Simplifying, Solving, and Operations 2.5, 2.9 Complex Numbers Simplifying, Solving, and Operations

WHY??? Solutions to many real-world problems, such as classifying a shock absorber spring system in your car, involve complex numbers.

Complex numbers do not have order Complex Numbers Who uses them in real life? The navigation system in the space shuttle depends on complex numbers! -2 Who goes first? Complex numbers do not have order

What is a complex number? It is a tool to solve an equation. It has been used to solve equations for the last 200 years or so. It is defined to be i such that: Or, in other words:

Complex Numbers Imaginary Unit, i Imaginary Numbers If r > 0, then the imaginary number is defined as follows:

Worked Examples Simplify: −4 −4 = 4 −1 = 4 ∙ −1 =2𝑖

EXAMPLES: −81 −100

EXAMPLES: −33 −24

SOLVING!!!! Use the discriminant first to find out how many solutions exist. If there are NO REAL solutions, your answer should be complex (i)

𝑫𝒊𝒔𝒄𝒓𝒊𝒎𝒊𝒏𝒂𝒏𝒕= 𝒃 𝟐 −𝟒𝒂𝒄 Discriminant > 0 → Two real solutions Discriminant = 0 → One real solution Discriminant < 0 → No real solutions (Complex solutions)

Let’s solve a couple of equations that have complex solutions. 𝑥 2 +25= 0

Let’s solve a couple of equations that have complex solutions. 𝑥 2 −6𝑥+13= 0

Find the discriminant and determine the number of real solutions Find the discriminant and determine the number of real solutions. Then solve using any method. −2 𝑥 2 +5𝑥−3=0

Find the discriminant and determine the number of real solutions Find the discriminant and determine the number of real solutions. Then solve using any method. 𝑥 2 +3𝑥+9=0

Find the discriminant and determine the number of real solutions Find the discriminant and determine the number of real solutions. Then solve using any method. 4𝑥 2 +4=𝑥

Operations with Complex Numbers Combine like terms (real/complex) Answers will have a real part and an imaginary part: a + bi (4 + 2i) + (–6 – 7i)

Operations with Complex Numbers Combine like terms (real/complex) Answers will have a real part and an imaginary part: a + bi (5 –2i) – (–2 –3i)

Operations with Complex Numbers Combine like terms (real/complex) Answers will have a real part and an imaginary part: a + bi (1 – 3i) + (–1 + 3i)

You Try! Add or subtract. Write the result in the form a + bi. 2i – (3 + 5i) –3 – 3i (4 + 3i) + (4 – 3i) 8 (–3 + 5i) + (–6i) –3 – i

You can multiply complex numbers by using the Distributive Property and treating the imaginary parts as like terms. Simplify by using the fact i2 = –1. Multiply. Write the result in the form a + bi. –2i(2 – 4i)

Multiply. Write the result in the form a + bi. (3 + 6i)(4 – i) (2 + 9i)(2 – 9i)

Multiply. Write the result in the form a + bi. (–5i)(6i)

You Try! Multiply. Write the result in the form a + bi. (4 – 4i)(6 – i) 20 – 28i 2i(3 – 5i) 10 + 6i (3 + 2i)(3 – 2i) 13

Classwork/Homework P. 97 #2-9, 12, 13 P. 130 #12-26 even