## anonymous one year ago Find all three complex cube roots of (2i - 2)

1. anonymous

|dw:1440071028488:dw| Am I at least on the right track?

2. ganeshie8

Yes $$\sqrt{8}$$ will be the magnitude of roots, it will be same for all three roots

3. anonymous

So what is the next step? I'm lost at the moment :(

4. ganeshie8

what exactly are we trying to find here ?

5. ganeshie8

if you understand the question well, you can answer it easily

6. anonymous

"Find all three complex roots of (2i - 2)".

7. ganeshie8

Yes, what is a root ?

8. ganeshie8

the root is a number which when you multiply itself three times produces the number 2i-2

9. anonymous

I'm not sure how to find out those roots though. I did it previously in another question but I'm getting so confused here for some reason

10. ganeshie8

you must be knowing that angles add up when you multiply two complex numbers : $e^{i\theta_1} *e^{i\theta_2} = e^{i(\theta_1+\theta_2)}$

11. ganeshie8

we can use that trick to find the angles of roots

12. anonymous

I think I've seen that before :)

13. ganeshie8

whats the angle of 2i-2 ?

14. anonymous

I'm still trying to find it out. I'm at root8 at the moment

15. ganeshie8

in which quadrant does the number (-2, 2) lie in ?

16. anonymous

sine

17. ganeshie8

wat do you mean by sine

18. anonymous

I'll just show you what I've been trying to do:

19. anonymous

|dw:1440073112572:dw|

20. ganeshie8

again, in which quadrant does the number (-2, 2) lie in ?

21. anonymous

2nd

22. ganeshie8

good, can you show it in the graph |dw:1440073212144:dw|

23. anonymous

|dw:1440073236919:dw|

24. ganeshie8

right, what is its angle ? |dw:1440073291571:dw|

25. anonymous

135 degrees

26. ganeshie8

when you see the number, you can eyeball the angle, arctan formula and all is not needed

27. ganeshie8

yes 135 is correct

28. ganeshie8

so the magnitude of given complex number is $$\sqrt{8}$$ and the angle is $$135$$ we're ready to find the roots

29. anonymous

Alright cool :)

30. ganeshie8

lets find the angles first

31. ganeshie8

basically you want to find $$\theta$$ such that $\theta+\theta+\theta = 135 + 360k$

32. ganeshie8

which is same as : $3\theta = 135 + 360k$

33. anonymous

So is theta 165?

34. ganeshie8

or $\theta = \dfrac{135}{3} + \dfrac{360k}{3}$

35. ganeshie8

or $\theta = 45 + 120k$

36. ganeshie8

letting $$k=0,$$ you get $$\theta = 45+120*0 = \color{red}{45}$$ letting $$k=1,$$ you get $$\theta = 45+120*1 = \color{red}{165}$$ letting $$k=2,$$ you get $$\theta = 45+120*2 =\color{red}{ 285}$$

37. ganeshie8

the magnitude is same for all of the roots : $$\large (\sqrt{~8~})^{\frac{1}{3}}$$

38. ganeshie8

notice that $$\large (\sqrt{~8~})^{\frac{1}{3}}$$ simplifies to $$\large \sqrt{~2~}$$

39. anonymous

Oh ok

40. ganeshie8

so the 3 roots are $\sqrt{~2~}(\cos(45)+i\sin(45))$ $\sqrt{~2~}(\cos(165)+i\sin(165))$ $\sqrt{~2~}(\cos(285)+i\sin(285))$

41. ganeshie8

simplify them if you want to by plugging in the trig values

42. anonymous

I'm amazed by how much easier it was to understand when I just saw it graphed.

43. anonymous

Thank you so much for helping me to understand this. I've been finding it difficult but I'll keep working on it :)

44. ganeshie8

complex numbers are really fun in polar form, roughly speaking : multiplying two complex numbers is same as adding the angles dividing two complex numbers is same as subtracting the angles

45. anonymous

Hopefully I'll find them fun one day!

46. ganeshie8

sure, you're on right track :) watch this video when you're free http://ocw.mit.edu/resources/res-18-008-calculus-revisited-complex-variables-differential-equations-and-linear-algebra-fall-2011/part-i/lecture-1-the-complex-numbers/

47. anonymous

Will do, thanks again for the help! :)