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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))

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  1. anonymous
    • one year ago
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    @satellite73 @jim_thompson5910 @ganeshie8 @mathstudent55 @zepdrix @Nnesha @Luigi0210 @whpalmer4 @sleepyjess @mathmate @KyanTheDoodle @misty1212

  2. zepdrix
    • one year ago
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    Yes. Have you learned about the exponential form of the complex number?

  3. anonymous
    • one year ago
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    It sounds familiar, if I see it I will probably remember

  4. zepdrix
    • one year ago
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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.

  5. zepdrix
    • one year ago
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    \[\Large\rm \frac{r_1}{r_2}e^{i(\alpha-\beta)}=\frac{r_1}{r_2}(\cos(\alpha-\beta)+i \sin (\alpha-\beta))\]

  6. misty1212
    • one year ago
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    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 !

  7. anonymous
    • one year ago
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    So in this case 14.76/44.29(cis(0.49-1.92))?

  8. anonymous
    • one year ago
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    Oh wait, nvm

  9. anonymous
    • one year ago
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    sorry, I added the wrong equation

  10. anonymous
    • one year ago
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    14.76/3(cis(0.49-1.43))

  11. zepdrix
    • one year ago
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    mmm ya that looks better! :)

  12. anonymous
    • one year ago
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    So if I simplify this, it is the answer?

  13. zepdrix
    • one year ago
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    (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

  14. anonymous
    • one year ago
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    4.92(cis(-0.94)) Is a negative acceptable?

  15. zepdrix
    • one year ago
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    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).

  16. zepdrix
    • one year ago
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    But in general, a negative is fine. :o

  17. anonymous
    • one year ago
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    awesome! Is rectangular form a+bi?

  18. anonymous
    • one year ago
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    Not for this equation

  19. anonymous
    • one year ago
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    Just curious

  20. zepdrix
    • one year ago
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    yes, thats what i was referring to. i guess they call it "standard form" or something though, my bad :)

  21. anonymous
    • one year ago
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    Actually for this problem they wanted polar. I was curious, because for another problem I am asked for rectangular form.

  22. anonymous
    • one year ago
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    Also, is the complex conjugate of a r(cos(theta)+isin(theta)) r(cos(theta)-isin(theta))?

  23. zepdrix
    • one year ago
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    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.

  24. zepdrix
    • one year ago
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    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

  25. anonymous
    • one year ago
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    ah gotcha

  26. anonymous
    • one year ago
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    But if there was an r, it would stay the same?

  27. zepdrix
    • one year ago
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    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.

  28. anonymous
    • one year ago
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    do you think I should do \[8^2\]? Or just write 8?

  29. zepdrix
    • one year ago
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    For what? 0_O

  30. anonymous
    • one year ago
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    oh sorry, for complex conjugate if r=8 for the original

  31. anonymous
    • one year ago
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    before the complex conjugate

  32. zepdrix
    • one year ago
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    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,\[\Large\rm 8\cos \theta -8i \sin \theta=8(\cos \theta-i \sin \theta)\]

  33. xapproachesinfinity
    • one year ago
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    1.43 was tnat supposed to be root2 lol

  34. anonymous
    • one year ago
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    @xapproachesinfinity yea

  35. xapproachesinfinity
    • one year ago
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    hehe that's weird in decimal and it is 1.41 not 1.43

  36. anonymous
    • one year ago
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    Oh, wait, it isn't

  37. anonymous
    • one year ago
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    it is just 1.43, not root 2

  38. anonymous
    • one year ago
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    The problem gave the equation liek that

  39. xapproachesinfinity
    • one year ago
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    oh im not thinking that's angle in radians lol

  40. anonymous
    • one year ago
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    oh kk

  41. anonymous
    • one year ago
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    Thanks though

  42. anonymous
    • one year ago
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    @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?

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

  44. anonymous
    • one year ago
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    as in multiplying theta by 3 and cubing 4

  45. anonymous
    • one year ago
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    so de movire I think

  46. zepdrix
    • one year ago
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    ya that seems like the way to go :)

  47. anonymous
    • one year ago
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    awesome :D Thanks for the help!

  48. zepdrix
    • one year ago
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    yay team

  49. xapproachesinfinity
    • one year ago
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    good work :)

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