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October 19th, 2010, 01:38 PM   #1
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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.
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October 19th, 2010, 05:38 PM   #2
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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
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October 19th, 2010, 06:50 PM   #3
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Re: Finding roots (complex numbers)

See also de Moivre's theorem and roots of unity.
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October 19th, 2010, 07:13 PM   #4
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Re: Finding roots (complex numbers)

Quote:
Originally Posted by MarkFL
Nice. That's it
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October 19th, 2010, 08:48 PM   #5
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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.
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