Cancelling confusion

  • #1
Trail_Builder
149
0
hi

i'm confused at to how my textbook has done the following cancelling :S. hope you can clear things up for me :D

thnx

context: im looking at complex numbers and De Moivre's Theorem and its consequences Ill use \oslash as the "arguement".

1. [tex]z_{1}z_{2} = r_{1}r_{2}(cos\oslash_{1}cos\oslash_{2} - sin\oslash_{1}sin\oslash_{2} + i(cos\oslash_{1}sin\oslash_{2} + sin\oslash_{1}cos\oslash_{2}))[/tex]

which then cancels to

[tex]z_{1}z_{2} = r_{1}r_{2}(cos(\oslash_{1} + \oslash_{2}) + isin(\oslash_{1} + \oslash_{2}))[/tex]

I see how the [tex]cos(\oslash_{1} + \oslash_{2})[/tex] gets there, but not sure whats going on with the rest :S.

2. [tex]\frac{1}{z} = \frac{1}{r}*\frac{cos\oslash-isin\oslash}{(cos\oslash+isin\oslash)(cos\oslash-isin\oslash)}[/tex]

cancels to

[tex]\frac{1}{z} = \frac{1}{r}*(cos\oslash-isin\oslash)[/tex]

have no idea whats going on there lol.


hope you can help :D
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,140
74
hi

i'm confused at to how my textbook has done the following cancelling :S. hope you can clear things up for me :D

thnx

context: im looking at complex numbers and De Moivre's Theorem and its consequences Ill use \oslash as the "arguement".

1. [tex]z_{1}z_{2} = r_{1}r_{2}(cos\oslash_{1}cos\oslash_{2} - sin\oslash_{1}sin\oslash_{2} + i(cos\oslash_{1}sin\oslash_{2} + sin\oslash_{1}cos\oslash_{2}))[/tex]

which then cancels to

[tex]z_{1}z_{2} = r_{1}r_{2}(cos(\oslash_{1} + \oslash_{2}) + isin(\oslash_{1} + \oslash_{2}))[/tex]

I see how the [tex]cos(\oslash_{1} + \oslash_{2})[/tex] gets there, but not sure whats going on with the rest :S.
This is just using the double angle formula for sine: sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
2. [tex]\frac{1}{z} = \frac{1}{r}*\frac{cos\oslash-isin\oslash}{(cos\oslash+isin\oslash)(cos\oslash-isin\oslash)}[/tex]

cancels to

[tex]\frac{1}{z} = \frac{1}{r}*(cos\oslash-isin\oslash)[/tex]

have no idea whats going on there lol.
Expand the denominator, and use the identity sin^2x+cos^2x=1
 
  • #3
Trail_Builder
149
0
thnx buddy :D
 

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