Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Trigonometry Ch6Sec6Day2, Schemes and Mind Maps of Trigonometry

Harvard UniversityTrigonometry

The theorem for finding products and quotients of complex numbers in polar form. It also provides examples of how to apply the theorem to find the product of two complex numbers in polar form. relevant for students studying trigonometry and complex numbers.

Typology: Schemes and Mind Maps

2022/2023

Uploaded on 03/14/2023

shanti_122
shanti_122 🇺🇸

3.9

(17)

231 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Trigonometry
Ch6Sec6Day2 1 1/26/2019
Section 6.6 – Complex Numbers in Polar Form – Day 2
Objective:
2) Given complex numbers in polar form, find products, quotients, and powers.
The polar form of a complex number provides an alternative for finding products and quotients of complex numbers.
Theorem:
Let
)sincos(rz 1111 i
and
)sincos(rz 2222 i
be two complex numbers.
Then
)(sin)(cosrrzz 21212121 i
)(cisrr 2121
If
,0z2
then
)(sini)(cos
r
r
z
z2121
2
1
2
1
)(cis
r
r21
2
1
Example 5: If
)40sin40cos(2z i
and
find zw and
.
w
z
Leave your
answers in polar form.
)20sin20cos(6)40sin40cos(2wz ii
)2040(sin)2040(cos)6(2 i
)60sin60cos(12 i
)20sin20cos(6
)40sin40cos(2
w
z
i
i
)2040(sin)2040(cos
6
2 i
)20sin20cos(
3
1 i
Example 6: If
)340sin340cos(2z i
and
find zw and
.
w
z
Leave your answers in polar form.
)50sin50cos(6)340sin340cos(2wz ii
)50340(sin)50340(cos)6(2 i
)390sin390cos(12 i
Not polar form, since
is not between
360and0
)30360(sin)30360(cos12 i
)30sin30cos(12 i
by the periodic property
)50sin50cos(6
)340sin340cos(2
w
z
i
i
)50340(sin)50340(cos
6
2 i
)290sin290cos(
3
1 i
pf2

Partial preview of the text

Download Trigonometry Ch6Sec6Day2 and more Schemes and Mind Maps Trigonometry in PDF only on Docsity!

Trigonometry

Ch6Sec6Day2 1 1/26/

Section 6.6 – Complex Numbers in Polar Form – Day 2

Objective:

  1. Given complex numbers in polar form, find products, quotients, and powers.

The polar form of a complex number provides an alternative for finding products and quotients of complex numbers.

Theorem:

Let z 1  r 1 (cos 1  i sin 1 ) and z 2  r 2 (cos 2  i sin 2 )be two complex numbers.

Thenz 1 z 2  r 1 r 2  cos( 1  2 )isin( 1  2 )

r 1 r 2 cis( 1  2 )

If z 2  0 ,then  cos( ) isin( )

r

r

z

z

1 2 1 2 2

1

2

1      

cis( ) r

r

1 2 2

1   

Example 5: Ifz 2 (cos 40 sin 40 )

    i andw 6 (cos 20 sin 20 ),

    i find zw and. w

z Leave your

answers in polar form.

z w  2 (cos 40 sin 40 )  6 (cos 20 sin 20 )

      ii

2 ( 6 ) cos( 40 20 ) sin( 40 20 )

       i

12 (cos 60 sin 60 )

    i

6 (cos 20 sin 20 )

2 (cos 40 sin 40 )

w

z

 

 

i

i

 cos( 40 20 ) sin( 40 20 )

   i

(cos 20 sin 20 ) 3

  i

Example 6: Ifz 2 (cos 340 sin 340 )

 

 i andw 6 (cos 50 sin 50 ),

 

 i find zw and.

w

z

Leave your answers in polar form.

z w  2 (cos 340 sin 340 )  6 (cos 50 sin 50 )

   

 i i

2 ( 6 ) cos( 340 50 ) sin( 340 50 )

   

  i 

12 (cos 390 sin 390 )

 

 i Not polar form, since is not between

  0 and 360

12  cos( 360 30 ) sin( 360 30 )

   

  i 

12 (cos 30 sin 30 )

 

 i by the periodic property

6 (cos 50 sin 50 )

2 (cos 340 sin 340 )

w

z

 

 

i

i

 cos( 340 50 ) sin( 340 50 )

  i 

(cos 290 sin 290 ) 3

 i

Trigonometry

Ch6Sec6Day2 2 1/26/

Section 6.6 – Complex Numbers in Polar Form – Day 2 (continued)

De Moivre’s Theorem is a formula for raising a complex number z to the power n, where n  1 is a positive integer.

Theorem – De Moivre’s Theorem

If z  r(cosisin)is a complex number, then z r  cos(n) sin(n)

n n   i  where n  1 is a positive integer.

Example 7: Write 

3 2 ( cos 40 sin 40 )

   i in the standard forma b i.

 2 (cos 40 sin 40 ) 2  cos 3 ( 40 ) sin 3 ( 40 )

3 3      i   i by De Moivre’s Theorem

8 (cos 120 sin 120 )

    i

  i r

y

r

x 8 Now, 

   2

3 , 2

1 120 : (x,y)

 .

i 1

i 2

 4  4 3 i

Example 8: Use De Moivre’s Theorem to write

3

(  4  3 i ) in the standard forma b i.

 4  3 i: rectangular coordinates (x,y) ( 4 , 3 ) Quadrant II

2 2 r  x y

2 2  (  4 ) ( 3 )

Find the reference angleref: x

y tan

1 ref

   is in Quadrant II, thus, ref

tan

1

    180  36. 9

tan

(^1)   143. 1

  36. 86

  36. 9

Thus, the polar form of  4  3 i is  4  3 ir(cosisin)

5 (cos 143. 1 sin 143. 1 )

   i

So,  

3 3 ( 4 3 ) 5 (cos 143. 1 sin 143. 1 )

 

  i  i

5  cos 3 ( 143. 1 ) sin 3 ( 143. 1 )

3  

 i by De Moivre’s Theorem

125 (cos 429. 3 sin 429. 3 )

 

 i

    

   

 125 cos 69. 3  360 isin 69. 3  360

125 (cos 69. 3 sin 69. 3 )

 

 i by the periodic property

 125 ( 0. 353475  0. 935444 i)

 44. 184375  116. 9305 i

 44  117 i