Chapter 1

Complex Numbers

In this chapter we introduce a new number system which is the extension of the real number system.

1.1 Pure Imaginary Unit i

Since we have known that x2 ≥ 0 for every real number x, the equation x2 = -4 has no real solution. But if there is a pure imaginary unit ⅈ such that
i2 = -1
with acceptance of the usual operations on real numbers and ⅈ such as
(2ⅈ)2 = 4ⅈ2 = 4(-1) = -4 and (-2ⅈ)2 = 4ⅈ2 = 4(-1) = -4
then x2 = -4 has two solutions 2ⅈ and -2ⅈ.
If we try to solve x2 = -n, for n > 0, like x2 = -4, then √ n ⅈ and -√ n ⅈ are solutions because
(±√ n ⅈ)2 = nⅈ2 = n(-1) = -n

Example 1.

Solve x2 + 2x + 5 = 0.
Solution

x2 + 2x + 5 = 0
x2 + 2x + 1 = -4
(x + 1)2 = -4
= 4 (-1)
x + 1 = ±√ 4 √ i2
x + 1 = ±√ 4 i
x = -1 ±2i

So there are two solutions x = -1 + 2i and x = -1 - 2i.

ပံ့ပိုးကူညီသူ



Example 2.
Solve x2 + 2x + 3 = 0 and check your answers.

Solution
x2 + 2x + 3 = 0
x2 + 2x + 1 = -2
(x + 1)2 = -2
= 2 (-1)
= 2 i2
x + 1 = ±√ 2 √ i2
x + 1 = ± √ 2 i
x = -1 ± √ 2 i

So two solutions are x = -1 + √ 2 i and x = -1 - √ 2 i

For x = -1 + √ 2 i,
x2 + 2x + 3 = (-1 + √ 2 i)2 + 2(-1 + √ 2 i) + 3
= 1 - 2√ 2 i + 2i2 - 2 + 2√ 2 i + 3
= 1 - 2 √ 2 i + 2(-1) - 2 + 2 √ 2 i + 3
= 1 - 2√ 2 i - 2 -2 + 2√ 2 i + 3
= 0

For x = -1 - √ 2 i,
x2 + 2x + 3 = (-1 - √ 2 i)2 + 2(-1 - √ 2 i) + 3
= 1 + 2√ 2 i + 2i2 - 2 - 2√ 2 i + 3
= 1 + 2√ 2 i - 2 -2 - 2√ 2 i + 3
= 0

Exercise 1.1

  1. Solve the following equations.

    2. (a) x2 - 6x + 10 = 0
    (b) -2x2 + 4x - 3 = 0
    (c) 5x2 - 2x + 1 = 0
    (d) 3x2 + 7x + 5 = 0

  2. Solve the following equations and check your answers.

    3. (a) x2 - 2x + 4 = 0
    (b) x2 - 4x + 5 = 0

  3. Find the value of in for every positive integer n, where i2 = -1, i3 = i2i , i4 = i2i2, etc.

Answers

  1. Solve the following equations.

    2. (a) x2 - 6x + 10 = 0
    Solution

    x2 - 6x + 10 = 0
    x2 - 6x + 9 = -1
    (x - 3)2 = -1
    = i2
    x - 3 = ±√ i2
    x - 3 = ±i
    x = 3 ± i

    ∴ x = 3 + i and x = 3 - i     ⇦

    (b) -2x2 + 4x - 3 = 0
    Solution

    -2x2 + 4x - 3 = 0
    2x2 - 4x + 3 = 0
    x2 - 2x + 3 ⁄ 2 = 0
    x2 - 2x + 1 = - 1 ⁄ 2
    (x - 1)2 = - 1 ⁄ 2
    = 1 ⁄ 2 (-1)
    = 1 ⁄ 2 i2
    x - 1 $$ = \pm \sqrt{\frac{1}{2}}\ \sqrt{i^2} $$
    x - 1 = ± 1 ⁄ √ 2 i
    = ± √ 2 ⁄ 2 i
    x = 1 ± √ 2 ⁄ 2 i

    ∴ x = 1 + √ 2 ⁄ 2 i and x = 1 - √ 2 ⁄ 2 i     ⇦

    (c) 5x2 - 2x + 1 = 0

    Solution
    5x2 - 2x + 1 = 0
    x2 - 2 ⁄ 5 x + 1 ⁄ 5 = 0
    x2 - 2 ⁄ 5 x + 1 ⁄ 25 = - 4 ⁄ 25
    (x - 1 ⁄ 5)2 = - 4 ⁄ 25
    = 4⁄ 25 (-1)
    = 4⁄ 25 i2
    x - 1 ⁄ 5 $$= \pm \sqrt{\frac{4}{25}}\ i $$
    = ± 2⁄ 5 i
    x = 1 ⁄ 5 ± 2 ⁄ 5 i

    ∴ x = 1 ⁄ 5 + 2 ⁄ 5 i and x = 1 ⁄ 5 - 2 ⁄ 5 i       ⇦

    (d) 3x2 + 7x + 5 = 0

    Solution

    3x2 + 7x + 5 = 0
    x2 + 7 ⁄ 3 x + 5 ⁄ 3 = 0
    x2 + 7 ⁄3 x + 49 ⁄ 36 = - 11 ⁄ 36
    (x + 7 ⁄ 6)2 = - 11 ⁄36
    = 11⁄ 36 (-1)
    x + 7 ⁄ 6 = $$ = \pm \sqrt{\frac{11}{36}} \ i $$
    = ± √ 11 ⁄ 6 i
    x = - 7 ⁄ 6 ± √ 11 ⁄ 6 i

    ∴ x = - 7 ⁄ 6 + √ 11 ⁄ 6 i and x = - 7 ⁄ 6 - √ 11 ⁄ 6 i       ⇦

  2. Solve the following equations and check your answers.

    3. (a) x2 - 2x + 4 = 0
    Solution

    x2 - 2x + 4 = 0
    x2 - 2x + 1 = -3
    (x - 1)2 = -3 = 3 (-1) = 3 i2
    x - 1 = ± √ 3 i
    x = 1 ± √ 3 i
    ∴ x =1 + √ 3 i and x = 1 - √ 3 i       ⇦

    Check
    For x = 1 + √ 3 i ,
    x2 - 2x + 4 = (1 + √ 3 i)2 - 2(1 + √ 3 i) + 4
    = 1 + 2 √ 3 i + 3i2 - 2 - 2 √ 3 i + 4
    = 1 + 2 √ 3 i + 3 (-1) - 2 - 2 √ 3 i + 4
    = 1 - 3 - 2 + 4
    = 0


    For x = 1 - √ 3 i ,
    x2 - 2x + 4 = (1 - √ 3 i)2 - 2(1 - √ 3 i) + 4
    = 1 - 2 √ 3 i + 3i2 - 2 + 2 √ 3 i + 4
    = 1 - 2 √ 3 i + 3 (-1) - 2 + 2 √ 3 i + 4
    = 1 - 3 - 2 + 4
    = 0


    (b) x2 - 4x + 5 = 0

    Solution

    x2 - 4x + 5 = 0
    x2 - 4x + 4 = -1
    (x - 2)2 = -1 = i2
    x - 2 = ± √ i2
    x - 2 = ± i
    x = 2 ± i

    ∴ x = 2 + i and x = 2 - i       ⇦

    Check
    For x = 2 + i,
    x2 - 4x + 5 = (2 + i)2 - 4(2 + i) + 5
    = 4 + 4i + i2 - 8 - 4i + 5
    = 4 + 4i - 1 - 8 - 4i + 5
    = 4 - 1 - 8 + 5
    = 0


    For x = 2 - i,
    x2 - 4x + 5 = (2 - i)2 - 4(2 - i) + 5
    = 4 - 4i + (-1) - 8 + 4i + 5
    = 4 - 4i + i2 - 8 + 4i + 5
    = 4 - 1 - 8 + 5
    = 0


  3. Find the value of in for every positive integer n, where i2 = -1, i3 = i2i , i4 = i2i2, etc.

    4. Solution
    i2 = -1
    i3 = i2i = (-1)i = -i
    i4 = i2 i2 = (-1) (-1) = 1
    i5 = i2 i2 i = (-1) (-1) i = i

    For n = 4k,     i4k = 1
    For n = 4k + 1,     i4k + 1 = i
    For n = 4k + 2,     i4k + 2 = -1
    For n = 4k + 3,     i4k + 3 = -i
    ∴ in = (-1) n⁄ 2             if n is an even integer,

    in = (-1) n-1⁄ 2 i             if n is an odd integer.