15. By giving to r the values 0, 1, 2 q-1 we obtain

...

ever integral value we give to r, we cannot obtain more than

Take a circle, centre O and radius OR.

1

2

starting from Po, into q equal arcs, P ̧P1, P ̧P,, P ̧P ̧, etc. Then each of the angles POP, POP, POP, ... etc. is

etc.

volving line, starting from OR, turns first into the position OP, and then on through r of the angles POP1, POP„ Hence, whatever integral value r may have, OP must stop in one of the q positions OP, OP1, OP, etc. and it can stop in no other position.

Also no two of these q positions are equi-sinal and at

the same time equi-cosinal.

Therefore this expression has q different values.

...

Also by giving to r the values 0, 1, 2, (9-1), in succession, OP will be made to stop in each of the q possible positions in turn.

Therefore by giving to r the values 0, 1, 2...(q-1) in succession, we obtain the q different values of the above expression. Q. E. D.

16. An expression of the form A +/-1 B, where A and Bare arithmetical quantities, can always be put into the form r {cos a +/-1 sin a}.

It will be convenient to take r positive: then we must take a in that quadrant which makes cos a the same sign as A. [Cf. E. 148, 149.] EXAMPLE 1. Express 1+ √1 in the form r (cos a+ √−1 sin a). Here r sin a= 1 and rcos a=1, ... r2=2, tana=1.

1+√-1=√√2{cos 45o +/- 1 sin 45°}.

Example 2. Express (− a) in the form r (cos a+isin a).
Here r cos a = a, r sin a=0, .'. r2=a2, a=(
= (2n+1) π.

.. -a=a{cos (2n+1) π + i sin (2n+1) π}, where n is an integer.

form

EXAMPLES. VII.

(1) Express 1-√ − 1, No3+ √ −1, 1+√3-1 each in the r(cos a+1 sin a).

(2) Find all the values of (i) (4√2+4√2√ − 1)3, (ii) (4√3+4√−1)3, (iii) (√/3+√−1)3.

(3) Find all the values of (i) 1a, (ii) 323, (iii) 273.

17. If we express any arithmetical quantity a in the form of a De Moivre's expression we obtain

a (cos 2rπ + √- 1 sin 2rπ),

i.e. the product of a by the De Moivre's expression for unity. Therefore the n nth roots of any arithmetical quantity a are found by multiplying the arithmetical nth root of a by each of the n nth roots of unity in succession.

The nth roots of unity are therefore important, and are discussed in the following examples.

EXAMPLE 1. Solve the equation x2-1-0. In other words, find all the values of /1, or, find the factors of x1 – 1.

It follows that x= cos 2rπ+1 sin 2r, where r is an integer,

This result is best discussed by means of a figure.

centre O and radius OR, measure off arcs P,P2, PP, etc. each equal to RP1. Then since n. ROP1=2π, n of these arcs will occupy the whole circumference, and OP will coincide with OR. Also, if r be a

2TT

whole number, in describing the angle the revolving line, starting

n

from OR, must stop in one of the positions OP1, OP, etc. and in no other. No two of these positions are both equi-sinal and equi-cosinal.

has n different values, and no more; and these values can be found by giving r in succession the values 0, 1, 2...n - 1.

When r=0, x=1: when n is odd this is the only arithmetical value; when n is even, there are two arithmetical values; for let n=2m, then when r=m, x= -1.

In any case, the angles ROP, and ROP, n-1 are equi-cosinal, and sin ROP1= − sin ROPɲ-1• The same thing is true of ROP2, ROP and of ROP, ROP, and so on.

Hence

n-3,

2-2,

Therefore the roots of the equation x-1=0 are 1, a, a2, a3... a^~1.]

II. When n is a fraction in its lowest terms?. Then x-1=0,

q

or x2-1o=0, or x2-1=0. This is the same as the case already

discussed.

III. When n is incommensurable (e.g. √2). Then as before

In this case, r being an integer and n incommensurable,

never be an exact multiple of 2. The angles will therefore not recur geometrically and the equation will have one arithmetical root, viz. 1, and an unlimited number of symbolical roots.

(1) Find the roots of the equation x5 – 1=0.

(2) Find the quadratic factors of x3 – 1.

(3) Write down the quadratic factors of 13 — 1. (4) Solve the equation îo—1=0.

(5) Give the general quadratic factor of 20 - a20. (6) Find all the values of 1/1.

EXAMPLE 2. To find the Quadratic factors of xn+1=0.

In the figure ROP。:

=

π

n

2π n

POOP1= and n angles each equal to

POP, make up 2; OR bisects POP-1. Also ROP, and ROP n-1 are equi-cosinal, while sin ROP。=-sin ROP the same relation holds good for any two angles equi-distant from OR,

n-1,