When dividing with the de moivre's, how do I do it?
Do I divide the r values and subtract the theta values?
For example if I have: 14.76(cos(0.49)+isin(0.49))/3(cos(1.43)+isin(1.43))

- anonymous

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- schrodinger

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- anonymous

@satellite73 @jim_thompson5910 @ganeshie8 @mathstudent55 @zepdrix @Nnesha @Luigi0210 @whpalmer4 @sleepyjess @mathmate @KyanTheDoodle @misty1212

- zepdrix

Yes.
Have you learned about the exponential form of the complex number?

- anonymous

It sounds familiar, if I see it I will probably remember

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## More answers

- zepdrix

That certainly sheds some light on these rules.\[\Large\rm r( \cos \theta + i \sin \theta)= r e^{i \theta}\]So if we have:\[\large\rm \frac{r_1(\cos \alpha+i \sin \alpha)}{r_2(\cos \beta+i \sin \beta)}=\frac{r_1 e^{i \alpha}}{r_2e^{i \beta}}=\frac{r_1}{r_2}e^{i(\alpha-\beta)}\]When you convert to exponential form, you can see that it's just an exponent rule being applied to the angles.

- zepdrix

\[\Large\rm \frac{r_1}{r_2}e^{i(\alpha-\beta)}=\frac{r_1}{r_2}(\cos(\alpha-\beta)+i \sin (\alpha-\beta))\]

- misty1212

sure is a damn sight easier than using the "addition angle" formula isn't it? just the laws of exponents is all, a much quicker explanation !

- anonymous

So in this case 14.76/44.29(cis(0.49-1.92))?

- anonymous

Oh wait, nvm

- anonymous

sorry, I added the wrong equation

- anonymous

14.76/3(cis(0.49-1.43))

- zepdrix

mmm ya that looks better! :)

- anonymous

So if I simplify this, it is the answer?

- zepdrix

(14.76/3) cis(0.49-1.43)
^ Just making sure that you know the cis is not in the denominator with the 3
Yes.
Just make sure you have the correct form, polar or rectangular or whatever :p

- anonymous

4.92(cis(-0.94)) Is a negative acceptable?

- zepdrix

Hehe, you tell me :)
If you're supposed to have a positive angle, you could can take a trip around the circle. Add 2pi or 360 degrees to your angle (Depending on whether or not that was in radians or degrees).

- zepdrix

But in general, a negative is fine. :o

- anonymous

awesome! Is rectangular form a+bi?

- anonymous

Not for this equation

- anonymous

Just curious

- zepdrix

yes, thats what i was referring to.
i guess they call it "standard form" or something though, my bad :)

- anonymous

Actually for this problem they wanted polar. I was curious, because for another problem I am asked for rectangular form.

- anonymous

Also, is the complex conjugate of a r(cos(theta)+isin(theta)) r(cos(theta)-isin(theta))?

- zepdrix

mmm yes!
you can actually solve this problem by multiplying top and bottom by the complex conjugate of the denominator. That's another approach to try sometime :) It's kinda neat.

- zepdrix

err i guess we shouldn't include the r's when we're talking about complex conjugate.
It's more accurate to say (cos theta- i sin theta) is the conjugate of (cos theta+ i sin theta).
Let's leave the r's out of there :3

- anonymous

ah gotcha

- anonymous

But if there was an r, it would stay the same?

- zepdrix

Well it's just that umm...
When you multiply complex conjugates,
you end up with `the sum of squares`.
Example:
\(\Large\rm (\cos x+i\sin x)(\cos x-i\sin x)=\cos^2x+\sin^2x\)
If we threw some r's into the mix though,
\(\Large\rm r(\cos x+i\sin x)\cdot r(\cos x-i\sin x)=r^2(\cos^2x+\sin^2x)\)
We get the sum of squares, but a little bit more garbage.
The r's are not part of that rule, that's all im trying to say.

- anonymous

do you think I should do \[8^2\]? Or just write 8?

- zepdrix

For what? 0_O

- anonymous

oh sorry, for complex conjugate if r=8 for the original

- anonymous

before the complex conjugate

- zepdrix

So you're asking, what is the complex conjugate of this?\[\Large\rm 8(\cos \theta+ i \sin \theta)\]Well I guess I'm forgetting about the fact that you can distribute the 8,\[\Large\rm 8\cos \theta+8i \sin \theta\]So the complex conjugate would be,\[\Large\rm 8\cos \theta -8i \sin \theta=8(\cos \theta-i \sin \theta)\]

- xapproachesinfinity

1.43 was tnat supposed to be root2 lol

- anonymous

@xapproachesinfinity yea

- xapproachesinfinity

hehe that's weird in decimal and it is 1.41 not 1.43

- anonymous

Oh, wait, it isn't

- anonymous

it is just 1.43, not root 2

- anonymous

The problem gave the equation liek that

- xapproachesinfinity

oh im not thinking that's angle in radians lol

- anonymous

oh kk

- anonymous

Thanks though

- anonymous

@zepdrix
If I need to turn (4(cos(7pi/9)+isin(7pi/9)))^3 into the form a+bi, would I just distribute the 3?

- zepdrix

Distribute the 3 as in....
Expanding out three sets of brackets? No.
Or do you mean by apply De'Moivre's Rule? Yes.

- anonymous

as in multiplying theta by 3 and cubing 4

- anonymous

so de movire I think

- zepdrix

ya that seems like the way to go :)

- anonymous

awesome :D Thanks for the help!

- zepdrix

yay team

- xapproachesinfinity

good work :)

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