4A: 1: Number Systems

Question 1

Convert recurring decimal a.bcbcbcbc… into a fraction.

Answer 1

(Here abc represents a three-digit number with digits a, b, and c.)
let x = a.bcbcbc…
100x = abc.bcbcbc…
99x = (abc-a)
x = (abc-a)/99 or a + bc/99

Question 2

Suppose a and b are two irrational numbers. Can (a + √2)(b + √2) be a rational number? If so, provide an example.

Answer 2

Yes.
(a + √2)(b + √2) = ab + (√2)(a+b) +2
Taking a = b = (√2)/2, ab = 1/2, (√2)(a+b) = 2.

Question 3

What is the definition of i?

Answer 3

i = √(-1)

Question 4

√[(-16)²] + [√(-16)]²

Answer 4

(-16)² = 256
√[(-16)²] = 16
√(-16) = 4i
[√(-16)]² = 16i² = -16
√[(-16)²] + [√(-16)]² = 0

Question 5

(revision exercise)
If a and b are positive numbers, which of the statement(s) are correct?
1. √(-a) x √(-b) = √ab
2. √(-a) + √(-b) = √[-(a+b)]
3. √(-a)[√(-a) + √(-b)] = -a + √ab
4. √(-a) x √(-a) = -a

Answer 5

Only (4)
Note that (1) and (3) contradict (4).
√(-a) = √a x i
√(-a) x √(-a) = √a x √a x i x i = -a

Question 6

i + i² + i³ +…+ i²⁰ =?

Answer 6

i² = -1
i³ = -i
i⁴ = 1
i + i² + i³ + i⁴ = 0
i + i² + i³ +…+ i²⁰ = 0

Question 7

What is the real part of (i+2)² + (i-4)²?

Answer 7

(i+2)² = 4-1+4i = 3+4i
(i-4)² = 16-1-8i = 15-8i
real part = 3+15 = 18

Question 8

What is the imaginary part of (i - i³ + i⁵ - i⁷ + i⁹) x (i² - i⁴ + i⁶ - i⁸)?

Answer 8

5i x (-4) = -20i
imaginary part = -20

Question 9

Simplify 1/i.

Answer 9

i x (-i) = 1
1/i = -i

Question 10

Simplify (1+i)/(1-i) + (1-i)/(1+i).

Answer 10

(1+i)/(1-i) = (1+i)²/(1+i)(1-i) = (1-1+2i)/(1+1) = 2i/2 = i
(1-i)/(1+i) = 1/i = -i
(1+i)/(1-i) + (1-i)/(1+i) = 0

Question 11

Let u = 1/(a+bi) and v = 1/(a-bi). Determine if u+v and uv are real, purely imaginary, or complex. Hence determine if u³ + v³ is real, purely imaginary, or complex.

Answer 11

u³ + v³ = (u+v) (u² + v² - uv) = (u+v) [(u+v)² - 3uv]
u + v = [(a-bi)+(a+bi)]/(a+bi)(a-bi) = 2a/(a² + b²), which is real
uv = 1/(a+bi)(a-bi) = 1/(a² + b²), which is real
Since both u+v and uv are real, u³ + v³ is real.

Question 12

Suppose a² + b² = 1.
Define u = 1/(a+bi) and v = 1/(a-bi). Simplify (u-1)(v+1) and determine if it is real, purely imaginary, or complex.

Answer 12

(u-1)(v+1) = uv + u - v - 1
uv = 1/(a+bi)(a-bi) = 1/a² + b² = 1
u - v = [(a-bi)-(a+bi)]/(a+bi)(a-bi) = -2bi/a² + b² = -2bi
(u-1)(v+1) = 1 -2bi - 1 = -2bi, which is purely imaginary.

Question 13

Simplify i²(a₁+b₁i)i³(b₂-a₂i)i⁶(a₃+b₃i)i⁷(₄b-a₄i)…i²⁷(a₁₄+b₁₄i)i³⁰(b₁₅-a₁₅i)i³¹(a₁₆+b₁₆i)/i¹(-b₁₆+a₁₆i)i⁴(-a₁₅-b₁₅i)i⁵(-b₁₄+a₁₄i)i⁹(-a₁₃-b₁₃i)…i²⁸(-b₃+a₃i)i²⁹(-a₂-b₂i)i³²(-b₁+a₁i),
where a₁, a₂…a₁₆, b₁, b₂ …b₁₆ are distinct real numbers ≠ 0.

Answer 13

terms like i²i³/i i⁴ cancel out.
(a+bi)/(-b+ai) = (a+bi)(b+ai)/-(b-ai)(b+ai) = -(ab-ab+a²i+b²i)/a²+b²) = -i
(b-ai)/(-a-bi) = 1/(-i) = i
the expression simplifies to (-i)(i)(-i)…(i) = i¹⁶ = 1.