Page:EB1911 - Volume 27.djvu/293

(a+¢b)'“/'* are obtained by giving s the values 0, 1, 2, ...nin the expression r"'/" (cos m.0 + 2s1r/n + L sin m .0 + 2S1I'/71), where r=(a2-I-b”)} and 0=arc tan b/a. We now return to the geometrical 'representation of the complex numbers. If the points B, , B2, B3, ...B, , represent the expression x+¢y. (x+»y)”. (x+¢y)“. B# . . (x-I-¢y)" respectively, the triangles OAB1, OBIB2, . . OB, , 1B, , are all similar. Let (x-+-Ly)"=a-l-vb, then the converse problem of finding the nth root of a~|-Lb is equivalent to the geometrical problem of describing such a series of triangles that OA is the first side of the first triangle and OB., the second side of the nth. Now it is obvious that this geometrical problem has more solutions than one, since any number of complete revolutions round O may be made in travelling from B1 to B". The first solution is that in which the vertical angle of each triangle is B, ,OA/n; the second is that FIG in which each is (B, .0A +2-/r)/n, 9' in this case one complete revolution bein made round O the third has B OA r h g; <.. -l-41)/"fo r e vertical angle of each triangle; and so on. There are n sets of triangles which satisfy the required conditions. For simplicity we will take the case of the determination of the values of (cos 04-L sin 0)?s. Suppose B to represent the expression cos 0+ r. sin 0. If the angle AOP, is § 0, P1 represent the root cos § 0-Q-L sin § 0; the angle AOB is filled up by the angles of the three similar triangles AOP1, P1Op1, PIOB. Also, if P¢, P3 be such that the angles P10P¢ P10193 are § r, $77 respectively, the two sets of triangles AOP2, P20p., p30B and AOP3, PsOp2, pzOB satisfy the conditions of similarity and of having OA, OB for the bounding sides; thus P2, P, represent the roots cos § (6+21r)-I-L sin § (0+21r), cos § (6-{-41;-) +L sin § (0+41r) respectively. If B coincides with A, the problem is reduced to that of finding the three cube roots of unity. One will be represented by A and the others by the two angular points of an equilateral triangle, with A as one angular point, inscribed in the circ e. The problem of determining the values of the nth roots of unity is equivalent to the geometrical problem of inscribing a regular The nth polygon of n sides in a circle. Gauss has'shown in his Roots of Dzsquzsztzones amthmetzcae that this can always be done Unity by the compass and ruler only when n is a prime of the form 2f'+I The determination of the nth root of Q B 1; 1% R 0 A P. fa FIG. ro. V any complex number requires in addition, for its geometrical solution, the division of an angle into n equal parts. 19. We are now in a sition to factorize an expression of F, ¢, -¢0|-|, , g- the form x"-(Ii10+ib). Usin the values which we tions. have obtained above for (a +tb§ I /", we have s= -1 3 x”-(a+Lb)=P n [x-1“(cos¥&r+isin@)]- (1) s=0 If b =o, a=1, this becomes ="'1 2s1r 2s1r x”-I=1;0 x-cos?-rsins=i" 1 2S1r . 2S7r =(x-I)(x-|-I)P 1 x-cos-7T=»=tsmT = s=%n-1 251. =(x-I)(x+I)P x2-2x cos-I-+I (n even). (2) 5=I =l(n-1) x"- 1 =(x-1)€' 1 x2-2x cos?%'+I (n odd). (3) If in (I) we put a= vi, b=o, and therefore 0=1r, we have =n~1 T-' °"" x*-{»-1:12 x c0S LSin =0 n n =i(-2) =i;' 0 n x'-2x cos 23-1%-r+I (n even). (4) s=i(n~3) x~+i =(x+1>P0 x2-2x cos il;-¥'+r (n odd)- (5) = Also x2"-2x”y" cos n0-l-y2" = (x"-yn cos n6+¢ sin n0)(x"-y" cos n0-L sin nt?) ="'1' "i-T"""f"T;'; =P x-ycos-i-lr=¢=1.sin-:Qls=0 77' n "="“'1 2 2S1r (6) Airy and Adams have given proofs of this theorem which do not involve the use of the symbol 1, (see Cafnb. Phil. Trans., vol. xi). A large number of interesting theorems may be derived from De Moivre's theorem and the factorizations which we have ==P x -2xycos0-—-y2 . s=0 M deduced from it; we shall notice one of them. Example if In equation (6) put y=I/x, take logarithms, and then?;:;?;:;1re 3 differentiate each side with respect to x, and we get 2n x2"'1-x'2"'1) § =""1 2(x-x'3) x2"-2 cos n0-1-x“2" s:0 x2-2 cos 6+-774-x" Put x2 =a/b, then we have the expression n(a2n b2n> (112 - b2l(a2"— 2a"b" cos n0+b2 ) for the sum of the series s=n- 1 I Z, 5:0 a2-2ab cos 0-|-%;1-r-I-b2 20. Denoting the complex number x-I-iy by z, let us consider the series 1+z+z2/2!+...+z"/nl+. . This series uniformly and absolutely for all values of z whose converges moduli do not exceed an arbitrarily chosen positive Theejfgl number R. Consequently the function E(z), defined QZZEZ as the limiting sum of the above series, is continuous in every finite domain. The two series representing E(z1) and E(z2), when multiplied together give the series represented by E(z1+z1). In accordance'with a known theorem, since the series for E(z1) E(z2) are absolutely convergent, we have E(z1)><E(z»1) =E(z, +z2). From this fundamental relation, we deduce at once that {E(z)1" =E(nz), where n is any positive integer. The number E(1), , the sum of the convergent series 1 +1-1-1/2!+1/gl., ., is usually denoted by e; its value can be shown to be 2-718281828459 .... It is known to be a transcendental number, Le. it cannotbe the root of any algebraical equation with rational coefficients; this was 'first established by Hermite. Writing z=I, we have E(n)=e", where n is a positive integer. If z has as a value a ositive fraction p/q, we find that{E(/5/g)}q=E(p) =e"; hence E(p/gg is the real positive value of ef'/'1. gain E(-11/g)><E(p/9) =E(0)=I, hence E(-p/g) is the real positive value of e'1'/41. It has been thus shown that for any real and rational number x, the value of E(x) is the principal value of ez. This result can be extended to irrational values of x, if we assume that e' 'is for such a value of x defined as the limit of the sequence e", e”, , where xl, x2, . is a sequence of rational numbers of wluph x is the limit, since E(x1), E(x2) . ., then converges to E x Next consider (I -1-z/m)'", where m is a positive integer. We have by the binomial theorem, m 2 on ~+Z+<~.f.>a+~+<1~f:.> <1~.a>~. s-I za z '" “M H+ "tial 1 2 S-I Also . . (I-W » lies between 1 and 1+ 5-ki-}- -}-5%;, hence the product equals I-085.5-I/2711 where 0, is 0 < 0, < I We have now such that m 2 (14%) =r-|-2+ (1-i) E-, +. . + 1-o, § 77l I m “ +V i-0, ,, i-1%= I-}-z-l-z2/2!-i- +28/s!+R, ,, where Zs+l Zm Z2 2 R, =E;- *Z-Y!-l- I-lf-03?-l-. Zs-2 Zm-2 -l-Hsgjm-|'~. . . +0m . Since the series for E(z) converges, 5 can be fixed so that for all values of m >s the modulus of z*+1/(s+I)!+. +z'"/ml is less than an arbitrarily chosen number és. Also the modulus of I-|-Gaz/I+. -l'9mZ""'2/(1')1"2)l is less than that of I-i-'I |z| /I! + [z[”/25 -l- , or of e'11°d=, hence mod R.<%e-l-(I/zm). mod (z”e=)<f, if m be chosen sufficiently great. It follows that 11mm-, ,°(I +z/m)"'=E(z), 'where z is any complex number. To evaluate E(z), write H-x/m=p cos ¢, y/m=p sin 41, then