Primitive roots of unity

  • Thread starter Driessen12
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  • #1
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Homework Statement


I must show that cos(2pi/n) + isin(2pi/n) is a primitive root of unity


Homework Equations


a primitive root of unity is an nth root of unity that does not equal 1 when raised to the kth power for k less than n and great than or equal to 1


The Attempt at a Solution


If we set z = cos(2pi/n) + isin(2pi/n) then z^k cannot equal 1. we can use de moivres theorem to make z^k = cos(2kpi/n) + isin(2kpi/n) and then i'm not certain what fact to use next
 

Answers and Replies

  • #2
lanedance
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how about writing it as
[tex] cos(\frac{2 \pi}{n}} ) + i sin(\frac{2 \pi}{n}} ) = e^{\frac{2 \pi}{n}} [/tex]
 
  • #3
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so then e^(2kpi/n) would be what i get. From there all i would need to show is that 2kpi/n cannot be zero, correct?
 
  • #4
Char. Limit
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how about writing it as
[tex] cos(\frac{2 \pi}{n}} ) + i sin(\frac{2 \pi}{n}} ) = e^{\frac{2 \pi}{n}} [/tex]
But that's not true. You're off by a factor of i in the exponential.
 
  • #5
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right, but to prove that z^k is not equal to zero i would have e^(i2πk/n) and this cannot equal 1 if we restrict k to be greater than or equal to 1 and less than n. So n cannot be zero and k cannot be zero, so i2(pi)k/n cannot be zero, thus z^k cannot equal 1 and is therefore primitive. and k is not equal to n so we can never have e^(i2(pi)). but how would i prove that the exponent can never be pi
 
Last edited:
  • #6
lanedance
Homework Helper
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yeah cheers mis-typed the i

well its a given that k<n, so k/n < 1, so (k/n)2.pi <2.pi, isn't that enough
 
Last edited:
  • #7
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no because sin(pi) is also zero and cos(pi) is 1 giving us 1 which is exactly what I have to prove cannot happen. For example choose k to be 2 and n to be 4, then we have e^(i(pi)) which is 1. I am not sure how to prove that this cannot happen though. I need to show that it is impossible to have any multiple of pi as my argument. any ideas?
 
  • #8
Char. Limit
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no because sin(pi) is also zero and cos(pi) is 1 giving us 1 which is exactly what I have to prove cannot happen. For example choose k to be 2 and n to be 4, then we have e^(i(pi)) which is 1. I am not sure how to prove that this cannot happen though. I need to show that it is impossible to have any multiple of pi as my argument. any ideas?
You're doing something wrong if you think that cos(pi)=1. cos(pi)=-1.
 
  • #9
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I wasn't thinking of course, you're right. I have it all proved now
 

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