/153 Chapter VII On imaginary expressions and their moduli. _______ § I – General considerations on imaginary expressions In analysis, we call a symbolic expression or symbol any combination of algebraic signs that do not mean anything by themselves or to which we attribute a value different from that which they naturally have. Likewise, we call symbolic equations all those that, taking the letters and the interpretations according to the generally established conventions, are inexact or do not make sense, but from which we can deduce exact results in modifying and altering them according to fixed rules either the equations themselves or the symbols which comprise them. The use of symbolic expressions or equations is often a means of simplifying calculations and of writing in a short form results that appear quite complicated. We have already seen this in the second paragraph of the third chapter where formula (9) gives a very simple symbolic value to the unknown x satisfying the equations (4). Among the symbolic expressions or equations the consideration of which is of some importance in analysis, we should especially distinguish those which we call imaginary. We are going to show how we can put them to good use. We know that the sine and the cosine of the arc a + b are given as functions of the sines and cosines of the arcs a and b by the formulas (1) cos a b cos a cosb sin asinb sin a b sin a cosb sinb cos a /154 But, without taking the pain to remember these formulas, we have a very simple means of recovering them at will. It suffices, in fact, to consider the following remark. Suppose that we multiply together the two symbolic expressions cos a 1sin a, cosb 1sinb, Cours d’analyse – Chap. VII: Imaginary expressions 1 by applying the known rules of algebraic multiplication and knowing that -1 is a real [?? Actual?] quantity the square of which is equal to -1. The resulting product is composed of two parts. The one real and the other has a factor -1. The real part gives the value of cos(a + b) and the coefficient of -1 gives the value of sin(a + b). To observe this remark, we write the formula (2) cos a b 1sin a b cosa 1sina cosb 1sinb The three expressions that make up the preceding equation, namely cos a + -1 sin a, cos b + -1 sin b, cos(a + b) + -1 sin(a + b), are three symbolic expressions that cannot be interpreted using the generally established conventions, and they do not represent anything real. For this reason, they are called imaginary expressions. The equation (2) itself, taken literally, inexact and it does not make sense. To get exact results, first we must expand its second part by algebraic multiplication, and this reduces the expression to (3) cos a b 1sin a b cosacosb sinasinb 1 sinacosb sinbcosa Secondly, we must equate the real part of the first member (left hand side) of equation (3) with the real part of the right hand side, then the coefficient of -1 on the left hand side with the coefficient of -1 on the right. Thus we recover the equations (1) both of which we ought to consider as implicitly contained in the formula (2) In general, we call an imaginary expression any symbolic expression of the form + -1, where and denote real quantities. We say that two expressions + -1 and + -1 are equal to each other when there is equality between corresponding parts, that is between the real parts and , and between the coefficients of -1, namely and . We indicate the equality between two imaginary expression in the same way that we it between two real quantities, by the symbol =, and this results in what we call an imaginary equation. This said, any imaginary expression is just the symbolic representation of two equations between real quantities. For example, the symbolic equation + -1 = + -1 Cours d’analyse – Chap. VII: Imaginary expressions 2 is just equivalent to the two real equations = and = . When, in the imaginary expression + -1, the coefficient of -1 vanishes, the term -1 supposed to reduce to zero, and the expression itself reduces to the real quantity . By virtue of this convention, imaginary expressions include, as special cases, real quantities. Imaginary expressions and those involving real quantities can be supposed to have different algebraic operations. In particular, if we do addition, subtraction, or multiplication /156 of two imaginary expressions, they operate according to the established rules for real quantities, and we will obtain as a result a new imaginary expression that we call the sum, the difference or the product of the given expressions, and the ordinary notations will serve to indicate that sum, difference or product. For example, if we are given two imaginary expressions, + -1 and + -1 we will find (4) (5) (6) 1 1 1 1 1 , 1 1 1 , 1 . It is good to remark that the product of two or more imaginary expressions, like that of two or more real binomials, remains the same regardless of the order in which we multiply the different factors. To divide a first imaginary expression by a second means to find an imaginary expression which, when multiplied by the second, reproduces the first. The result of this operation is the quotient of the two given expressions. To indicate this, we use the ordinary symbol for division. So, for example, 1 1 represents the quotient of the two imaginary expressions + -1 and + -1 Cours d’analyse – Chap. VII: Imaginary expressions 3 To raise an imaginary expression to the power m (where m denotes a integer) means to form the product of m factors equal to that expression. We write the mth power of + -1 with the notation 1 . m /158 In the particular case when is reduced to zero, one or two of these powers can become real. [What does this mean?] To raise the imaginary expression + -1 to a negative power, – m, or –1/n or –m/n is to divide 1 by the power m or 1/n or m/n of the same expression. The problem will have a unique solution in the first case, and several solutions in the two others. We denote the power of degree – m with the simple notation 1 m while the two notations 1 , 1 1 n m n represent, in the first case, any of the powers of degree =1/n, and in the second case, any of the powers of degree – m/n. We say that two imaginary expressions are conjugate to each other when the two expressions differ only in the signs of the coefficient of -1. The sum of two such expressions is always real, as is their product. Indeed, the two conjugate imaginary expressions + -1, - -1 have as their sum 2 and as their product 2 2 . The final part of this observation contained in a theorem about numbers, expressed as Theorem I: If we multiply together two whole numbers that are each the sum of two squares, then the product will always be a sum of two squares. Proof: Let the numbers be 2 + 2 and ’2 + ’2 /159 where 2, 2, ’2, ’2 denote the perfect squares. We obviously have the two equations 1 ' ' 1 ' ' ' ' 1 Cours d’analyse – Chap. VII: Imaginary expressions 4 1 ' ' 1 ' ' ' ' 1 and, by multiplying these, term by term, we will obtain (7) 2 2 a'2 '2 ' ' ' ' . 2 2 If we interchange the letters ’ and ’ in this last expression, we get (8) 2 2 a'2 '2 ' ' ' ' . 2 2 Thus, in general we have two ways to decompose into two squares the product of two whole numbers each of which is the sum of two squares. Thus, for example, one draws from equations (7) and (8) 2 2 1 32 2 2 4 2 7 2 12 8 2 . We see from these examples that the use of imaginary expressions can be of great use, not only in ordinary algebra but also in number theory. Sometimes we represent an imaginary expression by a single letter. It is an artifice which helps us use the resources of analysis and we will make use of it in what follows. 159 § II – On the moduli of imaginary expressions and on reduced expressions A remarkable property of any imaginary expression + -1 is that it can be put into the form cos 1sin , /160 denoting a positive quantity and a real arc. Indeed, if we write the symbolic equation 1 cos 1sin which amounts to the same as the two real equations cos sin Cours d’analyse – Chap. VII: Imaginary expressions 5 and we find that 2 2 2 cos2 sin2 2 2 2 (3) and, having thus determined the value of the number , all that remains to verify completely the equations (2) is to find an arc such that its cosine and sine are respectively cos 2 2 sin 2 2 (4) This last problem is always solvable because each of the quantities 2 2 2 2 and have numerical values less than one and the sum of their squares is equal to 1. Also, it has infinitely many different solutions because, having calculated one convenient value of the arc , we can, without changing the value of the sine or the cosine, increase or decrease this arc by any number of circumferences. When the imaginary expression + -1 is put into the form cos 1sin , the positive quantity is called the modulus of this imaginary expression, and what remains after the suppression of the module, that is /161 to say the factor cos 1sin we will call the reduced expression. Since we take the quantities and to be known quantities, we get only one unique value for the modulus as determined by equation (3), and as a result the modulus remains the same for any two imaginary expressions that are equal. Thus we can state the following theorem: Theorem I: The equality of two imaginary expressions always implies the equality of their moduli, and as a consequence, the equality of their reduced expressions. Cours d’analyse – Chap. VII: Imaginary expressions 6 If we compare two conjugate imaginary expressions, we find that their moduli are equal. The square of their common modulus is just their product. When in the imaginary expression + -1, the second term vanishes, the expression reduces to a real quantity . In this case, we get from equations (3) and (4), first, when is positive, 2 cos = 1, sin = 0 and so 2k , where k denotes any whole number; or second, when is negative, 2 cos = – 1, sin = 0 and so 2k 1 . Thus the modulus of a real quantity is just its numeric value 2 , and the reduced expression that corresponds to such a quantity is always +1 or – 1, since 1 cos 2k 1sin 2k /162 which always gives a positive quantity, and 1 cos 2k 1 1sin 2k 1 ,1 which gives a negative quantity. Any imaginary expression that has modulus zero itself reduces to zero, since its two terms both must vanish. Conversely, since the cosine and the sine of an arc are never zero at the same time, it follows that an imaginary expression cannot be reduced to zero unless its modulus vanishes. Any imaginary expression which has one as its modulus is necessarily a reduced expression. Thus, for example, 1 Here, rather than nesting parentheses, Cauchy uses the old notation 1 cos 2k 1 1sin 2k 1 Cours d’analyse – Chap. VII: Imaginary expressions 7 cosa 1sina , cosa 1sina , cosa 1sina , cosa 1sina are four reduced expressions, two conjugate pairs. Indeed, to derive these four expressions from the formula cos 1sin it is enough to take successively 2k a , 2k 1 a 2k a , 2k 1 a , where k denotes any whole number. Calculations involving imaginary expressions can be simplified by using reduced expressions. It is important to take note of their properties. These properties are contained in the theorems that I am about to state. Theorem II: To multiply two reduced expressions cos 1sin and cos ' 1sin ' , it suffices to add the arcs and ’ to which they correspond. /163 Proof: Indeed, we have (5) cos 1sin cos ' 1sin ' cos ' 1sin '. Corollary: If in the previous theorem we take ’ = – , we find, as expected, (6) cos 1sin cos 1sin 1 . Theorem III: To multiply together several reduced expressions, cos 1sin , cos ' 1sin ' , cos '' 1sin '' , …, it suffices to add the arcs , ’, ’’, … to which they correspond. Proof: Indeed, we get successively, [block of formulas] Cours d’analyse – Chap. VII: Imaginary expressions 8 and, continuing in the same way, we find generally that whatever the number of arcs, , ’, ’’, … may be, cos (7) 1sin cos ' 1sin ' cos '' 1sin '' L cos ' '' L 1sin ' '' L . Corollary: If we expand by ordinary multiplication the left hand side of equation (7), the expansion will consist of two parts, one real and the other having a factor -1. This said, the real part will take on the value cos ( + ’ + ’’ + …), /164 and the coefficient of -1 in the second part will have the value sin ( + ’ + ’’ + …). Let us suppose, for example, that we are considering only three arcs, , ’ and ’’. Then equation (7) becomes cos 1sin cos ' 1sin ' cos '' 1sin '' cos ' '' 1sin ' '', and, after expanding the left hand side of this last equation by algebraic multiplication, we conclude that cos ' '' cos cos 'cos '' cos sin 'sin '' sin cos 'sin '' sin sin 'cos '' sin ' '' sin cos 'cos '' cos sin 'cos '' cos cos 'sin '' sin sin 'sin '' Theorem IV: To divide the reduced expression cos 1sin by another one cos ' 1sin ' , it suffices to subtract the arc ’ corresponding to the second expression from the arc that corresponds to the first. Proof: Let x be the quotient we are seeking, so that Cours d’analyse – Chap. VII: Imaginary expressions 9 x cos 1sin . cos ' 1sin ' This quotient ought to be a new imaginary expression chosen so that when it is multiplied by cos ' 1sin ' it makes cos 1sin . In other words, x ought to satisfy the equation cos ' 1sin ' x cos 1sin . To solve this equation for x, it works to multiply both sides by cos ' 1sin ' . /165 In this way we reduce the coefficient of x to unity (see Theorem II, Corollary I), and we find that cos 1sin cos ' x cos 1sin cos ' 1sin ' 1sin ' cos ' 1sin '. Finally we will get (8) cos 1sin cos ' 1sin '. cos ' 1sin ' Corollary: If in equation (8) we take = 0, it gives (9) 1 cos ' 1sin '. cos ' 1sin ' Theorem V: To raise the imaginary expression cos 1sin to the power m (where m denotes any whole number), it suffices to multiply the arc in this expression by m. Proof: Indeed, the arcs , ’, ’’, … can be taken as in formula (7), and if we suppose that they are all equal to , and that there are m of them, we find Cours d’analyse – Chap. VII: Imaginary expressions 10 cos (10) 1sin cos m m 1sin m . Corollary: If in equation (10) we take successively = z and then = – z, we get the following two equations: cos z 1sin z cos z 1sin z (11) cos mz cos mz m 1sin mz m 1sin mz. The left hand sides of each of these is always the product of m equal factors, can be expanded by multiplication [what is “immediate multiplication” vs “algebraically”?] of these factors, or, what amounts to the same thing, by the /166 formula of Newton.2 If, after expanding the equation, if we equate corresponding parts of each equation, first the real parts and second the coefficients of -1, we will conclude (12) m m 1 m 2 m cos z sin 2 z cos mz coz z 1 2 m m 1m 2 m 3 m 4 cos z sin 4 z L 1 2 3 4 m sin mz cos z m 1 sin z 1 m m 1m 2 m 3 cos z sin 3 z L 1 2 3 Taking m = 2, for example, we will find cos 2z cos 2 z sin 2 z sin 2z 2 sin z cos z and supposing m = 23, cos 3z cos 3 z 3cos z sin 2 z sin 3z 3cos 2 z sin z sin 3 z, etc. Theorem VI: To raise the imaginary expression cos 1sin 2 That is, Newton’s binomial formula. Cours d’analyse – Chap. VII: Imaginary expressions 11 to the power – m, (where m denotes any whole number), it suffices to multiply in this expression the arc by the degree – m. Proof: Indeed, from the definition we have given of negative powers (see § I), we get cos 1sin m cos 1 1sin m 1 cos m 1sin m Consequently, using formula (9), we get (13) cos 1sin cos m 1sin m cos m 1sin m . m /167 or, what amounts to the same thing, (14) cos 1sin m After establishing, as we have just done, the principle properties of reduced expressions, it becomes easy to multiply or divide one by the other of two or more imaginary expressions, if we know their moduli, as well as to raise any imaginary expression to a power m or – m, (where m denotes a whole number). Indeed, we can easily perform these different operations with the aid of the following theorems: Theorem VII: To obtain the product of two or more imaginary expressions, it suffices to multiply the product of the reduced expressions to which they correspond by the product of the moduli. Proof: The stated theorem follows immediately from the principle that the product of several factors, real or imaginary, remains the same regardless of the order in which one multiplies them. Indeed, let (cos + -1 sin ), (cos ‘+ -1 sin ’), (cos ’’ + -1 sin ’’), … be several imaginary expressions, where , ’, ’’, … denote their moduli. When we want to multiply these expressions together, where each expression is the product of a modulus and a reduced expression, we can, by virtue of the principle just mentioned, form as one part the product of the moduli, and as another part that of all the reduced expressions, then multiply together these two products. We find in this way there results the final expression (15) ' ''L cos ' ''L 1sin ' ''L . Cours d’analyse – Chap. VII: Imaginary expressions 12 Corollary I: The product of several imaginary expressions is a new imaginary expression which has as its modulus the product of the moduli of all the others. Corollary II: Since only the modulus of an imaginary expression can ever vanish, /168 in order to make the product of several moduli vanish, it is necessary that one of them reduces to zero, and it is clear that one may draw from theorem VII the following conclusion: Theorem VIII: To obtain the quotient of two imaginary expressions, it suffices to multiply the quotient of their corresponding reduced expressions by the quotient of their moduli. Proof: Suppose that it is a question of dividing the imaginary expression (cos + -1 sin ) where the modulus is, by the following ’(cos ’ + -1 sin ’) where the modulus is ’. If we denote by x the desired quotient, then x must be a new imaginary expression satisfying the equation ’(cos ’ + -1 sin ’) x = (cos + -1 sin ) To solve this equation for the value of x, we will multiply both sides by the product of the two factors 1 and cos ' 1sin ' ' and in this way we will find, writing x 1 in place of , that ' ' cos ' 1sin ' . ' In the final analysis we will have (16) ' cos ' cos ' 1sin ' ' cos 1sin 1sin ' /169 and since, by virtue of theorem IV, Cours d’analyse – Chap. VII: Imaginary expressions 13 cos ' 1sin ' is exactly the quotient of the two reduced expressions cos + -1 sin and cos ’ + -1 sin ’, it is clear that, having established formula (16), we ought to consider proved theorem VIII. Corollary: If in equation (16) we take = 0, it gives (17) 1 ' cos ' 1sin ' 1 ' cos ' 1sin ' . Theorem IX: To obtain the mth power of an imaginary expression (where m denots any whole number) it suffices to multiply the mth power of the corresponding reduced expression by the mth power of the modulus. Proof: Indeed, if in theorem VII we take the imaginary expressions (cos + -1 sin ), (cos ‘+ -1 sin ’), (cos ’’ + -1 sin ’’), … all to be equal to each other and to be m in number, their product will be equivalent to the mth power of the first one, that is to say, equal to m cos 1sin , and, with this hypothesis, the expression becomes m cos m 1sin m (18) m cos 1sin m cos m 1sin m . /170 The reduced expression cos m + -1 sin m is equal to (by virtue of theorem V) Cours d’analyse – Chap. VII: Imaginary expressions 14 cos 1sin , m and, having established formula (10), as a result we ought to consider theorem IX to be proved. Theorem X: To raise an imaginary expression to the power – m (where m denotes a whole number), it suffices to form the same powers of the modulus and of the reduced expression, then multiplying the two parts together. Proof: Suppose that it is a question of raising the following imaginary expression to the power – m (cos + -1 sin ) where the modulus is, by virtue of the definition of negative powers cos 1sin m 1 cos 1sin 1 m cos m 1sin m . Consequently, making use of formula (17),we find cos 1sin m 1 m cos m 1sin m or, what amounts to the same thing, (19) cos 1sin m m cos m 1sin m This last formula, together with equation (13), furnish the complete proof of theorem X. 171 § III – On real and imaginary roots of the two quantities + 1 and –1 and on their fractional powers Suppose that m and n denote two whole numbers relatively prime to each other. If we use the notations adopted in § I, the nth roots of unity, or what amounts to the same 1 thing, the powers of degree will have the different values of the expression n Cours d’analyse – Chap. VII: Imaginary expressions 15 1 n 1 1 , n or, in the same way, the fractional positive or negative powers of unity of degree m or n m are the various values of n m n 1 or 1 m n . We will conclude that, to determine these roots and powers, it suffices to solve, one after another, the three following problems. Problem I: To find the various real and imaginary values of the expression 1 n 1 . Solution: Let x be one of these values, and in order to present it in a form that includes the real quantities and the imaginary quantities at the same time, let us suppose that x r cost 1sint where r denotes a positive quantity and t denotes a real arc. We have, because of the 1 definition of the expression (1) 1n , that xn = 1 or, in other words r n cosnt 1sinnt 1 . /172 We can draw from this last equation (with the aid of theorem I § II) rn = 1 cos nt + -1 sin nt = 1 and so, cos nt = 1, r=1 sin nt = 0, 2k t , n nt = 2kp Cours d’analyse – Chap. VII: Imaginary expressions 16 where k represents any whole number. The quantities r and t are also determined, their various values that satisfy equation (1) are obviously contained in theformula (2) x cos 2 k 2 k 1sin . n n In other words, the various values of (3) 1 n 1 cos 1 n 1 are given by the equation 2k 2k . 1sin n n Now let h be the whole number closest to the ratio k . The difference between n k will be at most equal to ½, and so we will have n k k' h , n n the numbers h and k' denotes a fraction less than or equal to ½, and, as a consequence, k’ will be a n n whole number less than, or at most equal to . From this we will concluce 2 where 2k 2k ' 2k 2k 2k ' 2k ' 2h , cos 1sin cos 1sin . n n n n n n Consequently, all the values of cos 1 n 1 will be contained in the formula 2k ' 2k ' 1 sin , n n /173 n , or, what amounts to the 2 same thing, in formula (3) if we suppose that k is contained between the same limits. if we suppose that k’ is contained between the limits of 0 and Corollary I: When n is even, the various values that the number k can assume n without leaving the limits 0 and are respectively 2 0, 1, 2, …, n2 n , . 2 2 Cours d’analyse – Chap. VII: Imaginary expressions 17 For each of these values of k, formula (3) gives in general two conjugate imaginary 1 1n , that is to say, two conjugate imaginary roots of unity of n degree n. However, we find for k = 0 but a single real root, +1, and for k another 2 real root, – 1. In summary, when n is even, the expression values of the expression 1 1n admits two real values, namely +1 and – 1, along with n – 2 imaginary values, conjugate two by two, namely (4) [block of equations.] The total number of these values, real and imaginary, is equal to n. Suppose, for example, that n = 2. We find that there exist two values of the expression 1 n 1 , /174 where, what amounts to the same thing, the two values of x are those that satisfy the equaton x2 = 1, and these values, both real, are respectively +1 and – 1. Now suppose that n = 4. We find that there are four values of the expression 1 4 1 , where, what amounts to the same thing, the four values of x are those that satisfy the equation x4 = 1. Among these four values, two of them are real, namely Cours d’analyse – Chap. VII: Imaginary expressions 18 +1 and – 1. The two others are imaginary and are respectively equal, first to cos 2 1sin 2 1 , and second to cos 2 1sin 2 1 . Corollary II: When n is odd, the various values that the number k can assume n without leaving the limits 0 and are respectively 2 0, 1, 2, …, n 1 . 2 For each of these values of k, formula (3) gives in general two conjugate imaginary 1 values of the expression 1n , that is to say, two conjugate imaginary roots of unity of degree n. However, we find for k = 0 but a single real root, +1. In summary, when n is odd, the expression 1 n 1 /175 admits a single real value, namely +1, along with n – 1 imaginary values, conjugate two by two, namely (5) [block of equations.] The total number of these values, real and imaginary, is equal to n. Suppose, for example, that n = 3. We find that there exist three values of the expression 1 3 1 , where, what amounts to the same thing, the three values of x are those that satisfy the equaton x3 = 1, Cours d’analyse – Chap. VII: Imaginary expressions 19 and these values, of which one is real, are respectively 2 2 cos 1 sin , 3 3 +1, cos 2 2 1 sin . 3 3 Also, the side of the hexagon is, as we know, equal to its radius, and the supplement of 2 the arc subtended by this side has for its measure , so we can easily obtain the 3 equations cos 2 1 , 3 2 1 2 sin 2 3 , 3 2 1 and, by virtue of these equations, the imaginary values of the expression 1 2 1 3 2 2 1 , 1 2 1 3 2 2 13 reduce to 1 . /176 [4092] Cours d’analyse – Chap. VII: Imaginary expressions 20