20 Chapter 2
If x ∈ [2
k
, 2
k+1
[ then Tlog
2
x U = k. If a, b are integers, the interval [a; b[ contains b − a integers. Thus
Tlog
2
1 U + Tlog
2
2 U + Tlog
2
3 U + ··· + Tlog
2
1000 U = 2
1
− 2 0 + 2
2
− 2
1
1 +2
3
− 2
2
2 + ···
+2
9
− 2
8
8 +1000 − 2
9
9 =
0 + 2 · 1 + 4 · 2 + 8 · 3
+16 · 4 + 32 · 5+
+64 · 6 + 128 · 7
+256 · 8 + 489 · 9
= 7987
the last interval has 1000 − 512 + 1 = 489 integers.
Practice
149 Problem Find the exact value of
1 log
2
1996 +
1 log
3
1996 +
1 log
4
1996 +
··· + 1
log
1996
1996 .
150 Problem Shew that log
1 2
x log
1 3
x only when 0 x 1.
151 Problem Prove that log
3
π
+ log
π
3 2.
152 Problem Let a
1. Shew that 1
log
a
x 1 only when 1 x a.
153 Problem Let A = log
6
16 , B = log
12
27. Find integers a , b, c such that A + aB +
b = c.
154 Problem Solve the equation
log
1 3
cos x + √
5 6
+ log
1 3
cos x − √
5 6
= 2 .
155 Problem Solve
log
2
x + log
4
y + log
4
z = 2 ,
log
3
x + log
9
y + log
9
z = 2 ,
log
4
x + log
16
y + log
16
z = 2 .
156 Problem Solve the equation
x
.5 log√ x
x2 −x
= 3
log9 4
.
157 Problem Given that log
ab
a = 4, find log
ab 3
√ a
√ b
.
2.6 Complex Numbers
We use the symbol i to denote i = √
−1. Then i
2
= −1 . Clearly i
= 1 , i
1
= 1 , i
2
= −1 , i
3
= −i , i
4
= 1 , i
5
= i , etc., and so the
powers of i repeat themselves cyclically in a cycle of period 4.
158 Example Find i
1934
. Solution: Observe that 1934 = 4483 + 2 and so i
1934
= i
2
= −1. Complex numbers occur naturally in the solution of quadratic equations.
159 Example
Solve 2x
2
+ 6x + 5 = 0
Complex Numbers 21
Solution: Completing squares, 2x
2
+ 6x + 5 =
2x
2
+ 6x + 9
2 +
1 2
= √
2x + 3
√ 2
2
− i 1
√ 2
2
= √
2x + 3
√ 2
− i 1
√ 2
√ 2x +
3 √
2 + i
1 √
2 .
Then x = − 3
2 ± i
1 2
. If a
, b are real numbers then the object a + bi is called a complex number. If a + bi, c + di are complex numbers, then the sum of them is naturally defined as
a + bi + c + di = a + c + b + di 2.29
The product of a + bi and c + di is obtained by multiplying the binomials: a + bic + di = ac + adi + bci + bdi
2
= ac − bd + ad + bci 2.30
160 Definition If a
, b are real numbers, then the conjugate a + bi of a + bi is defined by a + bi = a − bi
2.31 The norm
|a + bi| of a + bi is defined by |a + bi| =
È
a + bia + bi =
p
a
2
+ b
2
2.32
161 Example Find
|7 + 3i|. Solution:
|7 + 3i| =
È
7 + 3i7 − 3i =
p
7
2
+ 3
2
= √
58.
162 Example
Express the quotient 2 + 3i
3 − 5i in the form a + bi.
Solution: We have 2 + 3i
3 − 5i =
2 + 3i 3 − 5i
· 3 + 5i
3 + 5i ==
−9 + 19i 34
= −9
34 +
19i 34
If z
1
, z
2
are complex numbers, then their norms are multiplicative. |z
1
z
2
| = |z
1
||z
2
| 2.33
163 Example Write 2
2
+ 3
2
5
2
+ 7
2
as the sum of two squares. Solution: The idea is to write 2
2
+ 3
2
= |2 + 3i|
2
, 5
2
+ 7
2
= |5 + 7i|
2
and use the multiplicativity of the norm. Now 2
2
+ 3
2
5
2
+ 7
2
= |2 + 3i|
2
|5 + 7i|
2
= |2 + 3i5 + 7i|
2
= | − 11 + 29i|
2
= 11
2
+ 29
2
164 Example
Find the roots of x
3
− 1 = 0.
22 Chapter 2
Solution: x
3
− 1 = x − 1x
2
+ x + 1. If x 6= 1, the two solutions to x
2
+ x + 1 = 0 can be obtained using the quadratic formula, getting x = 1
2 ± i √
3 2. Traditionally one denotes
ω
= 1 2 + i
√ 3
2 and hence
ω
2
= 1 2 − i
√ 3
2. Clearly
ω
3
= 1 and
ω
2
+
ω
+ 1 = 0 .
165 Example AHSME 1992
Find the product of the real parts of the roots of z
2
− z = 5 − 5i .
Solution: By the quadratic formula, z
= 1
2 ±
1 2
√ 21 − 20i
= 1
2 ±
1 2
È
21 − 2 √
−100 =
1 2
± 1
2
q
25 − 2
È
25−4 − 4 =
1 2
± 1
2
È
5 − 2i
2
= 1
2 ±
5 − 2i 2
The roots are thus 3 − i and −2 + i. The product of their real parts is therefore 3−2 = −6 .
☞
Had we chosen to write 21 − 20i = −5 + 2i
2
, we would have still gotten the same values of z.
Practice 23
Practice
166 Problem Simplify
1 + i
2004
1 − i
2000
.
167 Problem Prove that
1 + 2i + 3i
2
+ 4i
3
+ ··· + 1995i
1994
+ 1996i
1995
= −998 − 998i .
168 Problem Let
1 + x + x
2 1000
= a + a
1
x + ··· + a
2000
x
2000
. Find
a + a
4
+ a
8
+ ··· + a
2000
.
Chapter
3
Arithmetic
3.1 Division Algorithm
