# C10 Group, 10th roots unity with complex number multiplication

1. Sep 23, 2008

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. Sep 24, 2008

### HallsofIvy

Staff Emeritus
If n+ k> 9, then n+ k= 10a+ b where a and b are positive integers and $b\le 9$. 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 $2\pi$ rather than 360.