Page:Encyclopædia Britannica, Ninth Edition, v. 23.djvu/589

Rh has $$\frac{1}{n}(B_nOA + 4\pi)$$ for the 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 $$(\cos\theta + i\sin\theta)^{\frac{1}{3}}$$. Suppose B to represent the expression $$\cos\theta + i\sin\theta$$. If the angle $$AOP_1$$ is $$\frac{1}{3}\theta$$, $$P_1$$ represents the root $$\cos\frac{\theta}{3} + i\sin\frac{\theta}{3}$$; the angle AOB is filled up by the angles of the three similar triangles $$AOP_1, P_1Op_1, p_1OB$$. Also, if $$P_2$$, $$P_3$$ be such that the angles $$P_1OP_2$$, $$P_1OP_3$$ are $$\frac{2\pi}{3}$$, $$\frac{4\pi}{3}$$ respectively, the two sets of triangles $$AOP_2$$, $$P_2Op_3$$, $$p_3OB$$ and $$AOP_3$$, $$P_3Op_2$$, $$p_2OB$$ satisfy the conditions of similarity and of having OA, OB for the bounding sides; thus $$P_2$$, $$P_3$$ represent the roots $$\cos\frac{\theta + 2\pi}{3} + i\sin\frac{\theta + 2\pi}{3}$$, $$\cos\frac{\theta + 4\pi}{3} + i\sin\frac{\theta + 4\pi}{3}$$ 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 circle.

The problem of determining the values of the nth roots of unity is equivalent to the geometrical problem of inscribing a regular polygon of n sides in a circle. Gauss has shown in his Disquisitiones arithmeticæ that this can always be done by the compass and ruler only when n is a prime of the form $$2^p + 1$$. The determina tion of the nth root of any complex quantity requires in addition, for its geometrical solution, the division of an angle into n equal parts.

We are now in a position to factorize an expression of the form $$x^n - (a + ib)$$. Using the values which we have obtained above for ib) n, we have s=0 If& = 0, a=l, this becomes + 2S7T cos H 0J-2S7T n ...(1). s=i-l x n -l = P s=0 X-( !&quot;~V 2STT,. 2STT [ x cos i sin =l V 5t n ) n r2 ~ (a?-2xcos^ + l eveu)(2). 5= o / 9cir aj-l = (a?-l)P (s-2xcoa- --+11 (ra odd) ...(3). s=l V / If in (1) we put a= - 1, 6 = 0, and therefore = 7r, we have x n + 1 = P a: - cos 1 sn s =o L n = P x 2 -2xcos hi I (n even) (4). s=0 I l J s = Also x- n - 2x&quot;y n cos n6+y- n fa? - 2x cos %2l + 1~| ( B odd) (5). L n J = (x n - y n cos n6 + 1 sin 7i D =7l ~Y = P [x- s=o V - y&quot; cos nd - 1 sin ?i0) o S = n-l = P s=0 + 2*77-, . ycos - rttrin 5&quot;! - + y 2 n J (6). Airy and Adams have given proofs of this theorem which do not involve the use of the symbol i (see Camb. Phil. Trans., vol. xi.). A large number of interesting theorems may be derived from De Moivre's theorem and the factorizations which we have deduced from it; we shall notice one of them.
 * = n - l r 2s + ITT

In equation (6) put $$y = \frac{1}{x}$$, take logarithms, and then differentiate each side with respect to x, and we get

Put $$x^2 = \frac{a}{b}$$, then we have the expression

for the sum of the series

We shall now consider what meaning can be assigned to the symbol $$e^{x + iy}$$. The quantity e is defined as the limit of $$(1 + \frac{1}{n})^n$$ where n is a positive quantity, and is increased indefinitely; then, for a real value of x, $$e^x$$ is the limit of $$(1 + \frac{1}{n})^{nx}$$ or $$(1 + \frac{x}{m})^{\frac{1}{m}}$$ where $$m = nx$$, when m is increased indefinitely. We may define $$e^{x + iy}$$ as the limit of $$(1 + \frac{x + iy}{m})^{\frac{1}{m}}$$ when m is increased indefinitely. To determine the value of this limit put $$1 + \frac{x}{m} = r\cos\theta, \frac{y}{m} = r\sin\theta$$; then e x+ly is the limit of r m (cos70 + t sin7?z0), and r m is equal to S9^. ^4-2 / 9v - 1 + f + 1 ( 2 or ultimately to ( 1 + ) 2 which has f for m m 2 J J n/ its limiting value. Also is arc tan in the limit ; x + m x + m hence 7n0 is ultimately equal to y, and thus the equation e x+l y=e c (cosy + i sin?/) follows from our definition. It may be shown at once that e*+ i y x e*i+ t 2 i= e x + x i+ l &amp;lt;y+yi an d, if we suppose that a x+iy denotes e^ +l ^ loga, we may show that complex expon ents defined thus obey the same laws as real ones.

When the exponent is entirely imaginary we have, in accordance with the above definition,

$$e^{iy} = \cos y + i\sin y$$ and $$e^{-iy} = \cos y - i\sin y$$; we thus obtain the exponential values of the sine and cosine—

If we give imaginary or complex values to the variables in algebraical expansions we obtain analogous trigonometrical theorems; it is, however, necessary to consider the convergency of the series so obtained in order to determine within what limits the values of the variables must lie. If we expand e?U and e~ y by putting iy ifl i/ 3 and their and-ty in the series l + y+ jS+s o~ 5+ we obtain the series sin y = y - ^~- + .-^- - ^=- I I

These series are convergent for all finite values of y. They may also be got from the expressions which we have obtained for the cosine and sine of a multiple of an angle in terms of the cosine and sine of the angle, and would thus be made to rest upon a basis independent of the symbol ı.

Consider the binomial theorem /_ i T. -_n i n n ~*-i~n-iT.i , 7^(7^-l)... (n-r+l) n _ rbr ^ E Putting a=e ıθ, b=e -ıθ , we obtain (2 cos 0)&quot; = 2 cos 710 + 2 cos 71 - 20 + j~ ^ 2 cos 71 - 40 + ... 71(71 - 1) ... (n - r - cos(n - 2r)0 + . ..

When n is odd the last term is 2 cos e ... n(n-).. . and when n is even it is J - ,, - , I In If we put a=e e , b= -e~^, we obtain the formula n (-if (2 sin 0)&quot; = 2 cos ?i0 - 2n cos(n - 2)0 + - ( ^ ~^2 cos(n - 4)0 - ... when 7i is even, and n-l (-1) 2 (2 sin 0)&quot;= 2 sin 710 -?i. 2 sin (TI -2)0 + 71-1 1.2 -2sin(?i-4)0... when ?i is odd. These formulae enable us to express any positive integral power of the sine or cosine in terms of sines or cosines of multiples of the argument. There are corresponding formulse when 71 is not a positive integer. XXIII.—72