Download 111 Ch3-3

3.3 Complex Numbers
Properties of the Imaginary Unit i (p.189)
𝑖 = √−1
𝑖 2 = −1
The definition of the number i allows us to find solutions to equations such as
x2 + 1 = 0. The solutions to this equation are i and –i.
Complex Numbers (p.189)
A complex number can be written in standard form as a + bi where a and b are
real numbers. The real part is a and the imaginary part is b. Every real number a
is a complex number because it can be written as a + 0i.
Imaginary Numbers (p.189)
A complex number a + bi with b ≠ 0 is an imaginary number. A complex
number a + bi with a = 0 and b ≠ 0 is sometimes called a pure imaginary
number. Examples of pure imaginary numbers include 3i and –i.
The Expression √−𝒂 (p.189)
If a > 0, then √−𝑎 = 𝑖√𝑎 .
Example 2 (p.190): Simplifying expressions
Simplify each expression.
a) √−3 ∙ √−3 = 𝑖 √3 ∙ 𝑖 √3 = 𝑖 2 √9 = −1 ∙ 3 = −3
b) √−2 ∙ √−8 = 𝑖√2 ∙ 𝑖 √8 = 𝑖 2 √16 = −1 ∙ 4 = −4
Example 3 (p.191): Performing complex arithmetic
a) (−3 + 4𝑖) + (5 − 𝑖) = −3 + 4𝑖 + 5 − 𝑖 = 2 + 3𝑖
b) (−7𝑖) − (6 − 5𝑖) = −7𝑖 − 6 + 5𝑖 = −6 − 2𝑖
c) (−3 + 2𝑖)2 = (−3 + 2𝑖)(−3 + 2𝑖) =
9 − 6𝑖 − 6𝑖 + 4𝑖 2 = 9 − 6𝑖 − 6𝑖 − 4 = 5 − 12
d)
17
4+𝑖
4−𝑖
∙ 4−𝑖 =
68−17𝑖
16+1
=
68−17𝑖
17
=4–i
Quadratic Equations with Complex Solutions (p.192)
We can use the quadratic formula to solve quadratic equations when the
solutions are complex (discriminant is negative). There are no real solutions,
and the graph does not intersect the x-axis. The solutions are expressed as
imaginary numbers.
Example 5a (p.193): Solving a quadratic with imaginary solutions
Solve the quadratic equation x2 + 3x + 5 = 0. Support your answer graphically.
Solution: a = 1, b = 3, c = 5
𝑥=
−3 ± √32 −4∙1∙5
2∙1
=
−3 ± √−11
2
=
−3 ±𝑖 √11
2
=−
3
2
±
𝑖 √11
2
Standard Form of a
Complex Number
The graph does not intersect the x-axis
so there are no real solutions. There are
two complex solutions that are imaginary.
Example 5b (p.193): Solving a quadratic with imaginary solutions
Solve the quadratic equation
1
2
𝑥 2 + 17 = 5𝑥 Support your answer graphically.
Solution: Rewrite the equation in standard form
1
2
𝑥 2 − 5𝑥 + 17 = 0
a = 1/2, b = –5, c = 17
𝑥=
−(−5) ± √(−5)2 −4∙0.5∙17
2∙0.5
=
5 ± √25−34
1
= 5 ± √−9 = 5 ± 3𝑖
To show this graphically, graph each side of the equation separately,
1
𝑦 = 𝑥 2 + 17 and 𝑦 = 5𝑥. The graphs do not intersect each other, so there are
2
no real solutions. There are two complex solutions that are imaginary.
Example 5c (p.193): Solving a quadratic with imaginary solutions
Solve the quadratic equation –2x2 = 3. Support your answer graphically.
Solution: Use the Square Root Property.
−2𝑥 2 = 3
𝑥2 = −
3
2
3
3
2
2
𝑥 = ±√− = ±𝑖√ = ±𝑖
√6
2
To show this graphically, graph each side of the equation separately, 𝑦 = −2𝑥 2
and 𝑦 = 3. The graphs do not intersect each other, so there are no real solutions.
There are two complex solutions that are imaginary.