Complex Numbers 5-9

8/8/2019 Complex Numbers 5-9

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Consider the quadratic equation x2 + 1 = 0.

Solving forx, givesx2 = 1

12

!x

1!x

We make the following definition:

1!i

Complex Numbers


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1!i

Complex Numbers

12

!iNote that squaring both sides yields:

therefore

and

so

and

iiiii !!! *1* 132

1)1(*)1(* 224 !!! iii

iiiii !!! *1*45

1*1* 2246 !!! iiii

And so on


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Real NumbersImaginary

Numbers

Real numbers and imaginary numbers are

subsets of the set of complex numbers.

Complex Numbers


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Definition of a Complex NumberDefinition of a Complex Number

Ifa and b are real numbers, the numbera + bi is acomplex number, and it is said to be written instandard form.

Ifb = 0, the numbera + bi= a is a real number.

If a = 0, the numbera + bi is called an imaginarynumber.

Write the complex number in standard form

81 !! 81 i !y 241 i 221 i


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Addition and Subtraction of ComplexAddition and Subtraction of Complex

NumbersNumbers

If a + biand c +diare two complex numbers writtenin standard form, their sum and difference are

defined as follows.

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

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

Sum:

Difference:


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Perform the subtraction and write the answerin standard form.

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

3 + 2i 6 13 i

3 11i

234188 i

234298 ii y

234238 ii

4


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Multiplying Complex NumbersMultiplying Complex Numbers

Multiplying complex numbers is similar tomultiplying polynomials and combining like terms.

Perform the operation and write the result instandard form.( 6 2i )( 2 3i )

F O I L

12 18i 4i + 6i2

12 22i + 6 ( -1 )

6 22i


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Consider ( 3 + 2i )( 3 2i )

9 6i + 6i 4i2

9 4( -1 )

9 + 4

13

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

This concept can be used to divide complex numbers.


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Complex Conjugates and DivisionComplex Conjugates and Division

Complex conjugates-a pair of complex numbers ofthe form a +bi and abi where a and b arereal numbers.

( a +bi )( abi )

a 2abi+abib 2i 2

a 2b 2( -1 )

a 2 +b 2

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


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To find the quotient of two complex numbersmultiply the numerator and denominator

by the conjugate of the denominator.

dic

bia

dic

dic

dic

bia

y

!

22

2

dc

bdibciadiac

!

22

dc

iadbcbdac

!


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Perform the operation and write the result instandard form.

i

i

21

76

i

i

i

i

21

21

21

76

y

!

22

2

21

147126

!

iii

41

5146

!

i

5

520 i!

5

5

5

20 i! i! 4


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ii

i

4

31 i

i

ii

i

i

i

y

y

!

4

4

4

31

Perform the operation and write the resultin standard form.

222

2

14312

! ii

ii116

312

1

1

!

ii

ii 17

3

17

12

1 ! ii 17

3

17

12

1 !

i

17

317

17

1217

!i

17

14

17

5

!


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Expressing Complex NumbersExpressing Complex Numbers

in Polar Formin Polar FormNow, 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


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in Polar Formin Polar FormRemember these relationships between polar andrectangular form:

x

y!Utan 222 ryx !

Ucosrx !Usinr

y !

So any complex number, X +Yi, can be written in

polar form: irrYiX UU sincos !

)sin(cossincos UUUU irirr !

Urcis

Here is the shorthand way of writing polar form:


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in Polar Formin Polar FormRewrite the following complex number in polar form:4 - 2i

Rewrite the following complex number in

rectangular form: 0307cis


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in Polar Formin Polar FormExpress the following complex number inrectangular form:

)

3

sin

3

(cos2TT

i


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in Polar Formin Polar FormExpress the following complex number in polar form: 5i


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Products and Quotients ofProducts and Quotients of

Complex Numbers in Polar FormComplex Numbers in Polar Form)sin(cos 111 UU ir

The product of two complex numbers,

and

Can be obtained by using the following formula:

)sin(cos 222 UU ir

)sin(cos*)sin(cos 222111 UUUU irir

)]sin()[cos(* 212121 UUUU ! irr


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Complex Numbers in Polar FormComplex Numbers in Polar Form)sin(cos 111 UU ir

The quotient of two complex numbers,

and

Can be obtained by using the following formula:

)sin(cos 222 UU ir

)sin(cos/)sin(cos 222111 UUUU irir

)]sin()[cos(/ 212121 UUUU ! irr


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Complex Numbers in Polar FormComplex Numbers in Polar FormFind the product of5cis30 and 2cis120

Next, write that product in rectangular form


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Complex Numbers in Polar FormComplex Numbers in Polar FormFind the quotient of36cis300 divided by4cis120

Next, write that quotient in rectangular form


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Complex Numbers in Polar FormComplex Numbers in Polar FormFind the result ofLeave your answer in polar form.

Based on how you answered this problem,what generalization can we make about

raising a complex number in polar form to

a giv

en power?

4))120sin120(cos5( i


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De Moivres TheoremDe Moivres Theorem

De Moivre'sTheorem is the theorem which

shows us how to take complex numbers to any

power easily.

De Moivre's Theorem Let r(cos *+isin *) be acomplex 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.


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De Moivres TheoremDe Moivres TheoremTry a sample problem: What is [3(cos 45r+isin45)]5?

To do this take 3 to the 5thpower, then multiply 45 times 5

and plug back into trigonometric form.

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

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

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


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Find the result of:

Because of the power involved, it would easier to change this

complex number into polar form and then use De Moivres Theorem.

4)1( i


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De Moivre'sTheorem also works not only for

integer values of powers, but also rational values

(so we can determine roots of complex numbers).

pprcisyix

11

)()( U!

)()

1

*(

11

pcisrpcisrpp U

U !!


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Simplify the following: 31

)344( i


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De Moivres TheoremDe Moivres TheoremEv

ery complex numberh

as p distinct pt

h 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 ofDe MoivresTheorem

)360

()(11

p

ncisryix pp

!

U

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 roots

would be evenly spaced about 360 degrees (360/p would

tell us how far apart the roots would be).


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


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Solve the following equation for all complex

number solutions (roots): 0273 !x

Complex Numbers 5-9

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