The Real And Complex Number Systems

In addition,
∑
a k b k + a j b j
1≤j≤k≤n
=
= n
n∑
a k b k + na 1 b 1 +
k=1
n∑
a k b k + (n − 1) a 2 b 2 + ... +
k=2
n∑
a k b k + a 1 b 1 + a 2 b 2 + ... + a n b n
k=1
= (n + 1)
n∑
a k b k
k=1
which implies that, by (**),
( n∑
) ( n∑
) ( n∑
)
a k b k ≤ n a k b k .
k=1 k=1
k=1
n∑
k=n−1
a k b k + 2a n−1 b n−1 + ∑ k=n
a k b k
Complex numbers
1.27 Express the following complex numbers in the form a + bi.
(a) (1 + i) 3
Solution: (1 + i) 3 = 1 + 3i + 3i 2 + i 3 = 1 + 3i − 3 − i = −2 + 2i.
(b) (2 + 3i) / (3 − 4i)
Solution: 2+3i
3−4i = (2+3i)(3+4i)
(3−4i)(3+4i) = −6+17i
25
= −6
25 + 17
25 i.
(c) i 5 + i 16
Solution: i 5 + i 16 = i + 1.
(d) 1 2 (1 + i) (1 + i−8 )
Solution: 1 2 (1 + i) (1 + i−8 ) = 1 + i.
1.28 In each case, determine all real x and y which satisfy the given
relation.
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