## 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))

1. anonymous

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

2. zepdrix

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

3. anonymous

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

4. 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.

5. zepdrix

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

6. misty1212

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

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

8. anonymous

Oh wait, nvm

9. anonymous

sorry, I added the wrong equation

10. anonymous

14.76/3(cis(0.49-1.43))

11. zepdrix

mmm ya that looks better! :)

12. anonymous

So if I simplify this, it is the answer?

13. 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

14. anonymous

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

15. 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).

16. zepdrix

But in general, a negative is fine. :o

17. anonymous

awesome! Is rectangular form a+bi?

18. anonymous

Not for this equation

19. anonymous

Just curious

20. zepdrix

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

21. anonymous

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

22. anonymous

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

23. 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.

24. 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

25. anonymous

ah gotcha

26. anonymous

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

27. 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.

28. anonymous

do you think I should do $8^2$? Or just write 8?

29. zepdrix

For what? 0_O

30. anonymous

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

31. anonymous

before the complex conjugate

32. 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,$\Large\rm 8\cos \theta -8i \sin \theta=8(\cos \theta-i \sin \theta)$

33. xapproachesinfinity

1.43 was tnat supposed to be root2 lol

34. anonymous

@xapproachesinfinity yea

35. xapproachesinfinity

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

36. anonymous

Oh, wait, it isn't

37. anonymous

it is just 1.43, not root 2

38. anonymous

The problem gave the equation liek that

39. xapproachesinfinity

oh im not thinking that's angle in radians lol

40. anonymous

oh kk

41. anonymous

Thanks though

42. 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?

43. zepdrix

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

as in multiplying theta by 3 and cubing 4

45. anonymous

so de movire I think

46. zepdrix

ya that seems like the way to go :)

47. anonymous

awesome :D Thanks for the help!

48. zepdrix

yay team

49. xapproachesinfinity

good work :)