Complex Numbers

1.2 Complex Number (x, y)

As we have just seen in section 1.1, there are numbers like x₁ + y₁i and x₂ + y₂i such that
(x₁ + y₁ i) + (x₂ + y₂ i) = (x₁ + x₂) + (y₁ + y₂)i and

(x₁ + y₁i) ( x₂ + y₂i)	= x₁x₂ + (x₁y₂ + y₁x₂)i + y₁y₂i²
	= x₁x₂ + (x₁y₂ + y₁x₂)i + y₁y₂(-1)
	= (x₁x₂ - y₁y₂) + (x₁y₂ + y₁x₂)i

for real numbers x₁, x₂, y₁, y₂, so we define the number x + yi as follows.

A complex number is an ordered pair (x, y) of real numbers with equality and operations— sum and product— of two complex numbers (x₁, y₁), (x₂, y₂) are defined as follows:

Equality (x₁, y₁) = (x₂, y₂) if and only if x₁ = x₂ and y₁ = y₂
Sum (x₁, y₁) + (x₂, y₂) = (x₁ + x₂, y₁ + y₂)
Product (x₁, y₁)(x₂, y₂) = (x₁x₂ - y₁ y₂, x₁ y₂ + y₁ x₂)

Then we have

(x, 0) + (y, 0) = (x + y, 0) and (x, 0)(y, 0) = (xy, 0)

From these facts, all real numbers can be considered as complex numbers as shown in the following table.

Real Numbers	Complex Numbers
x	(x, 0)
y	(y, 0)
x + y	(x, 0) + (y, 0)
xy	(x, 0)(y, 0)

Since (0, 1)(0, 1) = (0 - 1, 0 + 1) = (-1, 0), let us denote

i = (0, 1)

and with the convention i² = ii, i³ = i²i, etc., we have

i² = (0, 1)(0, 1) = (0 - 1, 0 + 0) = (-1, 0) = -1

And we have
x + yi = (x, 0) + (y, 0)(0, 1)
= (x, 0) + (0 - 0, y + 0)
= (x, 0) + (0, y)
x + yi = (x, y)

Note that sum and product of complex numbers satisfy commutative, associative and distributive properties.

ပံ့ပိုးကူညီသူများ

Example 3.
Compute (-2, 3)(1, -2) + (1, 1)(0, 1).

Solution
Method 1
(-2, 3)(1, -2) + (1, 1)(0, 1) = (-2 - (-6), 4 + 3) + (0-1, 1+0)
= (4, 7) + (-1, 1) = (3, 8)

Method 2
(-2, 3)(1, -2) + (1, 1)(0, 1) = (-2 + 3i)(1-2i) + (1 + i)i
= (-2 + 4i + 3i - 6i²) + (i + i²)
= (-2 + 7i - 6(-1)) + (i - 1)
= (4 + 7i) + (-1 + i)
= (3 + 8i)
= (3, 8)

Exercise 1.2

Compute:

Compute:

Answers
1. Compute:
(a) (2, 0)(3, 5) + (3, -2)(0, 1)

Solution

Method 1
(2, 0)(3, 5) + (3, -2)(0, 1) = (6-0, 10-0) + (0 - (-2), 3 + 0)
= (6, 10) + (2, 3)
= (8, 13) ⇦

Method 2
(2, 0)(3, 5) + (3, -2)(0, 1) = (2 + 0i)(3 + 5i) + (3 - 2i)(0 + 1i)
= 2 (3 + 5i) + (3 - 2i) i
= (6 + 10i) + (3i - 2i²)
= (6 + 10i) + (3i - 2(-1) )
= (6 + 10i) + (2 + 3i)
= (8 + 13i)
= (8, 13) ⇦

(b) (2, -5)(-1, 0) + (1, 0)(5, 1)

Solution

Method 1
(2, -5)(-1, 0) + (1, 0)(5, 1) = (-2 - 0, 0 + 5) + (5 - 0, 1 + 0)
= (-2, 5) + (5, 1)
= (3, 6)

Method 2
(2, -5)(-1, 0) + (1, 0)(5, 1) = (2 - 5i)(-1) + (1)(5 + i)
= (-2 + 5i) + (5 + i)
= (3 + 6i)
= (3, 6) ⇦

(c) (-3, -2)(-2, -3) + (-2, -3)(-3, -2)

Solution

Method 1
(-3, -2)(-2, -3) + (-2, -3)(-3, -2) = (6 - 6, 9 + 4) + (6 - 6, 4 + 9)
= (0, 13) + (0, 13)
= (0, 26)

Method 2
(-3, -2)(-2, -3) + (-2, -3)(-3, -2) = (-3, - 2i)(-2, - 3i) + (-2, - 3i)(-3 - 2i)
= (6 + 9i + 4i + 6i²) + (6 + 4i + 9i + 6i²)
=(6 + 13i + 6(-1) ) + (6 + 13i + 6(-1) )
= (6 - 6 + 13i) + (6 - 6 + 13i)
= (0 + 13i) + (0 + 13i)
= (0 + 26i)
= (0, 26) ⇦

(d) (1, 0)(0, 1) + (0, 1)(1, 0)

Solution

Method 1
(1, 0)(0, 1) + (0, 1)(1, 0) = (0 - 0, 1 + 0) + (0 - 0, 0 + 1)
= (0, 1) + (0, 1)
= (0, 2)

Method 2
(1, 0)(0, 1) + (0, 1)(1, 0) = (1 + 0i)(0 + 1i) + (0 + 1i)(1 + 0i)
= 1 (0 + i) + i(1 + 0)
= (0 + i) + (i + 0)
= (0 + i) + (0 + i)
= (0 + 2i)
= (0, 2) ⇦

2. Compute:
(a) (3 + 2i)(3 - 2i) + (-5 + 7i)(-1 - i)

Solution

(3 + 2i)(3 - 2i) + (-5 + 7i)(-1 - i) = (9 - 6i + 6i - 4i²) + (5 + 5i - 7i - 7i²)
= (9 - 4(-1) ) + (5 - 2i - 7(-1) )
= (9 + 4) + (5 + 7 - 2i)
= (13 + 12 - 2i)
= (25 - 2i) ⇦

(b) (-1 + i)(1 - i) + (2 + 3i)

Solution

(-1 + i)(1 - i) + (2 + 3i) = (-1 + i + i - i²) + (2 + 3i)
= (-1 + 2i - (-1) ) + (2 + 3i)
= (-1 + 2i + 1) + (2 + 3i)
= (2i + 2 + 3i)
= (2 + 5i) ⇦

(c) (1 + i)(1 - i) + (-2 + i)(-2 + i)

Solution
(1 + i)(1 - i) + (-2 + i)(-2 + i) = (1 - i + i - i²) + (4 - 2i - 2i + i²)
= (1 - (-1) ) + (4 - 4i + (-1) )
= (2 + 3 - 4i)
= (5 - 4i) ⇦

(d) (3 + 2i) + (7 - i)(-3 + 3i)

Solution

(3 + 2i) + (7 - i)(-3 + 3i) = (3 + 2i) + (-21 + 21i + 3i - 3i²)
= (3 + 2i) + (-21 + 24i - 3(-1) )
= (3 + 2i) + (-18 + 24i)
= (-15 + 26i) ⇦

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