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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. - PowerPoint PPT Presentation
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Complex Numbers - Day 1My introductionSyllabusStart with add/subtract like variables
(without any brackets) Introduce i and complex numbers
(include square root of negative numbers)
Add/subtract with iNow do distributing variables i equals ….Now distribute i
Complex Numbers - Day 2Review previous day in warm-up
(include conjugates)Now expand products of i to higher
powersDivision with i – can not have a square
root in denominator
Complex Numbers - Day 3Review previous day in warm-up
(include conjugates)Knowledge checkPre-test
Complex Numbers - Day 4 Intro to quadraticsAxis of symmetry
Complex Numbers
Standard MM2N1c: Students will add, subtract, multiply, and divide complex numbersStandard 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 NumberReal Number Imaginary NumberPure Imaginary NumberStandard Form
Review of old material
49
25
72
40
77*7
55*5
10210*2*2
262*6*6
Complex Numbers
36
81
12
125
i616)1(*6*6
i919)1(*9*9
55)1(*5*5*5 i
32)1(*3*2*2 i
How do you think we can reduce:
Complex NumbersThe “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 NumbersComplex 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 NumbersGive 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
Real Numbers(a + 0i)
-1 ⅜
Imaginary Numbers(a + bi, b ≠ 0)
2 + 3i 5 – 5i
Pure Imaginary Numbers(0 + bi, b ≠ 0)
-4i 8i
23
Complex NumbersWe 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 - 25x - 11
4a + 3
8 + 2i
Complex NumbersThe “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 NumbersSimplify 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 NumbersJust 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 + 28iAny questions as far as adding or
subtracting complex numbers?
x = 5, y = -3x = -3, y = 4
Complex Numbers – Guided PracticeDo problems 7 – 13 odd on page 9Do problems 35 – 39 odd on page 5
Warm Up:
Write in standard form:
Solve and write solution in standard form:
49
27
0502 x
i733i
25i
Warm-UpSimplify 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 – ApplicationApplications of Complex Numbers - Spri
ng/Mass Systemhttp://www.picomonster.com/complex-n
umbers rowing
Review of old material
49
25
72
40
77*7
55*5
10210*2*2
262*6*6
Complex Numbers
36
81
12
125
i616)1(*6*6
i919)1(*9*9
55)1(*5*5*5 i
32)1(*3*2*2 i
How do you think we can reduce:
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 NumbersHow do you think we would do the
following? (2 – 3i)(5 + 2i) Imaginary numbers may be multiplied
by the distributive rule.
Complex NumbersHow do you think we would do the
following? (2 – 3i)(5 + 2i) = 10 + 4i – 15i – 6i2= 10 – 11i – 6i2Can we simplify the i2? i2 = i * i = -1, so we get:= 10 – 11i – 6(-1)= 10 – 11i + 6 = 16 – 11i
Complex NumbersSimplify: (7 + 5i)(3 - 2i)= 21 -14i + 15i – 10i2= 21 + i – 10(-1)= 21 + i + 10 = 31 + i
Complex Numbers – Guided Practice – 5 minutesDo 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 NumbersDefinition: i 1
1i
1)1( 22 i
iiii 23
1)1)(1(224 iii
1)1)(1(448 iii
iiiii ))(1( 3347
1))(1( 2246 iiii
iiiii )1(45
Reducing Complex Numbers – Guided Practice – 5 minutesDo problems 7 – 13 odd on page 13
Complex Numbers – SummarySummarize :1. What are complex numbers?2. What does their standard form look
like??3. What does i equal?4. How do we add/subtract them?5. How do we take the square root of a
negative number?
Complex Numbers – SummarySummarize :1. How do we multiply them?2. How do we simplify higher order
imaginary numbers?
Complex Numbers – Ticket out the doorSimplify:1. (2 – 3i) – (-5 + 7i)2. (4 + i)(3 – 2i)
3. .
4. i23
32
Complex Numbers – Warm-upSolve 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 DenominatorConjugates
Review of past material: Simplify:
2124x
xxx
236 2
62212
24
xx
5.1323
26 2
xxx
xx
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
denominatorWe must “rationalize the denominator”,
which means we must eliminate all the square roots, including i.
Dividing by Imaginary Numbers
How can we solve ? ii
325
iiiii
ii
ii
321
32
352
325
3)25(
2
2
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
ii3225
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 youmade a
mistake!!!
ii3225
2
2
9664641510
)32()32(
32)25(
iiiiii
ii
ii
Dividing by Imaginary Numbers
How can we solve ? HINT: Look at the warm-up. i
i3225
2
2
9664641510
)32()32(
32)25(
iiiiii
ii
ii
iii1361
134
13194
)1(94)1(61910
PracticePage 13, # 29 – 37 odd