28 Proceedings of the Royal Irish Academy. 



and n the number ya must be multiplied by to give one mile. Then the 

 problem states 



mja = 1760 x 3 x 12 . a = (n+ 64) ;>:a = (n + 40) {x + 11 ) a ; 



1 1 24 _ 1 



■'' .«■ .r + 11 " 1760 x 3 x 12 ~~ 2640 ' 



x-+llx-U x2640 = 0; 

 .'. x = 165 or - 176. 



Thus, P is 165 inches measured in the direction OA, or 176 measured in the 

 opposite direction, and n = 320 or - 424. 



If we substitute the words " diminished by 11 inches" for " increased by 11," 

 the solution is complex. To interpret the complex solution, we take a, ft' to 

 represent an inch drawn in the direction OB inclined at an angle to OA. 

 With the four fundamental units a, a, ft, ft', the length of P will be denoted 

 by xa, where x is a complex number; similarly, y and n will be complex 

 numbers, and the problem is to find complex numbers to satisfy 



n>/a = 1760 x 3 x 12a = (n + 64) xa = (n + 40) (x - 11) a ; 



1 1_ _J_ 



'" x~ x-U~ 2640' 

 x- - lis + 29040 = 0, 

 x = ix (\ +4^/959). 



And n = — 52+ 12 .v/959 . i gives us two definite operations = n + 64 which, 

 when performed on the quantities - 1 / (1 + i y/959) a respectively, give one 

 mile. 



Briefly, to summarise further work with complex numbers, we develop the 

 properties of the absolutely convergent series 



l + z + Y~2 + ! 9 3 + &c -> 



which series we call <f, so that <r .c- = r +: . Thus e x *' v = « r (cosy + ismy). 

 We define log(^-+ iy) as being such that 



e ios(*«y) = x+ iy = r (cos9+isw8) = e 10 *''" 6 , 



so that log (.r + iy) = log r + id, and has an infinite number of values, as 6 

 may be replaced by 6 ± 2Ajr. We define i p , where z and p may both be 



complex, to be e p]og: , and so is many-valued. If p is of the form + — , where 



m and n are integers, it has the a values found before. Taking one of the 

 values of : p corresponding to 6 ± 21cir, and varying 



z = x + iy = >• (cos + i sin 6) = e l0 s r + ' ^""'), 



