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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 +
1. 2

n (n − 1) (n − 2)
sin no
= n cosN-1 A sin A.

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 + ...
cot no

n coth-10-in(n-1) (n-2) cosu-36 + ...
These results have already been obtained (Art. 359).
618. Let n be a negative integer, =- m say.
(1) Let tan 0<l. Then

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

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cos mo

secm

(2)

=

secm @{1

įm (m + 1) tan’ A +...}
sin mo 0 {m tan 0 įm (m + 1)(m + 2) tan? 6 + ...}
where m is positive.

Let tan 0>). Then
cos no + i sin n0 =(cos 8 + i sin 0)" =sin” (i + cot 6)".
.. * n cos no (n even) or i-n+1 sin no (n odd)

= sin" @{1 - In(n-1) cot? @ +...}
-n
-1+2 sin n0 (n even) or ;-n+2 cos no (n odd)

1

= sin" @ {n cot 0 In (n 1)(n- 2) cot: 0 +.. where n is negative. Or

m

m-1
2

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] cos2 9

4

9

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+

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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

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PT + i sin

621. Next let cos O be negative; and q even, but 19 odd.
Then

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
imaginary and the lower line wholly real,
Thus
P-9

P
(1
2

9

2

W/(-cos” 0) {1 -P(p = 0) (tam g)*+...}=(-1) sin ;

p-1

2

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9

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p+1

2

tan o p (P-9) (p 29) /tan 03 3/(-cos? e) {P

tone)

) + ]
a
3

= (-1) ? cos
** ?

P
cos? ; ф

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

;
and lower line = (-1)

POTT
a

2
Hence the general values of the upper and lower line are,
р 1 st

P )

0, where 0 is the

, 9

9 angle nearest to £97, whose cosine cos and whose tangent

tan A.

p

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p+1

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2 COS

p+1

respectively, (→ 1)7*sin 2 o and (-1)

COS

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)

COS

622. Let cos O be negative, and р

and

2
have
any

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

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= COS

2r +1 sin

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+ COS

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