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C10 Group, 10th roots unity with complex number multiplication

  1. Sep 23, 2008 #1
    This is for 10th root unity with complex number multiplication. I am working on closure. I have multiplied 2 elements of my set and I have so far that cos[(n+k)360/10] + isin[(n+k)360/10]. Thus I know that if n+k<=9 then there is an element in the set. Now I need to show for if n+k>9 and if n+k<0. If someone could please help me, I would appreciate it. Thanks.
    Last edited: Sep 23, 2008
  2. jcsd
  3. Sep 24, 2008 #2


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    If n+ k> 9, then n+ k= 10a+ b where a and b are positive integers and [itex]b\le 9[/itex]. In that case, cos[(n+k)360/10]= cos[(10a+b)360/10]= cos[360a+ b(360/10)]. Now use the fact that cosine is periodic with period 360 (degrees).

    If n+ k< 0, then thre exist an integer a such that 10a> -(n+k) so 10a+ n+ k> 0. Again, (10a)(360/10)= 360a, a multiple of 360 degrees.

    Is there any specific reason for working in degrees? For problems like this it is almost always standard to work in radians- that is, use [itex]2\pi[/itex] rather than 360.
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