Cancelling Confusion with Complex Numbers and De Moivre's Theorem

[Trail_Builder]

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: I am looking at complex numbers and De Moivre's Theorem and its consequences Ill use \oslash as the "argument".

1. $$z_{1}z_{2} = r_{1}r_{2}(cos\oslash_{1}cos\oslash_{2} - sin\oslash_{1}sin\oslash_{2} + i(cos\oslash_{1}sin\oslash_{2} + sin\oslash_{1}cos\oslash_{2}))$$

which then cancels to

$$z_{1}z_{2} = r_{1}r_{2}(cos(\oslash_{1} + \oslash_{2}) + isin(\oslash_{1} + \oslash_{2}))$$

I see how the $$cos(\oslash_{1} + \oslash_{2})$$ gets there, but not sure what's going on with the rest :S.

2. $$\frac{1}{z} = \frac{1}{r}*\frac{cos\oslash-isin\oslash}{(cos\oslash+isin\oslash)(cos\oslash-isin\oslash)}$$

cancels to

$$\frac{1}{z} = \frac{1}{r}*(cos\oslash-isin\oslash)$$

have no idea what's going on there lol.


hope you can help :D

 

[cristo]

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: I am looking at complex numbers and De Moivre's Theorem and its consequences Ill use \oslash as the "argument".

1. $$z_{1}z_{2} = r_{1}r_{2}(cos\oslash_{1}cos\oslash_{2} - sin\oslash_{1}sin\oslash_{2} + i(cos\oslash_{1}sin\oslash_{2} + sin\oslash_{1}cos\oslash_{2}))$$

which then cancels to

$$z_{1}z_{2} = r_{1}r_{2}(cos(\oslash_{1} + \oslash_{2}) + isin(\oslash_{1} + \oslash_{2}))$$

I see how the $$cos(\oslash_{1} + \oslash_{2})$$ gets there, but not sure what's 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. $$\frac{1}{z} = \frac{1}{r}*\frac{cos\oslash-isin\oslash}{(cos\oslash+isin\oslash)(cos\oslash-isin\oslash)}$$

cancels to

$$\frac{1}{z} = \frac{1}{r}*(cos\oslash-isin\oslash)$$

have no idea what's going on there lol.

Expand the denominator, and use the identity sin^2x+cos^2x=1

 

[Trail_Builder]

thnx buddy :D