COMPLEX
NUMBERS
Complex Roots of Quadratic Equations
If b2 – 4ac < 0, the equation has no real solution.
However, in the complex number system,
the equation will always have solutions.
This is because negative numbers have
square roots in this expanded setting.
If we will solve the equation x2 = –4 , we get:
x = −4.
However, this is impossible, since the square
of any real number is positive.
For example, (–2)2 = 4, a positive number.
Thus, negative numbers don’t have real
square roots.
Complex Number System
To make it possible to solve all quadratic
equations, mathematicians invented an expanded
number system—called the complex number
system.
Complex Number
First, they defined the new number
i = −1
This means i 2 = –1.
A complex number is then a number of the form
a + bi, where a and b are real numbers.
Complex Numbers
In the complex number system, every quadratic equation has
solutions.
The numbers 2i and –2i are solutions of x2 = – 4 because:
(2i)2 = 22i2 = 4(–1) = –4
and
(–2i)2 = (–2)2i2 = 4(–1) = –4
Quadratic Equations with
Complex Solutions
Solve each equation.
(a) x2 + 9 = 0
(b) x2 + 4x + 5 = 0
Complex Solutions
The equation x2 + 9 = 0 means x2 = –9.
So,
𝑥 = ± −9 = ± −1 ⋅ 9 = ±𝑖 9 = ±3𝑖
The solutions are therefore 3i and –3i.
OPERATIONS ON
COMPLEX NUMBERS
Complex Numbers (a + bi)
Natural (Counting) Numbers
Imaginary #’s
Complex Numbers are written in
the form a + bi, where a is the real
part and bi is the imaginary part.
a + bi
real part
imaginary part
ADDING &
SUBTRACTING
COMPLEX
NUMBERS
When adding complex numbers,
add the real parts together and
add the imaginary parts together.
imaginary part
(3 + 7i) + (8 + 11i)
real part
11 + 18i
When subtracting complex numbers,
be sure to distribute the subtraction
sign; then add like parts.
(5 + 10i) – (15 – 2i)
5 + 10i – 15 + 2i
–10 + 12i
i =1
4
i = −i
3
i
i = −1
2
Divide the exponent by 4
No remainder: answer is 1.
remainder of 1: answer is i.
remainder of 2: answer is –1.
remainder of 3:answer is –i.
Powers of i
1.) Find 𝑖23 = −i
2.) Find 𝑖2006 = −1
3.) Find 𝑖37 =i
4.) Find 𝑖828 =1
MULTIPLYING
& DIVIDING
COMPLEX
NUMBERS
When multiplying complex numbers,
use the distributive property and
simplify.
(3 – 8i)(5 + 7i)
15 + 21i – 40i – 56i 2
15 – 19i + 56 Remember,
i2 = –1
71 – 19i
REMEMBER: i² = -1
Multiply
1) 3i 4i = 12i = 12(−1) = −12
2
2) (7i )
2
= 7 i = 49(−1) = −49
2 2
You try…
3)
−7i 12i = −84i = −84(−1)
2
= 84
4) (− 11i ) = (− 11) (i ) = 121(−1)
2 2 2
= −121
Multiply
5) (4 + 3i )(7 + 2i )
= 28 +8i +21i +6i
2
= 28 + 29i + 6i 2
= 28 + 29i + 6(−1)
= 28 + 29i − 6
= 22 + 29i
You try…
6) (2 − i )(3 + 10i )
= 6 + 20i − 3i − 10i 2
= 6 + 17i − 10i 2
= 6 + 17i − 10(− 1)
= 6 + 17i + 10
= 16+ 17i
You try…
7)
(5 + 7i )(5 − 7i )
= 25−35i +35i −49i
2
= 25 − 49(−1)
= 25 + 49
= 74
