6-Root unity (polynomial roots):-
Suppose we want to find out n-the roots of unity
i.e. we want to solve zn=1
suppose , z=rei? , r>0
then , zn=1 means
(rei?)n=1=ei2k? , k=0,?1,?2,…..
rn ein?=ei2k? , k=0,?1,?2,…..
rn=1 ,n?=2k? , k=0,?1,?2,……
so ,
r=1 and ?=2k?/n , k=0,?1,?2,………….
Then the solutions of zn=1 are given by
Z= rei?=1.ei2k?/n= ei2k?/n , k=0,?1,?2,………….
=cos(2k?/n)+isin(2k?/n) , k=0,?1,?2,………….
Since |z|=1 ,here the roots lying on the unit circle having center at origin ,these roots are equally spaced on the boundary of the unit circle at each 2?/n radians .
Because of the circular structure it follows that
The roots are given by ei2k?/n , k=0,?1,?2,………….
Suppose k=n , then
ei2k?/n=ei2?=1
so, k=n case is same as k=0 case
if k=n+1
ei2k?/n=ei2((n+1)/n)? =ei2?.ei2?/n= ei2?/n
which means k=n+1 case is same as k=1 case
therefore ,the distinct n-roots of 1 are given by
ei2k?/n , k=0,1 ,…..,n-1
?these roots are the vertices of a regular polygon having n-sides in the unit circle having center at origin ,these polygon has always a vertex at 1 if wn= ei2?/n ,then the roots are given by
1, wn, wn2,……………, wnn-1
-In the general case suppose we want to find the roots of
Zn=z0 , z0?0
Suppose Z= rei? ,r>0 and Z0= r0ei?0 ,r0>0
(rei?)n=( r0ei?0)
? rnein? = r0ei?0
So that
rn= r0 while n?=?0+2k? , k=0,1,………n-1
thus
r=r01/n , ?=(?0+2k?)/n , k=0,1,………….,n-1
then the roots will be given by
z= r01/n.ei((?0+2k?)/n) , k=0,1,……….,n-1
definition:
the root obtained by taking into consideration the principal value of the argument is called the principal root.