Download The complex number system

Pre-Calculus Algebra.
Summary 6
The complex number system
Complex Numbers:
The imaginary unit.
The non-real number  1 is called the unit of the imaginary numbers and it is denoted by i in
mathematics and by j in engineering.
That is i = j =  1
The complex number system.
A complex number is any number of the form a + bi where a and b are real numbers and i=
1.
Example: 2 + 3i is a complex number;
Example: 5 +  4 = 5 + 4(1)  5  2  1  5  2i is a complex number
Real numbers.
If b = 0, the complex number a + bi = a, is a real number. Therefore, the set of real numbers is a
subset of the complex numbers. Every real number is complex but not every complex number
is real.
Example: the number 5 = 5 + 0i is a complex number and a real number.
Pure imaginary number.
If a = 0, the complex number a + bi = bi is a pure imaginary number.
Example 6: the number  50  25(2)( 1)  5 2i  5i 2 is a pure imaginary number.
Conjugate of a complex number.
The complex number a – bi where i=  1 is called the complex conjugate of a + bi.
Example: the conjugate of 5 – 3i is the number 5 + 3i
The conjugate of -3 -4i is the number -3 + 4i
Complex numbers. Any number of the form a + bi where i   1 is called a complex
number. “a” is called the real component and “b’ is called the imaginary component. a – bi is
called the conjugate of the complex number a + bi
Properties:


 1  i , i 2  1, i 3   i , i 4  1
For any natural number n greater than 4, divide n by 4 and let r=remainder.
n
r
Then, i  i where r=0, 1, 2, or 3. Simplify the result.
Exercises:
1) i 26  i 2  1 because 26 divided by 4 leaves a remainder of 2.
2) i 600  i 0  1 because 600 divided by 4 leaves a remainder of 0.
3) i 87  i 3  i because 87 divided by 4 leaves a remainder of 3
4) i 7892347568 917  i1  i because 789234756917 divided by 4 leaves a remainder of 1.
5) i 7  i 5  i  1  i 3  i  i  1  i  i  i  1  1  i
-1-

To operate with complex numbers use the rules of algebra: combining similar terms and
using the lows of exponents. Answers should be of the form a + bi.
Exercise: write each of the following in the a + bi form.
1) (2 – 3i) – (-3 -7i)
Solution: (2 – 3i) – (-3 -7i) = 2 – 3i +3 +7i = 5 + 4i
2) 2(-3+5i) +3(4 – 5i)
Solution: 2(-3+5i) +3(4 – 5i) = -6 +10i +12 – 15i = 6 – 5i
3) (3  2i )(5  4i )
Answer
1) 5 + 4i
2) 6 – 5i
3) 23 + 2i
Solution : (3  2i )(5  4i )  15  12i  10i  8i 2  15  2i  8( 1)  23  2i
4) (3  2i )(3  2i )
4) 13 + 0i
Solution : (3  2i )(3  2i )  9  6i  6i  4i 2  9  4( 1)  9  4  13  13  0i
5) 3i(5  7i )
5) 21+15i
Solution : 3i(5  7i )  15i  21i 2  15i  21(1)  15i  21  21  15i
6) 3i ( 2  3i )( 2  i )
6) -12-21i
Solution : 3i ( 2  3i )( 2  i )  3i( 4  2i  6i  3i 2 )  3i( 4  4i  3)
 3i( 7  4i )  21i  12i 2  21i  12  12  21i
7) 10 - 20i
7 ) (3  5i )  ( 2  3i )  (3  2i ) 2
Solution : (3  5i )  ( 2  3i )  (3  2i ) 2  (3  5i )  ( 2  3i )  (3  2i )(3  2i )
 (3  5i )  ( 2  3i )  (9  12i  4i 2 )  3  5i  2  3i  9  12i  4( 1)
 3  5i  2  3i  9  12i  4  10  20i
8) 46+9i
8) ( 2  3i ) 3
Solution : ( 2  3i ) 3  (  2  3i )( 2  3i )( 2  3i )  ( 2  3i )( 4  12i  9i 2 )
 ( 2  3i )( 5  12i )  10  9i  36i 2  46  9i
To divide complex number
Step 1: write as a fraction
Step 2: multiply numerator and denominator by the conjugate of the denominator
Step 3: simplify the resulting expression.
9)
9) Divide : 2  5i   (3  2i )
16 11
2  5i 2  5i 3  2i
 i


Solution : 2  5i  (3  2i ) 



13
13
3  2i 3  2i 3  2i
6  11i  10i 2
16  11i 16 11

 i
13
13 13
9  4i
10) Divide the complex number 0  10i by 2  i
2

Solution : (0  10i )  ( 2  i )
10i
10i 2  i 20i  10i 2 10  20i




 2  4i
2i 2i 2i
5
4  i2
-2-
10) 2 +4i
11) Divide the real number 20 by 1  3i
20
20 1  3i
20  60i
Solution : ( 20  0i )  (1  3i ) 



1  3i 1  3i 1  3i 1  3i  3i  9i 2
20(1  3i )

 2(1  3i )  2  6i
10
11) 2+6i
12) Find the solution set on the complex system, of the equation x 2  9  0
12)
0  3i
Solution: x 2  9  0  x 2  9  x    9  x  3i
13) Find all four solutions in the complex number system of the equation
x4  1
Solution:
x 4  1  x 4  1  0  ( x 2  1)( x 2  1)  0  ( x  1)( x  1)( x 2  1)  0
13)
x = -1,
x = 1,
x=0+i
x=0-i
 x  1  0 or x  1  0 or x 2  1  0  x  1 or x  1 or x 2  1
 x  1 or x  1 or x    1  x  1 or x  1 or x  i or x   i
14) Find the solution set on the complex system of the equation 5x 2  1  2x
Solution : 5x 2  1  2x  5x 2  2x  1  0
2
 b  b 2  4ac  ( 2)  ( 2)  4(5)(1)
a  5, b  2, c  1, x 

2a
2(5)
 x


2  4  20 2   16 2  4i 2(1  2i ) 1  2i 1 2




  i
10
10
10
10
5
5 5
14)
1 2
 i,
5 5
1 2
 i
5 5
Complex numbers are represented as vectors in the Complex number plane. The length of
the vector is called the modulus or absolute value of the complex number and it is written
as abs (a  bi )  a 2  b 2 .
The angle  that the vector makes with the positive part of the x-axis is called the argument
of the complex number.
b
Using Trigonometry of a right triangle, tan   where  is the argument of the
a
complex number a + bi.
Imaginary Axis
a + bi
2
a b
2

a
-3-
Real Axis
Examples:
Find the modulus and argument of the following complex numbers
Complex Number
Modulus
Argument
a) -3i = 0 - 3i
a) 3
a) 270 or  90
b) 5 = 5 + 0i
b) 5
b) 0
c) 180
c) -8 = -8 + 0i
c) 8
d) 90
d) 2.3 i = 0 + 2.3i
d) 2.3
e) -1 + i
 1 
e) ( 1) 2  12  2
e) tan 1    135 0
 1
f) -3 – 4i
f)
( 3) 2  ( 4) 2  25  5
 4
f ) 180   tan 1    233 .13 0 or  126 .87 
 3
If a, b, and c are real numbers and the discriminant b 2  4ac of the quadratic equation
ax 2  bx  c  0 is negative, then the two solutions of the equation are complex. This is
an indication that the quadratic function f ( x)  ax 2  bx  c has no real zeros or xintercepts.
Example
The quadratic equation x 2  4x  29  0 has discriminant equal to

b 2  4ac  ( 4) 2  4(1)( 29)  16  116  100
The two solutions are complex:
 b  b 2  4ac  (4)   100 4  10i
x


 2  5i
2a
2(1)
2
The solutions are 2 + 5i and 2 – 5i
-4-