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

Review of old material

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”

Do problems 21 – 27 odd on page 4 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!!!!

Complex Numbers Vocabulary Imaginary Numbers (a + bi, b ≠ 0) 2 + 3i 5 – 5i Real Numbers (a + 0i) -1 ⅜  Pure Imaginary Numbers (0 + bi, b ≠ 0) -4i 8i

Complex Numbers We are now going to add some complex numbers.

Review of past material: Add: 5x + 4 – 3 – 2x = 2x - 7 – 10x – 2 = 3x + 1 -8x - 9

Review of past material: 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) = 6x - 2 5x - 11 4a + 3 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

Write in standard form: Solve and write solution in standard form: Warm Up: Write in standard form: Solve and write solution in standard form:

Warm-Up Simplify and write in standard form: Solve for x and y: (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

Review of old material

Complex Numbers How do you think we can reduce:

Do problems 21 – 27 odd on page 4 Guided Practice: Do problems 21 – 27 odd on page 4

Review of past material: Multiply: (2x + 1)(4x -5) = (7x – 5)(2x + 6) = 8x2 - 6x - 5 14x2 +32x - 30

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

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