Mathematics Homework Solutions
#5005
Working with complex numbers.
Show that if z0 is an nth root of unity(z0 is not equal to 1) then
1+z0+z0^2+...+z0^(n-1)=0.
Hence show that
cos(2Pi/1998)+cos(4Pi/1998)+...+cos(2Pi*1997/1998)=-1
The solution contains a detailed proof of an equality involving the n-th root of unity.
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