C10 Group, 10th roots unity with complex number multiplication

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SUMMARY

This discussion focuses on the mathematical concept of 10th roots of unity using complex number multiplication. The user has established that for elements n and k in the set, the expression cos[(n+k)360/10] + isin[(n+k)360/10] holds true for n+k values less than or equal to 9. The user seeks assistance in proving the closure property for cases where n+k exceeds 9 or is less than 0, utilizing the periodicity of cosine. Additionally, a suggestion is made to consider using radians instead of degrees for standardization in mathematical problems.

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  • Knowledge of trigonometric functions, particularly cosine and sine
  • Familiarity with the concept of periodicity in trigonometric functions
  • Basic grasp of roots of unity in complex analysis
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  • Study the properties of complex numbers and their multiplication
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  • Explore the concept of roots of unity in greater detail, focusing on higher-order roots
  • Investigate the conversion between degrees and radians in trigonometric contexts
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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.
 
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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.
 

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