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Transcript
Complex Numbers
Section 3.2 Beginning on page 104
Imaginary Numbers and Complex
Numbers
𝑖 = −1
𝑖 2 = −1
Imaginary numbers allow us to simplify radicals with no real solutions.
−4
= −1 4
= 2𝑖
−3
= −1 3
=𝑖 3
−20 = −1 4 5
= 2𝑖 5
A complex number written in standard form is a number 𝑎 + 𝑏𝑖 where 𝑎 and 𝑏 are
real numbers. The number 𝑎 is the real part, and the number 𝑏𝑖 is the imaginary part.
Real Numbers:
5
−1
𝜋
Complex Numbers:
2
3
2 + 3𝑖
9 − 5𝑖
Pure Imaginary Numbers:
3𝑖
− 5𝑖
Examples 1 and 2
Example 1: Find the square root of each number
a) −25
= −1 25
= 5𝑖
b)
−72
c) −5 −9
= −1 36 2
= −5 −1 9
= −5 ∙ 3𝑖
= −15𝑖
= 6𝑖 2
Example 2: Find the values of x and y that satisfy the equation 2𝑥 − 7𝑖 = 10 + 𝑦𝑖
** Set the real parts equal to each other,
set the imaginary parts equal to each
other. Solve the resulting equations. **
2𝑥 = 10
𝑥=5
−7𝑖 = 𝑦𝑖
−7 = 𝑦
y = −7
The Sums and Differences of Complex
Numbers
To add (or subtract) two complex numbers, add (or subtract) their real parts and their
imaginary parts separately.
Example 3:
a) 8 − 𝑖 + (5 + 4𝑖)
= 8 + 5 + (−𝑖 + 4𝑖)
= 13 + 3𝑖
b) 7 − 6𝑖 − (3 − 6𝑖)
= 7 − 3 + (−6𝑖 − (−6𝑖))
= 4 + 0𝑖
=4
c)
13 − 2 + 7𝑖 + 5𝑖
= 13 − 2 − 7𝑖 + 5𝑖
= 11 − 2𝑖
Multiplying Complex Numbers
Example 5: Multiply. Write the answer in standard form.
a) 4𝑖(−6 + 𝑖)
** 𝑖 ∙ 𝑖 = 𝑖 2 = −1 **
b) (9 − 2𝑖)(−4 + 7𝑖)
= −24𝑖 + 4𝑖 2
= −36 + 63𝑖 + 8𝑖 − 14𝑖 2
= −24𝑖 + 4(−1)
= −36 + 71𝑖 − 14(−1)
= −24𝑖 − 4
= −36 + 71𝑖 + 14
= −4 − 24𝑖
= −22 + 71𝑖
Solving Quadratic Equations
Example 6: Solve
a) 𝑥 2 + 4 = 0
𝑥 2 = −4
𝑥 = −4
b) 2𝑥 2 − 11 = −47
2𝑥 2 = −36
𝑥 2 = −18
𝑥 = −1 4
𝑥 = −18
𝑥 = ±2𝑖
𝑥 = −1 9 2
𝑥 = ±3𝑖 2
Finding Zeros of Quadratic Functions
Example 7: Find the zeros of 𝑓 𝑥 = 4𝑥 2 + 20
0 = 4𝑥 2 + 20
−20 = 4𝑥 2
−5 = 𝑥 2
𝑥 = −5
𝑥 = −1 5
𝑥 = ±𝑖 5
The zeros of the function
are 𝑖 5 and −𝑖 5
Practice
Find the square root of the number.
1)
−4
2) −12
3) − −36
4) 2 −54
Find the values of x and y that satisfy the equation.
5) 𝑥 + 3𝑖 = 9 − 𝑖𝑦
6) 9 + 4𝑦𝑖 = −2𝑥 + 3𝑖
Perform the operation. Write the answer in standard form.
8) 9 − 𝑖 + (−6 + 7𝑖)
9) 3 + 7𝑖 − (8 − 2𝑖)
10) −4 − 1 + 𝑖 − (5 + 9𝑖)
11) (−3𝑖)(10𝑖)
12) 𝑖(8 − 𝑖)
13) (3 + 𝑖)(5 − 𝑖)
Answers:
1) 2𝑖
2) 2𝑖 6
6) 𝑥 = −4.5, 𝑦 = 0.75
11) 30
12) 1 + 8𝑖
3) −6𝑖
8) 3 + 6𝑖
13) 16 + 2𝑖
4) 6𝑖 6
5) 𝑥 = 9, 𝑦 = −3
9) −5 + 9𝑖
10) −10 − 10𝑖