Here putting n= 0, 1, 2...q-1, we get a different values, since 2ni is the smallest value of x for which the expansional value of all is 1. After this the values recur. Hence If x=%; p and q being real integers prime to one another, = 9 = P ; 9 am has 9 2μπί whose common ratio is a qth root of unity, viz. e (2) Let x be wholly imaginary = yi. (dea +2nni)yi are those of e iynea é e-2nya. This gives a different value for every value of n. Hence, if x is wholly imaginary, aw has an infinite number of values, which form a geometrical progression whose common ratio is the expansional value of e (3) Let a be complex = y + zi. (dea +2nni)(y+zi) are those of eenca pizdea ,—2nzt. e2nymi. The factor e has an infinite number of values in geometric progression ; and the number of values of e2nywi is equal to the denominator of y. 2επί - 2n ZT : 644. To trace the changes in the expansional value of eai, where x is real. Writing f (xi) = the expansional value of exi, we have f (xi) xf (yi) = f {(x+y) i}. Let x (a) and iy (a) be the real and imaginary parts of f(xi). Then {x (2) + 14 (x)} {x(y) + iy (y)}= x (x + y) + 2y (oC + y). Hence equating the real and imaginary terms x (x + y) = x (x). (y)-4 ().(y) (1), y (x + y) = x (x): x(y) + x(x). + (y) .........(2). . Now since x(Qc) contains only even and y (x) only odd powers of X, :: x(-x) = x (x) and (-x)=-4 () .(3), also and .(4), and x (x - y)= x (x). (y) + y (x). (y) (5), y (ac – y) = x (x) . x(y) – x(2). ¥ (y) .. (6). From (1), (2), (5), and (6) we have x(y)-x(x) = 24 {1 (x+y)}. {1 (x - y)}. .....(7), V (x) - (y) = 2x {} (x + y)}.4 {} (x - y)}.........(8). Putting in (5) y=x, we have 1 = {x (x)} + {v (x)}. ·(9). Hence x (oc) and y (a) are arithmetically not greater than 1, Now f (2+i) = 1, .. f (xi + 27i) = f (xi); i.e. the values of f (xi) recur whenever x is increased by 27. Again, since f (2mi) = 1, .: S (T2) = J1= + 1 or - 1. But f (tri) cannot be + 1, because, by definition, 21 is the smallest value of x for which f (xi)=1. :: f(ni)=-1; i.e. X (7)=-1 and v (T) = 0, x 1 , – 7 . x(37) = 0 and ¥ (17) = +i or - i. We must show that, from x = 0 to x = 37, x (x) and y (x) are X both positive. If x is a small positive quantity, x(x) and y (x) are clearly positive. Hence, being continuous, they cannot become negative without passing through the value zero. If possible, let x (y) or y(y) = 0 for some value of x between O and . Then, since = so that i.e. {x (y)} + {y (y)} = 1 and f (yi) = x (y) + iy (y), f (yi) would equal +1 or #i. Hence f (2ye) would be = 1 and + ( f (4yi) would be 1. But f (xi) is not 1, until x = 27, .. neither 4 (y) nor x(y) can vanish if y is between 0 and 17. :: y(x) and x (x) are always positive from x = 0 to x = }. 4 .: x(-) = 0 and 4 (-1) = -1. 1 :: x(47) =ų (IT) since {x (2)} + {y (x)} = 1: and so on. 12 Finally, by (7) and (8), as long as } (x + y) and } (x - y) lie between 0 and 3, 4 (x) must increase and x (2) must decrease as x increases. Thus, between 0 and 17, V(x) continually increases from 0 to 1; and x () continually decreases from 1 to 0. Hence the values of ų (x) and x (oc) cannot be repeated as x passes from 0 to 2. To find a value for de (1+y), i.e. a root of the 645. equation е + + =y; The value of de (1 + y) has to satisfy the condition {łe (1 + y)} {re (1 + y)}} 1. (1+y)+ 2 B Y 1. de (1 + y) If then de (1 + y) = co + C1y+ cuya + czy3 + ..., we must have co :0, and Ci 1; so that ...) we have = a Writing f (c) for the series 1 + x + 3x + frites + f (a) f (x) = f (x + x'); :: if a is a root of f(x) = 1 + y; and a' of f (x) = 1 +z; then a +a' is a root of f (c) = (1+y)(1 + 2) = 1 +y+(1+y) 2. That is, we may write 1. (1+y)+1, (1 + x) = 1, {1+ y + (1 + y) }; :: 1 (1+y)+z+cqx2 + ... ={y +(1+y)2}+ ... +cr{y +(1+y) z}" +... + =); ; Equating the coefficients of , we have 1=1 + y + ... + rcr y^-*(1 + y) + (r + 1) Cr+1 Y" (1+y)+... Equating to zero the coefficient of y”, we have e + Putting r successively equal to 1, 2, 3... we have (-1)" Cr+1 ; :: 1 (1+y)=y- }y + y – 444 + ... which is a convergent series if mod. y is not greater than 1. [For, by Art. 596, mod. y" = (mod. y)". See also Arts. 616, 453, 445.] We must here distinguish the cases according as y is wholly real, wholly imaginary, or complex. (1) Let y be wholly real. +11 ( 1 + zi .:. 21 (3 – 4* + 3x3 - ...) = 21. (1 + zi) – de (1 + x^)=de ** 2i 2 ) 1 1 + zi is wholly imaginary. Now when z equals 0, the 1 series 0; hence as z varies from 0 to 1 or from 0 to - 1, the series changes continuously. + Thus de Now between – ja and + 37 there is only one value of a for which the expansion of eit has any given value. 1+zi Hence the series 2i (2 – 3x3 + 3x –...)= the value of de 1 which lies between - and + žt. :.de (1 + zi) = } the real value of loge (1+za) 1 + zi 1 . 1 + zi/(1+y) A. (1 + y + 2) = x (1 + 2 + y + 2) + , 1 – zi/(1+y)' which reduces to cases (2) and (3). 646. Hence we can find an expansion for 7. 1 + i 1 + i Thus f (i)=i= ..de 1 i 1-i žti. Hence putting =1 in (2) of last article, 1 in = 1 - 3 +}-}+... 1 + x + y + xy 1 1 - 2 – y + xy Thus putting x = ji and y = ji, we have 1 + ži 1+ 3 +de = mi. 1 -2 1 1 5 1 + x 1 + Y = 1 + i he 1-i de 1-i 1 647. The indeterminateness of the powers and logarithms of quantities, which arise from our introducing imaginary indices, gives rise to some difficulties in the management of equations. Thus given that a=b, we have The general values of aare the expansional values of (teb+2017), for the several integral values of n. 1b + 2nti And the general values of loge a are 1 + 2mai |