Mathematics Homework Solutions
Problem
#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


Solution Summary

The solution contains a detailed proof of an equality involving the n-th root of unity.

Solution
What is this?
By OTA - Overall OTA Rating
Changping Wang, MA - 4.9/5
Purchase Cost Now
$2.19 CAD
Included in Download
  • Plain text response
$2.19 Instant Download
Add to Cart
Why you can trust BrainMass.com
  • Your Information is Secure
  • Best Online Academic Help Service
  • Students find real academic Success
Related Solutions
  • 500 level complex variable in undergraduate - The problems are from complex variable 500 level in undergraduate. Please specify your notation(if necessary) and explain clearly each step of your solution. Thank you very much.
  • Ring Unity - Let R be a ring with unity 1 and let S be a subring of R. Is it possible that S has unity e such that e does not equal 1?
  • Write answer in a + bi - Perform the operation. Write answer in the form a + bi 6 + principle square root -18 --------------------------------- 3 NOTE: I'm unsure how to graph the symbol for princip ...
  • Root Locus - The open loop transfer function of a unity feed-back control system is given: G(s) = K/s(s+3)^2 Sketch the root locus of the system. See attached file for the problem as well. Please show st ...
  • Groups - For a fixed positive integer n - ** decide whether each of the given sets is a group with respect to the indicated operation. 1- For a fixed positive integer n, the set of all complex numbers x such that x^n=1(that is, the set of ...
Browse