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 ƒ(m) may be itself complex. Then, if ƒ (m) ׃ (n) =ƒ (m+n), ƒ (n) is one of the values of ƒ (1)". This theorem applies, as shown in Arts. 465, 466, to any series of the form of the Binomial Expansion, whether x be complex or real, provided n is real. It will remain, therefore, to determine to which of the values. of ƒ (1)" the series must be equated, when n is fractional. 617. In De Moivre's theorem, let n be a positive integer. Then cos n✪ + i sin n✪ = (cos 0 + i sin 0)". Expanding, by the binomial theorem for a positive integer, and equating the real and imaginary terms, we have cot no = n (n − 1) (n − 2) n tan 0-1 n (n − 1) (n − 2) tan3 0 + 1 - 1 n (n − 1) tan2 0 + ... n cot"-10-n (n-1) (n-2) cos"-30 + ... These results have already been obtained (Art. 359). 618. Let n be a negative integer, =-m say. (1) Let tan < 1. Then cos no + i sin no = (cos 0 + i sin 0)" = cos” 0 (1 + i tan 0)”. We may expand in ascending powers of tan 6, since tan ◊ is <1. Thus, equating real and imaginary terms, cos n0 = cos2 0 {1 − n (n − 1) tan2 0 + .....} sin nê = cos” 0 {n tan 0 − ồn (n − 1) (n−2) tan30+..} where n is negative. Or cos no + i sin no = (cos 0 + i sin 0)” = sin” 0 (i + cot 0)”. .. ¿ ̄a cos n✪ (n even) or i-n+1 sin n✪ (n odd) =sin" 0 {1-n (n-1) cot30+...} ¿−n+2 sin no (n even) or i−" where n is negative. m = sin" 0 {n cot 0 -n (n-1) (n-2) cot3 0+...} Or m-1 (− 1) cos me (m even) or (− 1) 2 sin me (m odd) m+2 = cosecm 0 {1 – 1m (m + 1) cot2 0 +...} m-1 (-1) sin me (m even) or (-1) cos me (m odd) = cosecm 0{m cot 0 – 1m (m + 1) (m + 2) cot3 0 +...} where m is positive. Then, by De Moivre's theorem, the q values of (cos + i sin are given by cos2+isin 2 +isin where is any angle whose g Then, by the binomial theorem, since p/q is real so that the argument of Art. 466 holds although cos + i sin is complex, where, by Art. 606, the multiplier /(cos" ) has q values; and the series into which it is multiplied is finite and single-valued, since tan < 1. Now suppose cos 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. where is some angle whose cosine = cos and whose tangent tan 0. = It remains to determine which of the alternative values of is to be taken. To do this we apply the principle of continuity. Thus As cos approaches the limit + 1 and tan 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 20 and the lower line = 2 sin Hence the general values of the upper and lower line are, respectively, cose and i sin 20, where 0 is the (positive or negative) acute angle whose cosine: = cos 0 and whose tangent For, if any other values, e.g. cos Ρ Ρ (2π + 0) and i sin (2π + 0) were given, then as approached the limit 0, the two series Ρ would approach the limits cos 2 and i sin 2, and yet when P 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" 01-n(n-1) tan20 + 1m (n-1) (n − 2) (n − 3) tan1 0...} = cos no cos" {n tan 0 - } n (n − 1) (n − 2) tan3 0 +...} = sin n0 where 0 is acute, and cos" has its real positive value. 620. Next let cos @ be negative, and let q be odd. Then taking the real negative value of (cos") if p is odd, and the real positive value of (cos? 0) if p is even, the upper line is wholly real and the lower line wholly imaginary. Now as cos approaches the limit – 1 and tan the limit 0, the upper line approaches the limit +1 or 1 according as p is even or odd, and the lower line the limit 0. Hence, in the Hence, by the principle of continuity, the general values of the upper and lower line are, respectively, cos Ø and i sin Ρ q is the angle nearest to q whose cosine tangent = tan 0. P 0, where = cos 0 and whose 621. Next let cos 0 be negative; and q even, but 19 odd. Then Ρ and q being supposed prime to one another, p is odd. Then, by Art. 608, where has any integral value. Put 2x+1= 19. Then one p-1 value of 2/(cos? 0) is i2/(- cos? 0). (− 1), where 2/(— cos" 0) has its real positive value. Giving to /(coso 0) this value, the upper line becomes wholly imaginary and the lower line wholly real. = Now as cos approaches the limit 1 and tan the limit 0, the upper line here approaches the limit 1 and the lower line the limit 0. p-1 Hence, in the limit, upper line = (-1) sin ?. 2+1 and lower line = (-1) 2 cos P qπ q 2 Hence the general values of the upper and lower line are, respectively, (-1) sin? and (-1) p-1 p+1 cos? 0, where is the 2 9 = cos and whose tangent 622. Let cos be negative, and p and prime to one another. Then, by Art. 608, Hence giving to /(— cos 0)o its real positive value, we have 2+1 pw × 2 (− cos 6)p { 1 _ P (p ≥ 4) (tan 6)2 + ...} 12 (tane p(p-q) (p-2q)/tan0 Ρ 1 3 Չ 3 sin 3 |