1873_solutions

# 1 2 3 which yields z 1 also implies that z 2 z 2 2z

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Unformatted text preview:  cos  i sin   cos2 i sin2  cos   i sin  cos   sin  we see that the equation cos   i sin  n  cos n  i sin n also holds when n  1. Finally, if n is any positive integer then we have n  cos   i sin  n cos   i sin  n  cos   i sin  1  cos n   i sin n   cos n  i sin n and so the equation cos   i sin  n  cos n  i sin n does also hold when n is negative. Incidentally, this equation also holds when n  0. 8. Making use of Exercise 7, prove that cos   i sin  9 9 is a cube root of 1 3i 2 2 cos   i sin  9 9 3 3  cos   i sin   1  i. 3 3 2 2 9. Suppose that n is a natural number. Making use of Exercise 7, find a complex number w such that w n  1 but w k 1 whenever k  1, 2, , n 1. A number with these properties is called a primitive nth root of 1. The number cos 2n  i sin 2n will work. 10. Prove that if n is a natural number, w n  1, and w 1, then 1  w  w2  w3    wn This fact follows at once from the fact that 1  w  w2  w3    wn 1 1  0. n 1 w. 1w 11. Prove that if w is a cube root of 1 and w 1, then for any numbers a, b, and c we have a  b  c a  wb  w 2 c a  w 2 b  w 4 c  a 3  b 3  c 3 3abc. 94 If you know something about determinants, prove that these expressions are also equal to abc det bca cab Can you generalize this result to higher order determinants? This does generalise nicely to a general n by n determinant a1 a2 a3  an an 1 a2 a3 a4  a1 a3 a4 a5  det an a1 a2       an a1 a2  an If w is an nth root of 1 and w 1 an a2 a3 a4  an a1 a3 a4 a5  a1 a2     1 1 then a1 a2 a3  an det an 2 .  an a1 a2  an n  an n j1 n w 2j a j  wjaj j1  2 n aj j1 w n 1 ja j j1 1 It’s easy to find all of these factors by performing elementary row operations on the matrix a1 a2 a3  an 1 an a2 a3 a4  an a1 a3 a4 a5  a1 a2       an a1 a2  an 2 an 1 6 Elementary Topology of the Real Line Exercises on Neighborhoods 1. Complete the following sentence “A set U fails to be a neighborhood of a number x when for every number   0, ...” A set U fails to be a neighborhood of a number x when for every number   0 the interval x , x   contains at least one number that does not belong to U. 2. Explain carefully why the assertion Ý  1, 1 nn n1 0 that was made above is true. You will need to make use of an earlier result. Since the number 0 belongs to the interval 1 , 1 for every positive integer n we have nn Ý 0  n1 95 1, 1 . nn Ý Now we need to show that 0 is the only number in the set  n1 1 , 1 . Suppose that x 0. Using nn the fact that |x |  0 and that earlier result, choose a positive integer n such that 1  |x |. n 1 1 0 |x | |x | n n 11 Since we have found a value of n for which x fails to belong to the interval n , n we know that Ý  x Ý Thus 0 is the only number in the set  n1 n1 11 n, n 1, 1 . nn , which is what we wanted to prove. 3. Given that x is an interior point of U and that U V, explain why x must be an interior point of V. Using th...
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