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Complex Numbers - Day 1. My introduction Syllabus Start with add/subtract like variables (without any brackets) Introduce i and complex numbers (include square root of negative numbers) Add/subtract with i Now do distributing variables i equals …. Now distribute i.
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Complex Numbers - Day 1 • My introduction • Syllabus • Start with add/subtract like variables (without any brackets) • Introduce i and complex numbers (include square root of negative numbers) • Add/subtract with i • Now do distributing variables • i equals …. • Now distribute i
Complex Numbers - Day 2 • Review previous day in warm-up (include conjugates) • Now expand products of i to higher powers • Division with i – can not have a square root in denominator
Complex Numbers - Day 3 • Review previous day in warm-up (include conjugates) • Knowledge check • Pre-test
Complex Numbers - Day 4 • Intro to quadratics • Axis of symmetry
Complex Numbers Standard MM2N1c: Students will add, subtract, multiply, and divide complex numbers Standard MM2N1b:Write complex numbers in the form a + bi.
Complex Numbers Vocabulary(Name, Desc., Example) You should be able to define the following words after today's lesson: Complex Number Real Number Imaginary Number Pure Imaginary Number Standard Form
Complex Numbers • How do you think we can reduce:
Complex Numbers • The “i” has special meaning. • It equals the square root of negative 1. • We can not really take the square root of negative 1, so we call it “imaginary” and give it a symbol of “i”
Guided Practice: Do problems 21 – 27 odd on page 4
Complex Numbers • Complex numbers consist of a “real” part and an “imaginary” part. • The standard form of a complex number is: a + bi, where “a” is the “real” part, and “bi” is the imaginary part.
Complex Numbers • Give some examples of complex numbers. • Can “a” and/or “b” equal zero? • Give some examples of complex numbers when “a” and/or “b” equals zero. • Can you summarize this into a nice chart? Yes!!!!
Imaginary Numbers (a + bi, b ≠ 0) 2 + 3i 5 – 5i Real Numbers (a + 0i) -1 ⅜ Complex Numbers Vocabulary Pure Imaginary Numbers (0 + bi, b ≠ 0) -4i 8i
Complex Numbers • We are now going to add some complex numbers.
3x + 1 Review of past material: -8x - 9 Add: 5x + 4 – 3 – 2x = 2x - 7 – 10x – 2 =
6x - 2 Review of past material: 5x - 11 4a + 3 Add: (2x + 1) + (4x -3) = (7x – 5) – (2x + 6) = And we can change variables: (3a -2) + (a + 5) = And we could put them in different order: (6 + 5i) + (2 - 3i) = 8 + 2i
Complex Numbers • The “i” term is the imaginary part of the complex number, and it can be treated just like a variable as far as adding/subtracting like variables.
Complex Numbers • Simplify and put in standard form: • (2 – 3i) + (5 + 2i) = • (7 - 5i) – (3 - 5i) = • Any questions as far as adding or subtracting complex numbers? 7 - i 4
Complex Numbers • Just like you can not add variables (x) and constants, you can not add the real and imaginary part of the complex numbers. • Solve for x and y: • x – 3i = 5 + yi • -6x + 7yi = 18 + 28i • Any questions as far as adding or subtracting complex numbers? x = 5, y = -3 x = -3, y = 4
Complex Numbers – Guided Practice • Do problems 7 – 13 odd on page 9 • Do problems 35 – 39 odd on page 5
Warm Up: Write in standard form: Solve and write solution in standard form:
Warm-Up • Simplify and write in standard form: • (3x – 5) – (7x – 12) • Solve for x and y: • 2x + 8i = 14 – 2yi • 30 minutes to do the “Basic Skills for Math” NO CALCULATORS!! (add, subtract, multiply, divide)
Complex Numbers – Application • Applications of Complex Numbers - Spring/Mass System • http://www.picomonster.com/complex-numbers rowing
Complex Numbers • How do you think we can reduce:
Guided Practice: Do problems 21 – 27 odd on page 4
8x2 - 6x - 5 Review of past material: 14x2 +32x - 30 Multiply: (2x + 1)(4x -5) = (7x – 5)(2x + 6) =
Complex Numbers • How do you think we would do the following? • (2 – 3i)(5 + 2i) • Imaginary numbers may be multiplied by the distributive rule.
Complex Numbers • How do you think we would do the following? • (2 – 3i)(5 + 2i) • = 10 + 4i – 15i – 6i2 • = 10 – 11i – 6i2 • Can we simplify the i2? • i2 = i * i = -1, so we get: • = 10 – 11i – 6(-1) • = 10 – 11i + 6 = 16 – 11i
Complex Numbers • Simplify: • (7 + 5i)(3 - 2i) • = 21 -14i + 15i – 10i2 • = 21 +i– 10(-1) • = 21 + i + 10 • = 31 +i
Complex Numbers – Guided Practice – 5 minutes • Do problems 7 – 13 odd on page 13
Complex Numbers • If i2 = -1, what does i3 equal? • What does i4 equal? • How about i5: • Continue increasing the exponent, and determine a rule for simplifying i to some power. • What would i40 equal? • What would i83 equal? 1 -i
Imaginary Numbers • Definition:
Reducing Complex Numbers – Guided Practice – 5 minutes • Do problems 7 – 13 odd on page 13
Complex Numbers – Summary • Summarize : • What are complex numbers? • What does their standard form look like?? • What does i equal? • How do we add/subtract them? • How do we take the square root of a negative number?
Complex Numbers – Summary • Summarize : • How do we multiply them? • How do we simplify higher order imaginary numbers?
Complex Numbers – Ticket out the door • Simplify: • (2 – 3i) – (-5 + 7i) • (4 + i)(3 – 2i) • . • i23
Complex Numbers – Warm-up • Solve for x and y: • 27 – 8i = -13x + 3yi • Simplify • (2 + 3i)(2 – 3i) = • What happened to the “i” term? x = -2 1/13, y = -2 2/3 13
Complex Numbers Standard MM2N1c: Students will add, subtract, multiply, and divide complex numbers Standard MM2N1a:Write square roots of negative numbers in imaginary form.
Dividing by Imaginary Numbers Vocabulary(Name, Description, Example) You should be able to define the following words after today's lesson: Rationalizing the Denominator Conjugates
Review of past material: Simplify:
Dividing by Imaginary Numbers • Is there a problem when we try to divide complex numbers into real or complex numbers? HINT: yes, there is a problem • Problem – we can not leave a radical in the denominator • We must “rationalize the denominator”, which means we must eliminate all the square roots, including i.
Dividing by Imaginary Numbers • How can we solve ? Rationalize the denominator by multiplying by i/i
Dividing by Imaginary Numbers • How can we solve ? HINT: Look at the warm-up. • We can rationalize the denominator by multiplying by it’s conjugate. • In Algebra, the conjugate is where you change the sign in the middle of two terms, like (3x + 5) and (3x – 5) • Conjugates are (a + bi) and (a – bi), we just change the sign of the imaginary part
Dividing by Imaginary Numbers • How can we solve ? HINT: Look at the warm-up. Signs are opposite If these do not add to zero, then you made a mistake!!!
Dividing by Imaginary Numbers • How can we solve ? HINT: Look at the warm-up.
Practice • Page 13, # 29 – 37 odd