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October 19th, 2010, 01:38 PM  #1 
Member Joined: Apr 2010 Posts: 91 Thanks: 0  Finding roots (complex numbers)
Find the 8th roots of 1 and sketch the roots in the complex plane. I am not sure what to do. I know that the 8th roots of 1 is 1, but obviously thats not the case. 
October 19th, 2010, 05:38 PM  #2 
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Finding roots (complex numbers)
so ,you know j^4=1 (j)^4=1 what are the sqrare roots of j and j? we can express this more generally exp(j*2*n*pi)=1 and exp(j*2*n*pi)=1 ,where n are integers so the 8th roots of 1 are exp(j*n*pi/4) exp(j*n*pi/4) exp(j*n*pi/4) exp(j*n*pi/4) since exp(j*n*pi/4) and exp(j*n*pi/4) are functions with period=8 we can manage the values Only For n=0,1,2,3 (exp(j*(4+n)*pi/4)= exp(j*n*pi/4) and exp(j*(4+n)*pi/4)= exp(j*n*pi/4)) So, while n=0 exp(j*0*pi/4)=1 exp(j*0*pi/4)=1 exp(j*0*pi/4)=1 exp(j*0*pi/4)=1 while n=1 exp(j*1*pi/4)= cos(pi/4)+j*sin(pi/4)= sqrt(2)/2 +j* sqrt(2)/2 =sqrt(2) * (1+j)/2 exp(j*1*pi/4)=  sqrt(2) * (1+j)/2 exp(j*1*pi/4)= sqrt(2) * (1j)/2 exp(j*1*pi/4)= sqrt(2) * (1j)/2 while n=2 exp(j*2*pi/4)=cos(pi/2)+j*sin(pi/2) =j exp(j*2*pi/4)=  j exp(j*2*pi/4)=  j exp(j*2*pi/4)= j while n=3 exp(j*3*pi/4)=cos(3*pi/4)+j*sin(3*pi/4) =  sqrt(2) * (1j)/2 exp(j*3*pi/4)= sqrt(2) * (1j)/2 exp(j*3*pi/4)= sqrt(2) * (1+j)/2 exp(j*3*pi/4)=  sqrt(2) * (1+j)/2 therefor the 8th roots of 1 are: sqrt(2) * (1+j)/2 ; sqrt(2) * (1+j)/2 ; sqrt(2) * (1j)/2; sqrt(2) * (1j)/2 ; 1; 1; j ; j 
October 19th, 2010, 06:50 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Finding roots (complex numbers)
See also de Moivre's theorem and roots of unity.

October 19th, 2010, 07:13 PM  #4  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Finding roots (complex numbers) Quote:
 
October 19th, 2010, 08:48 PM  #5 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,767 Thanks: 5  Re: Finding roots (complex numbers)
Since Mark loves an echo... There are "n" nth roots of unity, and they form a regular ngon in the complex plane. A regular (2^n)gon (like an octagon) should be fairly simple to visualize. 

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