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))

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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))

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Yes. Have you learned about the exponential form of the complex number?
It sounds familiar, if I see it I will probably remember

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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.
\[\Large\rm \frac{r_1}{r_2}e^{i(\alpha-\beta)}=\frac{r_1}{r_2}(\cos(\alpha-\beta)+i \sin (\alpha-\beta))\]
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 !
So in this case 14.76/44.29(cis(0.49-1.92))?
Oh wait, nvm
sorry, I added the wrong equation
14.76/3(cis(0.49-1.43))
mmm ya that looks better! :)
So if I simplify this, it is the answer?
(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
4.92(cis(-0.94)) Is a negative acceptable?
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).
But in general, a negative is fine. :o
awesome! Is rectangular form a+bi?
Not for this equation
Just curious
yes, thats what i was referring to. i guess they call it "standard form" or something though, my bad :)
Actually for this problem they wanted polar. I was curious, because for another problem I am asked for rectangular form.
Also, is the complex conjugate of a r(cos(theta)+isin(theta)) r(cos(theta)-isin(theta))?
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.
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
ah gotcha
But if there was an r, it would stay the same?
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.
do you think I should do \[8^2\]? Or just write 8?
For what? 0_O
oh sorry, for complex conjugate if r=8 for the original
before the complex conjugate
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,\[\Large\rm 8\cos \theta -8i \sin \theta=8(\cos \theta-i \sin \theta)\]
1.43 was tnat supposed to be root2 lol
hehe that's weird in decimal and it is 1.41 not 1.43
Oh, wait, it isn't
it is just 1.43, not root 2
The problem gave the equation liek that
oh im not thinking that's angle in radians lol
oh kk
Thanks though
@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?
Distribute the 3 as in.... Expanding out three sets of brackets? No. Or do you mean by apply De'Moivre's Rule? Yes.
as in multiplying theta by 3 and cubing 4
so de movire I think
ya that seems like the way to go :)
awesome :D Thanks for the help!
yay team
good work :)

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