anonymous
  • anonymous
Find all three complex cube roots of (2i - 2)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1440071028488:dw| Am I at least on the right track?
ganeshie8
  • ganeshie8
Yes \(\sqrt{8}\) will be the magnitude of roots, it will be same for all three roots
anonymous
  • anonymous
So what is the next step? I'm lost at the moment :(

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ganeshie8
  • ganeshie8
what exactly are we trying to find here ?
ganeshie8
  • ganeshie8
if you understand the question well, you can answer it easily
anonymous
  • anonymous
"Find all three complex roots of (2i - 2)".
ganeshie8
  • ganeshie8
Yes, what is a root ?
ganeshie8
  • ganeshie8
the root is a number which when you multiply itself three times produces the number 2i-2
anonymous
  • 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
ganeshie8
  • ganeshie8
you must be knowing that angles add up when you multiply two complex numbers : \[e^{i\theta_1} *e^{i\theta_2} = e^{i(\theta_1+\theta_2)}\]
ganeshie8
  • ganeshie8
we can use that trick to find the angles of roots
anonymous
  • anonymous
I think I've seen that before :)
ganeshie8
  • ganeshie8
whats the angle of 2i-2 ?
anonymous
  • anonymous
I'm still trying to find it out. I'm at root8 at the moment
ganeshie8
  • ganeshie8
in which quadrant does the number (-2, 2) lie in ?
anonymous
  • anonymous
sine
ganeshie8
  • ganeshie8
wat do you mean by sine
anonymous
  • anonymous
I'll just show you what I've been trying to do:
anonymous
  • anonymous
|dw:1440073112572:dw|
ganeshie8
  • ganeshie8
again, in which quadrant does the number (-2, 2) lie in ?
anonymous
  • anonymous
2nd
ganeshie8
  • ganeshie8
good, can you show it in the graph |dw:1440073212144:dw|
anonymous
  • anonymous
|dw:1440073236919:dw|
ganeshie8
  • ganeshie8
right, what is its angle ? |dw:1440073291571:dw|
anonymous
  • anonymous
135 degrees
ganeshie8
  • ganeshie8
when you see the number, you can eyeball the angle, arctan formula and all is not needed
ganeshie8
  • ganeshie8
yes 135 is correct
ganeshie8
  • ganeshie8
so the magnitude of given complex number is \(\sqrt{8}\) and the angle is \(135\) we're ready to find the roots
anonymous
  • anonymous
Alright cool :)
ganeshie8
  • ganeshie8
lets find the angles first
ganeshie8
  • ganeshie8
basically you want to find \(\theta\) such that \[\theta+\theta+\theta = 135 + 360k\]
ganeshie8
  • ganeshie8
which is same as : \[3\theta = 135 + 360k\]
anonymous
  • anonymous
So is theta 165?
ganeshie8
  • ganeshie8
or \[\theta = \dfrac{135}{3} + \dfrac{360k}{3}\]
ganeshie8
  • ganeshie8
or \[\theta = 45 + 120k\]
ganeshie8
  • 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 \(\theta = 45+120*2 =\color{red}{ 285}\)
ganeshie8
  • ganeshie8
the magnitude is same for all of the roots : \(\large (\sqrt{~8~})^{\frac{1}{3}}\)
ganeshie8
  • ganeshie8
notice that \(\large (\sqrt{~8~})^{\frac{1}{3}}\) simplifies to \(\large \sqrt{~2~}\)
anonymous
  • anonymous
Oh ok
ganeshie8
  • ganeshie8
so the 3 roots are \[\sqrt{~2~}(\cos(45)+i\sin(45))\] \[\sqrt{~2~}(\cos(165)+i\sin(165))\] \[\sqrt{~2~}(\cos(285)+i\sin(285))\]
ganeshie8
  • ganeshie8
simplify them if you want to by plugging in the trig values
anonymous
  • anonymous
I'm amazed by how much easier it was to understand when I just saw it graphed.
anonymous
  • anonymous
Thank you so much for helping me to understand this. I've been finding it difficult but I'll keep working on it :)
ganeshie8
  • 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
anonymous
  • anonymous
Hopefully I'll find them fun one day!
ganeshie8
  • 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/
anonymous
  • anonymous
Will do, thanks again for the help! :)

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