Page 29 - Schaum's Outline of Theory and Problems of Advanced Calculus

20 NUMBERS [CHAP. 1

2
2

1.74. If z 1 and z 2 are complex numbers, prove (a) z 1 ¼ jz 1 j , (b) jz 1 j¼jz 1 j giving any restrictions.

z 2 jz 2 j
1.75. Prove (a) jz 1 þ z 2 j @ jz 1 jþjz 2 j, (b) jz 1 þ z 2 þ z 3 j @ jz 1 jþjz 2 jþjz 3 j, (c) jz 1 z 2 j A jz 1 j jz 2 j.
3
4
2
1.76. Find all solutions of 2x 3x 7x 8x þ 6 ¼ 0.
1
Ans.3, , 1 i
2
1.77. Let z 1 and z 2 be represented by points P 1 and P 2 in the Argand diagram. Construct lines OP 1 and OP 2 ,
where O is the origin. Show that z 1 þ z 2 can be represented by the point P 3 , where OP 3 is the diagonal of a
parallelogram having sides OP 1 and OP 2 . This is called the parallelogram law of addition of complex
numbers. Because of this and other properties, complex numbers can be considered as vectors in two
dimensions.
1.78. Interpret geometrically the inequalities of Problem 1.75.
p ffiffiffi p ffiffiffi
1.79. Express in polar form (a)3 3 þ 3i, (b) 2 2i, (c)1 3i, (d)5, (e) 5i.
ffiffiffi
p
Ans.(a)6 cis =6 ðbÞ 2 2 cis 5 =4 ðcÞ 2 cis 5 =3 ðdÞ 5 cis 0 ðeÞ 5 cis 3 =2
12 cis 168
1.80. Evaluate (a) ½2ðcos 258 þ i sin 258ފ½5ðcos 1108 þ i sin 1108ފ, (b) :
ð3 cis 448Þð2 cis 628Þ
p
ffiffiffi
ffiffiffi
Ans.(a) 5 2 þ 5 2i; ðbÞ 2i
p
1.81. Determine all the indicated roots and represent them graphically:
1=3 1=5 1=3 1=4
p ffiffiffi p ffiffiffi p ffiffiffi
(a) ð4 2 þ 4 2iÞ ; ðbÞð 1Þ ; ðcÞð 3 iÞ ; ðdÞ i .
Ans.(a)2 cis 158; 2 cis 1358; 2 cis 2558
(b) cis 368; cis 1088; cis 1808 ¼ 1; cis 2528; cis 3248
p ffiffiffi p ffiffiffi p ffiffiffi
(c) 3 2 cis 1108; 3 2 cis 2308; 3 2 cis 3508
(d) cis 22:58; cis 112:58; cis 202:58; cis 292:58
p ffiffiffi
1.82. Prove that 1 þ 3i is an algebraic number.
1.83. If z 1 ¼ 1 cis 1 and z 2 ¼ 2 cis 2 ,prove (a) z 1 z 2 ¼ 1 2 cisð 1 þ 2 Þ, (b) z 1 =z 2 ¼ð 1 = 2 Þ cisð 1 2 Þ.
Interpret geometrically.
MATHEMATICAL INDUCTION
Prove each of the following.

1.84. 1 þ 3 þ 5 þ þ ð2n 1Þ¼ n 2
1 1 1 1 n
1.85.
1 3 þ 3 5 þ 5 7 þ þ ð2n 1Þð2n þ 1Þ ¼ 2n þ 1
1.86. 1
2
a þða þ dÞþ ða þ 2dÞþ þ ½a þðn 1ÞdŠ¼ n½2a þðn 1ÞdŠ
1 1 1 1
1.87. þ þ þ þ ¼ nðn þ 3Þ
1 2 3 2 3 4 3 4 5 nðn þ 1Þðn þ 2Þ 4ðn þ 1Þðn þ 2Þ
n
2
1.88. a þ ar þ ar þ þ ar n 1 aðr 1Þ ; r 6¼ 1
r 1
¼
1.89. 3 3 3 3 1 2 2
4
1 þ 2 þ 3 þ þ n ¼ n ðn þ 1Þ
5 þð4n 1Þ5 nþ1
2
3
1.90. 1ð5Þþ 2ð5Þ þ 3ð5Þ þ þ nð5Þ n 1 ¼
16
1.91. x 2n 1 þ y 2n 1 is divisible by x þ y for n ¼ 1; 2; 3; ... .