Complex Analysis- Lecture 10 - Roots of Unity

Z_1=?2[cos??((-?)/18+2?/3)+i sin??((-?)/18+2?/3)]? ?
Z_2=?2[cos??((-?)/18+4?/3)+i sin??((-?)/18+4?/3)]? ?
n=2?k=0,1
?=tan^(-1)??((-2?3)/(-2))=tan^(-1)??(?3)=? ? ?/3
r=?(4+12)=?16=4
When k=0?Z_0=(4)^(1/2) (cos??(?/3+2(0)?)/2?+i sin??(?/3+2(0)?)/2? )
=2(cos???/6+i sin???/6)? ?
=2( ?3/2+i 1/2)
=?3+i
When k=1?Z_1=(4)^(1/2) (cos??(?/3+2(1)?)/2?+i sin??(?/3+2(1)?)/2? )
=2[cos??(?/6+?)+i sin??(?/6+?)]? ?
=2[- cos???/6-i sin???/6]? ?
=2[-?3/2-1/2 i ] =-?3-i
Thus the roots is:-
Z_0=?3+i , Z_1=-?3-i
1.11 The nth roots of unity:
We want to find the equation
?^n=1
Since k=0,1,…,n-1 Let Z=x+iy=1?x=1 ,y=0
Then ?=tan^(-1)??(0/1)=tan^(-1)??(0)=0? & r=?(x^2+y^2 )?

=?(1^2+0^2 )=1
So that ?=r^(1/n) (cos??(0+2k?)/n?+i sin??(0+2k?)/n? )
=(1)^(1/2) (cos??2k?/n?+i sin??2k?/n? )
Thus ?(?=(cos??2k?/n?+i sin??2k?/n? ) ,k=0,1,2,…n-1)
Since ?_0=cos??(2(0)?)/n?+i sin??(2(0)?)/n?=cos?0+i sin?0=1
?_1=cos??(2( 1)?)/n?+i sin??(2(1)?)/n=cos??2?/n?+i sin??2?/n? ?
?_2=cos??(2( 2)?)/n?+i sin??(2(2)?)/n=cos??4?/n?+i sin??4?/n? ?
=cos?(2?/n+2?/n)+i sin?(2?/n+2?/n)=(cos??2?/n?+i sin??2?/n? )(cos??2?/n?+i sin??2?/n? )
Thus ?_2=?_1 ?_1=?_1^2Also ?_3=?_1 ?_1 ?_1=?_1^3. Thus the nth roots of unity is
?(?(1,?_1,?_1^2,?_1^3,…….,?_1^(n-1)@where ?_1=cos??2?/n?+i sin??2?/n? ))
In complex plane the nth roots of unity are the vertices of regular polygon of n sides inscribed in circle |Z|=1 with one vertex at the point Z=1


EX:-Find all the 5th roots of unity
Sol:- we get n=5
?_0=1
? ??_1=cos??2?/5?+i sin??2?/5?
? ??_2=?_1^2=cos??4?/5?+i sin??4?/5?
? ??_3=?_1^3=cos??6?/5?+i sin??6?/5?
?_4=?_1^4=cos??8?/5?+i sin??8?/5?
H.W:-Find the square roots of unity?