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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 j1 n w 2j a j wjaj j1
2 n aj j1 w n 1 ja j j1 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 n1 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
n1 95 1, 1 .
nn Ý
Now we need to show that 0 is the only number in the set n1 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 n1 n1
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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 Fall '08
 STAFF
 Math, Calculus

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