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Simplifying, Solving, and Operations
2.5, 2.9 Complex Numbers Simplifying, Solving, and Operations
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WHY??? Solutions to many real-world problems, such as classifying a shock absorber spring system in your car, involve complex numbers.
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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
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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:
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Complex Numbers Imaginary Unit, i Imaginary Numbers If r > 0, then the imaginary number is defined as follows:
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Worked Examples Simplify: −4 −4 = −1 = 4 ∙ −1 =2𝑖
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EXAMPLES: −81 −100
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EXAMPLES: −33 −24
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SOLVING!!!! Use the discriminant first to find out how many solutions exist. If there are NO REAL solutions, your answer should be complex (i)
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𝑫𝒊𝒔𝒄𝒓𝒊𝒎𝒊𝒏𝒂𝒏𝒕= 𝒃 𝟐 −𝟒𝒂𝒄 Discriminant > 0 → Two real solutions Discriminant = 0 → One real solution Discriminant < 0 → No real solutions (Complex solutions)
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Let’s solve a couple of equations that have complex solutions.
𝑥 2 +25= 0
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Let’s solve a couple of equations that have complex solutions.
𝑥 2 −6𝑥+13= 0
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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
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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
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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
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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)
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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)
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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)
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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
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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)
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Multiply. Write the result in the form a + bi. (3 + 6i)(4 – i) (2 + 9i)(2 – 9i)
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Multiply. Write the result in the form a + bi. (–5i)(6i)
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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
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Classwork/Homework P. 97 #2-9, 12, 13 P. 130 #12-26 even
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