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anonymous
 one year ago
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
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 @jim_thompson5910 @ganeshie8 @mathstudent55 @zepdrix @Nnesha @Luigi0210 @whpalmer4 @sleepyjess @mathmate @KyanTheDoodle @misty1212

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Yes. Have you learned about the exponential form of the complex number?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It sounds familiar, if I see it I will probably remember

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5That 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
 one year ago
Best ResponseYou've already chosen the best response.5\[\Large\rm \frac{r_1}{r_2}e^{i(\alpha\beta)}=\frac{r_1}{r_2}(\cos(\alpha\beta)+i \sin (\alpha\beta))\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0sure 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
 one year ago
Best ResponseYou've already chosen the best response.0So in this case 14.76/44.29(cis(0.491.92))?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, I added the wrong equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.014.76/3(cis(0.491.43))

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5mmm ya that looks better! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if I simplify this, it is the answer?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5(14.76/3) cis(0.491.43) ^ 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
 one year ago
Best ResponseYou've already chosen the best response.04.92(cis(0.94)) Is a negative acceptable?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Hehe, 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
 one year ago
Best ResponseYou've already chosen the best response.5But in general, a negative is fine. :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome! Is rectangular form a+bi?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not for this equation

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5yes, thats what i was referring to. i guess they call it "standard form" or something though, my bad :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually for this problem they wanted polar. I was curious, because for another problem I am asked for rectangular form.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also, is the complex conjugate of a r(cos(theta)+isin(theta)) r(cos(theta)isin(theta))?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5mmm 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
 one year ago
Best ResponseYou've already chosen the best response.5err 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
 one year ago
Best ResponseYou've already chosen the best response.0But if there was an r, it would stay the same?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Well it's just that umm... When you multiply complex conjugates, you end up with `the sum of squares`. Example: \(\Large\rm (\cos x+i\sin x)(\cos xi\sin x)=\cos^2x+\sin^2x\) If we threw some r's into the mix though, \(\Large\rm r(\cos x+i\sin x)\cdot r(\cos xi\sin x)=r^2(\cos^2x+\sin^2x)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0do you think I should do \[8^2\]? Or just write 8?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry, for complex conjugate if r=8 for the original

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0before the complex conjugate

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5So 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,\[\Large\rm 8\cos \theta 8i \sin \theta=8(\cos \thetai \sin \theta)\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.01.43 was tnat supposed to be root2 lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@xapproachesinfinity yea

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hehe that's weird in decimal and it is 1.41 not 1.43

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is just 1.43, not root 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The problem gave the equation liek that

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0oh im not thinking that's angle in radians lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.5Distribute the 3 as in.... Expanding out three sets of brackets? No. Or do you mean by apply De'Moivre's Rule? Yes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0as in multiplying theta by 3 and cubing 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so de movire I think

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5ya that seems like the way to go :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome :D Thanks for the help!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0good work :)
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