50 



when it is counter clockwise, is the angle of the number. Thus, let (2, oO°) 

 denote a number that doubles length and turns 30° counter clockwise. Its 

 tensor is 2, its vei'sor is (1, 30°), and its angle is 30°. 



After multiplying a step by (2, oO°) multiply the result by (:>, 20°). 

 Plainly the final step is (6, 50°) times the first step. This example of a 

 product enables us to see at once that : 



The tensor of a product equals the product of the tensors of the factors ; 

 and the angle of a product equals the sum of the angles of the factors. 

 Hence the factors may be combined in any order without altering their 

 product. 



The definition of a sum of two numbers p and q is that (p + q) O B=: 

 p O B - q B. ■• Replacing O B by r O A we have that (p — q) r = p r + q r; 

 and since the factors of a product have been shown to be interchangeable, 

 therefore r(p-i-q) = {p — q)r=rp + rq. 



We thus find that these versi-tensors follow the ordinary laws of alge- 

 braic combination. To identify them with imaginaries, notice that (1, 90°)- 

 = (1, 180°) = — 1 = (1, —90°)-, These two square roots of —1 are nega- 

 tives of each other, for —1 (1, —90°) = (1, 180°) (1, —90°) = (1, 90°). So 

 — 1 has three cube roots, — 1 and (1, =b 60°); and so on. 



It is convenient to represent versi-tensors by steps. Some step O A is 

 taken to represent unity ; and then any other step represents its ratio to 

 the unit step O A. Thus, if B, O B^ are steps of the same length as A, 

 and make angles of 60° and — ()0° respectively with O A, they represent the 

 imaginary cube roots of — 1. AVe may use geometry to put these roots in 

 the standard form x y i, where x and y are real numbers and i = (1 , 90°). 

 Let BBi meet O A in C; then OC represents, or say =, i, and CB= 

 i l/7 i = — C Bi ; and from O B = O C ^ C B, O B^ = C + C B^ we have 

 (1, ± 60°) = i ± ^, 7 i. 



This example just given makes it plain that any imaginary number may 

 be put in the form x ^ y i, in one and only one way ; and from the right 

 triangle involved, we also see that the tensor of x + y i is v x- + y-, the 

 so-called modulus in imaginaries. It is easy to show by geometry how it 

 is that every equation with real or imaginary co-efficients has at least one 

 root, and therefore just as many roots as its degree and no more, or even to 

 show the whole directly. ♦ In fact, all the fundamental properties of imag- 



"To see that this does detiiie the sum, try it for the case of p = {2, 30 ), q= (2, 150 ), 

 which gives p -r q= (2. 90' i. Also compare with the verification that 2-j-i= 5. 



