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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) * (1-j)/2 -exp(-j*1*pi/4)=- sqrt(2) * (1-j)/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) * (1-j)/2 -exp(j*3*pi/4)= sqrt(2) * (1-j)/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) * (1-j)/2; -sqrt(2) * (1-j)/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
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Joined: Oct 2010
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Re: Finding roots (complex numbers)

Quote:
 Originally Posted by MarkFL See also de Moivre's theorem and roots of unity.
Nice. That's it 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 n-gon in the complex plane. A regular (2^n)gon (like an octagon) should be fairly simple to visualize. Tags complex, finding, numbers, roots ,
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