We have, secondly, to show how the proof of the Binomial Theorem is to be applied to complex quantities. The student will notice, then, that the index theorem required that the index m involved should be real (positive or negative) and that f(m) should be single-valued. If these conditions are fulfilled f (m) may be itself complex. Then, if f(m) xf (n)= f (m + n), f(n) is one of the values of f (1)". This theorem applies, as shown in Arts. 465, 466, to any series of the form of the Binomial Expansion, whether a be complex or real, provided n is real. It will remain, therefore, to determine to which of the values of f (1)" the series must be equated, when n is fractional. 617. In De Moivre's theorem, let n be a positive integer. Then cos no + i sin n0 =(cos 0 + i sin 6)". Expanding, by the binomial theorem for a positive integer, and equating the real and imaginary terms, we have n (n-1) cos no = cos” COS”—2 A sin 0 + n (n − 1) (n − 2) cos"–3 A sin3 0 + ... 1.2.3 Dividing we have n tan 0 - n(n-1) (n − 2) tans 0 + tan no = 1-in (n-1) tano 0 + cot" 0 - 1 n(n − 1) coth-20 + ... n coth-10-in(n-1) (n-2) cosu-36 + ... cos no + i sin n6=(cos 0 + i sin 6)" =cos" A (1 + i tan A)". We may expand in ascending powers of tan 7, since tan 0 is <l. Thus, equating real and imaginary terms, cos no=cos" 0{1 - {n(n − 1) tano 0 + ...} sin n0 = cosO {n tan 6 – jn (n − 1)(n- 2) tan: 0 +...} where n is negative. Or -3 2 n cos mo secm (2) = secm @{1 įm (m + 1) tan’ A +...} Let tan 0>). Then = sin" @{1 - In(n-1) cot? @ +...} 1 = sin" @ {n cot 0 – In (n − 1)(n- 2) cot: 0 +.. where n is negative. Or m m-1 m+2 m-1 (-1)2 cos mo (m even) or (-1) ? sin mo (m odd) = cosecm 0 {1 - m (m + 1) cot 0 +...} (-1)** sin mo (m even) or (- 1)^2 cos mo (m odd) ? m = cosecm 0 {m cot 0 Žm (m + 1) (m + 2) cot: 0 +...} where m is positive. 619. Let n be fractional=plq say. P Then, by De Moivre's theorem, the q values of (cos 6+ i sin 6) р P are given by cos $+isin ¢ where is any angle whose $ $ 9 9 cosine = cos 0 and whose sine = sin 0. Now suppose tan 0 <1. Then, by the binomial theorem, since p/q is real so that the argument of Art. 466 holds although cos 6+ i sin 6 is complex, the q values of (cos 6 + i sin 6) are given by the series P-) V] cos” 2 9 4 9 + where, by Art. 606, the multiplier V(cos? A) has q values; and the series into which it is multiplied is finite and single-valued, since tan 0 < 1. Now suppose cos 0 is positive. Then (cos”) has one real positive value. Using this value in the above series, the upper line is wholly real and the lower line is wholly imaginary. Hence the р line P upper cos - and the lower line = i sin = 0, 9 9 where ø is some angle whose cosine = cos 0 and whose tangent = tan 8. It remains to determine which of the alternative values of 0 is to be taken. To do this we apply the principle of continuity. Thus As cos 0 approaches the limit + 1 and tan 0 the limit 0, the upper line approaches the limit +1 and the lower line the limit 0. Hence, in the limit, the upper line = cos Lo and the lower line 2 20. Hence the general values of the upper and lower 9 line are, respectively, cos?o and i sin 20, where 0 is the (positive P р 9 9 or negative) acute angle whose cosine = cos 0 and whose tangent - tan . = 2 sin P For, if any other values, e.g. cos P (27 +0) and sin (2+0) 9 a were given, then as @ approached the limit 0, the two series would approach the limits cos? 27 and i sin 2 2m, and yet when 6 a 9 reached its limit, the two series would be 1 and 0 respectively. This would involve a breach of continuity, which we have shown to be impossible in a convergent series. J. T. 28 Hence we conclude as follows: If n is fractional, and if tan 0 <1, and if cos 0 is positive, then cos" O{1- In(n-1)tan 0 + n(n-1)(n-2)(n-3) tan* 0...} = cos no cos" {n tan 6- In (n-1) (n - 2) tan8 +...}=sin no where 0 is acute, and cos” 0 has its real positive value. n 620. Next let cos 6 be negative, and let q be odd. . Then taking the real negative value of y(cos”) if p is odd, and the real positive value of /(cos() if p is even, the upper ? P , line is wholly real and the lower line wholly imaginary. Now as cos 0 approaches the limit -1 and tan the limit 0, the upper line approaches the limit +1 or - 1 according 1 as p is even or odd, and the lower line the limit 0. Hence, in the , , P limit, the upper line р at and the lower line = = i sin T. 9 9 Hence, by the principle of continuity, the general values of the and lower line are, respectively, cos P upper O and i sin !0, where р a 9 0 is the angle nearest to att whose cosine = cos 0 and whose tangent = tan 0. = COS PT + i sin 621. Next let cos O be negative; and q even, but 19 odd. р and q being supposed prime to one another, p is odd. Then, by Art. 608, 22 +1 22+1 PTT 9 9 where I has any integral value. Put 21 + 1 = 19. Then one p-1 value of \(cos? A) is i 3/(- cos? 6). (-1), where Y(-cos” ) has its real positive value. Giving to /(cos” ) this value, the upper line becomes wholly P 9 2 W/(-cos” 0) {1 -P(p = 0) (tam g)*+...}=(-1) sin ; p-1 2 9 p+1 2 tan o p (P-9) (p – 29) /tan 03 3/(-cos? e) {P tone) ) + ] = (-1) ? cos P 9 where ® is some angle whose cosine cos O and whose tan = tan 0. Now as cos 0 approaches the limit - 1 and tan 0 limit 0, the upper line here approaches the limit 1 and the lower line the limit 0. Hence, in the limit, upper line =(-1) ? sin -1. P IT 2 2 ; POTT 2 P ) 0, where 0 is the , 9 9 angle nearest to £97, whose cosine cos and whose tangent tan A. p p+1 2 COS p+1 respectively, (→ 1)7*sin 2 o and (-1) COS ) COS 622. Let cos O be negative, and р and 2 values prime to one another. Then, by Art. 608, 2r + 1 2r + 1 (cosP ) =(- cos 6)" (cos pa + i sin рп 9 9 Hence giving to (- cos @)P its real positive value, we have 2r +1 P 9 12 2r + 1 p(p-1)(p-29) (tan 3 - sin a 3 9 P 0; and 9 -90,2 9 2 2r+1 p(1( ) a 9 3 = COS 2r +1 sin + COS |
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