2, y 2 ) R 2 and let f 1, f 2 be a bijection from R onto o 1 and o 2, respectively, such that f 1 (x i ) = A i, f 2 (y i ) = B i, i = 1, 2. 21

18 3.3 Further extensions of the real numbers 22 o 2 B 1 B 2 S 1 y 2 =: B 2 Z 2 y 1 =: B 1 Z 1 O x 2 =: A 2 x 1 =: A 1 A 1 A 2 o 1 Figure 3.5: Let then g(x 1, y 1 ) = Z 1 o 1 o 2, g(x 2, y 2 ) = Z 2 o 1 o 2, s R, Z 1 Z 2 = S 1 o 1 o 2, sz 1 = S 2 o 1 o 2. The mapping g is then the requested isomorphism. If the axes o 1 and o 2 from Theorem are perpendicular then their cartesian product o 1 o 2 is called the complex plane. The axis o 1 is called real axis and the axis o 2 is called the imaginary axis. It is exactly Theorem that allows for considering the set of all complex numbers C and the complex plane to be equal from the algebraic point of view. We will write for any (x, y) C. (x, y) = Z(= g(x, y)), Polar form of a complex number From the above mentioned geometrical interpretation of a complex number one easily obtains the following polar form of a complex number. THEOREM For any z = (x, y) C, z 0 there exists just one φ [0, 2π) - called the argument or the phase of the complex number z - and just one r > 0 - called the absolute 22

19 3.3 Further extensions of the real numbers 23 o 2 sb 1 =: E S 2 S y 1 =: B 1 Z 1 O J x 1 =: A 1 s =: S sa1 =: D o 1 Figure 3.6: value or the modulus of the complex number z - such that Proof. The proof is evident from Figure 3.7. z = (r cos(φ), r sin(φ)) = r(cos(φ) + i sin(φ)). o 2 y Z r φ x o 1 It holds true that the numbers: Figure 3.7: r =È x2 + y 2 R +, φ [0, 2π) 23

20 3.3 Further extensions of the real numbers 24 are dened uniquely by equations: x = r cos(φ), y = r sin(φ). The number r is often denoted by z. An exact denition of used goniometric functions is given in section 4.4 of this book. Problems 1. Prove in details Theorems and Formulate and explain the algorithm of construction of the image of: a) product of complex numbers b) symmetric element z 1 to an element z = (x, y) C in the complex plane by using the isomorphism g from Theorem Show that for z 1 = r 1 (cos(φ 1 ) + i sin(φ 1 )) C and z 2 = r 2 (cos(φ 2 ) + i sin(φ 2 )) C, for any n N we have: a) z 1 z 2 = r 1 r 2 (cos(φ 1 + φ 2 ) + i sin(φ 1 + φ 2 )), b) z n 1 = r n (cos(nφ) + i sin(nφ)) - Moivre theorem, c) n z1 = z C; z = n r cos φ + 2kπ n Œ + i sin φ + 2kπ n Œ, for k = 0, 1, 2,..., n 1«, (let us remind that n z 1 is a set of all complex numbers z for which z n = z 1 ), d) z 1 0 and z 1 = 0 z 1 = 0 C, e) z 1 z 2 = z 1 z 2, f) z 1 z 2 z 1 ± z 2 z 1 + z 2. 24

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