Page 24 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 1] NUMBERS 15

ffiffiffi
p 10 1=3
1.28. Evaluate (a) ð 1 þ 3iÞ ,(b) ð 1 þ iÞ .
(a)By Problem 1.27(b) and De Moivre’s theorem,
10 10 10
p ffiffiffi
ð 1 þ 3iÞ ¼½2ðcos 2 =3 þ i sin 2 =3ފ ¼ 2 ðcos 20 =3 þ i sin 20 =3Þ
¼ 1024½cosð2 =3 þ 6 Þþ i sinð2 =3 þ 6 ފ ¼ 1024ðcos 2 =3 þ i sin 2 =3Þ
1 1
p ffiffiffi p ffiffiffi
2
¼ 1024 þ 2 3i ¼ 512 þ 512 3i
p ffiffiffi p ffiffiffi
2½cosð1358 þ k 3608Þþ i sinð1358 þ k 3608ފ. Then
(b) 1 þ i ¼ 2ðcos 1358 þ i sin 1358Þ¼
1358 þ k 3608 1358 þ k 3608

ffiffiffi
p
1=3 1=3
3 3 P 1
ð 1 þ iÞ ¼ð 2Þ cos þ i sin
The results for k ¼ 0; 1; 2are
165
P 2
6
p ffiffiffi
2ðcos 458 þ i sin 458Þ; 45
ffiffiffi
2ðcos 1658 þ i sin 1658Þ;
p
6 6
285 √2
6
p ffiffiffi
2ðcos 2858 þ i sin 2858Þ
The results for k ¼ 3; 4; 5; 6; 7; .. . give repetitions of these. These
complex roots are represented geometrically in the complex plane
by points P 1 ; P 2 ; P 3 on the circle of Fig. 1-5. P 3
Fig. 1-5
MATHEMATICAL INDUCTION
2 2 3 2 2 1
1.29. Prove that 1 þ 2 þ 3 þ 4 þ þ n ¼ nðn þ 1Þð2n þ 1Þ.
6
1
2
The statement is true for n ¼ 1 since 1 ¼ ð1Þð1 þ 1Þð2 1 þ 1Þ¼ 1.
6
Assume the statement true for n ¼ k. Then
2 2 2 2 1
6
1 þ 2 þ 3 þ þ k ¼ kðk þ 1Þð2k þ 1Þ
2
Adding ðk þ 1Þ to both sides,
2 2 2 2 2 1 2 1
6 6
1 þ 2 þ 3 þ þ k þðk þ 1Þ ¼ kðk þ 1Þð2k þ 1Þþ ðk þ 1Þ ¼ðk þ 1Þ½ kð2k þ 1Þþ k þ 1Š
1 2 1
6
6
¼ ðk þ 1Þð2k þ 7k þ 6Þ¼ ðk þ 1Þðk þ 2Þð2k þ 3Þ
which shows that the statement is true for n ¼ k þ 1 if it is true for n ¼ k. But since it is true for n ¼ 1, it
follows that it is true for n ¼ 1 þ 1 ¼ 2 and for n ¼ 2 þ 1 ¼ 3; .. . ; i.e., it is true for all positive integers n.
n
n
1.30. Prove that x y has x y as a factor for all positive integers n.
1
1
The statement is true for n ¼ 1 since x y ¼ x y.
k
k
Assume the statement true for n ¼ k, i.e., assume that x y has x y as a factor. Consider
k
k
x kþ1 y kþ1 ¼ x kþ1 x y þ x y y kþ1
k k k
¼ x ðx yÞþ yðx y Þ
The first term on the right has x y as a factor, and the second term on the right also has x y as a factor
because of the above assumption.
k
k
Thus x kþ1 y kþ1 has x y as a factor if x y does.
3
1
3
2
2
1
Then since x y has x y as factor, it follows that x y has x y as a factor, x y has x y as a
factor, etc.