complex numbers - day 1

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