Consider the quadratic equation x 2 + 1 = 0. Solving for x, gives x 2 = – 1 We make the following...

Consider the quadratic equation x2 + 1 = 0. Solving for x , gives x2 = – 1

12 x

1x

We make the following definition:

1i

Complex Numbers

1i

Complex Numbers

12 iNote that squaring both sides yields:therefore

and

so

and

iiiii *1* 13 2

1)1(*)1(* 224 iii

iiiii *1*45

1*1* 2246 iiii

And so on…

Real NumbersImaginary Numbers

Real numbers and imaginary numbers are subsets of the set of complex numbers.

Complex Numbers

Definition of a Complex Number Definition of a Complex Number

If a and b are real numbers, the number a + bi is a complex number, and it is said to be written in standard form.

If b = 0, the number a + bi = a is a real number.

If a = 0, the number a + bi is called an imaginary number.

Write the complex number in standard form

81 81 i 241 i 221 i

Addition and Subtraction of Complex Addition and Subtraction of Complex Numbers Numbers

If a + bi and c +di are two complex numbers written in standard form, their sum and difference are defined as follows.

i)db()ca()dic()bia(

i)db()ca()dic()bia(

Sum:

Difference:

Perform the subtraction and write the answer in standard form.

( 3 + 2i ) – ( 6 + 13i ) 3 + 2i – 6 – 13i –3 – 11i

234188 i

234298 ii

234238 ii

4

Multiplying Complex NumbersMultiplying Complex Numbers

Multiplying complex numbers is similar to multiplying polynomials and combining like terms.

Perform the operation and write the result in standard form. ( 6 – 2i )( 2 – 3i )

F O I L12 – 18i – 4i + 6i2

12 – 22i + 6 ( -1 )6 – 22i

Slide 8-8

Multiplication of Complex Multiplication of Complex NumbersNumbersFor complex numbers a + bi and c + di,

The product of two complex numbers is found by multiplying as if the numbers were binomials and using the fact that i2 = 1.

(a bi)(c di) (ac bd ) (ad bc)i.

Examples: MultiplyingExamples: Multiplying (2 4i)(3 + 5i) (7 + 3i)2

Examples: MultiplyingExamples: Multiplying (2 4i)(3 + 5i) (7 + 3i)2

2

2(3) 2(5 ) 4 (3) 4 (5 )

6 10 12 20

6 2 20( 1)

26 2

i i i i

i i i

i

i

2 2

2

7 2(7)(3 ) (3 )

49 42 9

49 42 9( 1)

40 42

i i

i i

i

i

Consider ( 3 + 2i )( 3 – 2i )9 – 6i + 6i – 4i2

9 – 4( -1 )9 + 4 13

This is a real number. The product of two complex numbers can be a real number.

This concept can be used to divide complex numbers.

Complex Conjugates and Division Complex Conjugates and Division

Complex conjugates-a pair of complex numbers of the form a + bi and a – bi where a and b are real numbers.

( a + bi )( a – bi )a 2 – abi + abi – b 2 i 2

a 2 – b 2( -1 )a 2 + b 2

The product of a complex conjugate pair is a positive real number.

Advanced Algebra 2Advanced Algebra 2

Agenda: November 9, 2011 Do Now: (2 4i)(3 - 5i)And (5+ 2i)2

(be ready to present)

Class work: Lecture1. Dividing Complex

numbers2. Determine the complex

conjugate!

We Will Define and determine the

COMPLEX CONJUGATE! Multiply complex numbers Divide complex numbers

HW: OK, NOW complete the packet and hand into me

And I have another practice worksheet

13

To find the quotient of two complex numbers multiply the numerator and denominator by the conjugate of the denominator.

dic

bia

dic

dic

dic

bia

22

2

dc

bdibciadiac

22 dc

iadbcbdac

DIVIDING COMPLEX NUMBERS!

(mild) Perform the operation and write the result in standard form. We DO NOT want to leave i in denominator and we know that the product of a complex number and its conjugate is always a real number.

What is the complex conjugate of (1-2i) ?

i

i

21

76

(mild) Perform the operation and write the

result in standard form.

i

i

21

76

i

i

i

i

21

21

21

76

(mild) Perform the operation and write the

result in standard form.

i

i

21

76

i

i

i

i

21

21

21

76

22

2

21

147126

iii

41

5146

i

(mild) Perform the operation and write the

result in standard form.

i

i

21

76

i

i

i

i

21

21

21

76

22

2

21

147126

iii

41

5146

i

5

520 i

5

5

5

20 i i4

(medium) Dividing complex Numbers(medium) Dividing complex Numbers

For real numbers a and b, (a + bi)(a bi) = a2 + b2.

…the product of a complex number and its conjugate is always a real number.

What’s the complex conjugate of (2-i)?

Example 2:

i

i

2

35

(medium) Dividing complex Numbers(medium) Dividing complex Numbers

For real numbers a and b, (a + bi)(a bi) = a2 + b2.

The product of a complex number and its conjugate is always a real number.

Example

2

2

5 3

2(5 3 )(2 )

(2 )(2 )

10 5 6 3

47 11

57 11

5 5

i

ii i

i i

i i i

ii

i

(medium) Dividing complex Numbers(medium) Dividing complex Numbers

For real numbers a and b, (a + bi)(a bi) = a2 + b2.

The product of a complex number and its conjugate is always a real number.

Example

2

2

5 3

2(5 3 )(2 )

(2 )(2 )

10 5 6 3

47 11

57 11

5 5

i

ii i

i i

i i i

ii

i

(medium) Dividing complex Numbers(medium) Dividing complex Numbers

For real numbers a and b, (a + bi)(a bi) = a2 + b2.

The product of a complex number and its conjugate is always a real number.

Example

2

2

5 3

2(5 3 )(2 )

(2 )(2 )

10 5 6 3

47 11

57 11

5 5

i

ii i

i i

i i i

ii

i

ii

i

4

31

(SPICY) Perform the operation and write the result in standard form.We have to find 2 complex conjugates this time!!

ii

i

4

31 i

i

ii

i

i

i

4

4

4

31

(SPICY) Perform the operation and write the result in standard form.

ii

i

4

31 i

i

ii

i

i

i

4

4

4

31

(SPICY) Perform the operation and write the result in standard form.

222

2

14

312

i

i

ii

ii

i

4

31 i

i

ii

i

i

i

4

4

4

31

(SPICY) Perform the operation and write the result in standard form.

222

2

14

312

i

i

ii116

312

1

1

ii

ii

i

4

31 i

i

ii

i

i

i

4

4

4

31

(SPICY) Perform the operation and write the result in standard form.

222

2

14

312

i

i

ii116

312

1

1

ii

ii17

3

17

121

ii

i

4

31 i

i

ii

i

i

i

4

4

4

31

(SPICY) Perform the operation and write the result in standard form.

222

2

14

312

i

i

ii116

312

1

1

ii

ii17

3

17

121 ii

17

3

17

121

ii

i

4

31 i

i

ii

i

i

i

4

4

4

31

(SPICY) Perform the operation and write the result in standard form.

222

2

14

312

i

i

ii116

312

1

1

ii

ii17

3

17

121 ii

17

3

17

121

i17

317

17

1217

ii

i

4

31 i

i

ii

i

i

i

4

4

4

31

(SPICY) Perform the operation and write the result in standard form.

222

2

14

312

i

i

ii116

312

1

1

ii

ii17

3

17

121 ii

17

3

17

121

i17

317

17

1217

i

17

14

17

5

Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form

Now, any Complex Number can be expressed as:X + Y i That number can be plotted as on ordered pair in rectangular form like so…

6

4

2

-2

-4

-6

-5 5

Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form

Remember these relationships between polar and rectangular form:

x

ytan 222 ryx

cosrx sinry

So any complex number, X + Yi, can be written inpolar form: irrYiX sincos

)sin(cossincos irirr

rcisHere is the shorthand way of writing polar form:

Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form

Rewrite the following complex number in polar form: 4 - 2i

Rewrite the following complex number inrectangular form: 0307cis

Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form

Express the following complex number inrectangular form: )

3sin

3(cos2

i

Expressing Complex NumbersExpressing Complex Numbers in Polar Form in Polar Form

Express the following complex number inpolar form: 5i

Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form

)sin(cos 111 ir

The product of two complex numbers, and

Can be obtained by using the following formula:)sin(cos 222 ir

)sin(cos*)sin(cos 222111 irir

)]sin()[cos(* 212121 irr

Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form

)sin(cos 111 ir

The quotient of two complex numbers, and

Can be obtained by using the following formula:)sin(cos 222 ir

)sin(cos/)sin(cos 222111 irir

)]sin()[cos(/ 212121 irr

Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form

Find the product of 5cis30 and –2cis120

Next, write that product in rectangular form

Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form

Find the quotient of 36cis300 divided by 4cis120

Next, write that quotient in rectangular form

Products and Quotients of Products and Quotients of Complex Numbers in Polar FormComplex Numbers in Polar Form

Find the result ofLeave your answer in polar form.

Based on how you answered this problem, what generalization can we make aboutraising a complex number in polar form toa given power?

4))120sin120(cos5( i

De Moivre’s TheoremDe Moivre’s Theorem

De Moivre's Theorem is the theorem which shows us how to take complex numbers to any power easily.

De Moivre's Theorem – Let r(cos +isin ) be a complex number and n be any real number. Then

[r(cos +isin ]n = rn(cos n+isin n)

What is this saying?

The resulting r value will be r to the nth power and the resulting angle will be n times the original angle.

De Moivre’s TheoremDe Moivre’s Theorem

Try a sample problem: What is [3(cos 45+isin45)]5?

To do this take 3 to the 5th power, then multiply 45 times 5 and plug back into trigonometric form.

35 = 243 and 45 * 5 =225 so the result is 243(cos 225+isin 225)

Remember to save space you can write it in compact form.

243(cos 225+isin 225)=243cis 225

De Moivre’s TheoremDe Moivre’s Theorem

Find the result of: Because of the power involved, it would easier to change this complex number into polar form and then use De Moivre’s Theorem.

4)1( i

De Moivre’s TheoremDe Moivre’s Theorem

De Moivre's Theorem also works not only for integer values of powers, but also rational values (so we can determine roots of complex numbers).

pp rcisyix11

)()(

)()1

*(11

pcisr

pcisr pp

De Moivre’s TheoremDe Moivre’s Theorem

Simplify the following: 3

1

)344( i

De Moivre’s TheoremDe Moivre’s Theorem

Every complex number has ‘p’ distinct ‘pth’ complex roots (2 square roots, 3 cube roots, etc.)

To find the p distinct pth roots of a complex number, we use the following form of De Moivre’s Theorem

)360

()(11

p

ncisryix pp

…where ‘n’ is all integer values between 0 and p-1. Why the 360? Well, if we were to graph the complexroots on a polar graph, we would see that the p rootswould be evenly spaced about 360 degrees (360/p wouldtell us how far apart the roots would be).

De Moivre’s TheoremDe Moivre’s Theorem

Find the 4 distinct 4th roots of -3 - 3i

De Moivre’s TheoremDe Moivre’s Theorem

Solve the following equation for all complexnumber solutions (roots):

0273 x