Lecture 9 Complex Analysis

1.10 Extraction of Roots:-
The problem of extracting the nth roots of a complex number Z is that of solving the equation Z_0^n=Z for Z_0 , when Z and positive integer n are given.
Let the polar form of Z be Z=r(cos??+i sin?? ) and Let Z_0=r_0 (cos???_0 ?+i sin???_0 ? ) where r_0 and ?_0 are as unknown. Then equation Z_0^n=Z becomes
??[r_0 (cos???_0 ?+i sin???_0 ? )]?^n=r(cos??+i sin?? )
?r_0^n (cos???_0 ?+i sin???_0 ? )^n=r(cos??+i sin?? )
?r_0^n (cos???n??_0 ?+i sin???n??_0 ? )=r(cos??+i sin?? )
?r_0^n=r & ?n??_0=?+2k? where k=0,1,…,n-1
?r_0=r^(1/n) & ?_0=(?+2k?)/n Where k=0,1,2,…,n-1
Thus
Z_0=r^(1/n) (cos??(?+2k?)/n?+i sin??(?+2k?)/(n )? ) where k=0,1,2,…,n-1
Let Z_1^(n/m)=Z then
??[r_1 (cos???_1 ?+i sin???_1 ? )]?^(n/m)=r(cos??+i sin?? )
?r_1^(n/m) (cos???_1 ?+i sin???_1 ? )^(n/m)=r(cos??+i sin?? )
?r_1^(n/m) (cos???n/m ??_1+i sin???n/m ??_1 ? ? )=r(cos??+i sin?? )
?r_1^(n/m)=r & n/m ?_1=?+2k? , k=0,1,…,n-1
?r_1=r^(m/n) & ?_1=m(?+2k?)/n , k=0,1,…,n-1
Thus
Z_1=r^(m/n) (cos??m(?+2k?)/n?+i sin??m(?+2k?)/n? ), k=0,1,…,n-1
Example:-
Find the cube roots of (?6-i?2)
Find the square roots of (-2-i2?3)
Find the cube roots of (-1-i)
SOL:-
n=3?k=0,1,2 ,?=tan^(-1)??((-?2)/?6)=-tan^(-1)??(1/?3)=-?/(6 )? ?
r=?((?6)^2+(?2)^2 )=?8=2?2
When k=0?Z_0=(2?2)^(1/3) (cos??(-?/6+2(0)?)/3?+i sin??(-?/6+2(0)?)/3? )
=?2 (cos??(-?/6)/3?+i sin??(-?/6)/3? )=?2 (cos??(-?)/18+i sin??(-?)/18? ? )
When k=1?Z_1=(2?2)^(1/3) (cos??(-?/6+2(1)?)/3?+i sin??(-?/6+2(1)?)/3? )
=?2[cos?((-?)/18+2?/3)??+isin?((-?)/18+2?/3)]?
When k=2?Z_2=(2?2)^(1/3) (cos??(-?/6+2(2)?)/3+i sin??(-?/6+2(2)?)/3? ? )
=?2[cos??((-?)/18+4?/3)+i sin??((-?)/18+4?/3)]? ?
Thus the roots is Z_0=?2 (cos??(-?)/18+i sin??(-?)/18? ? )