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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440071028488:dw Am I at least on the right track?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes \(\sqrt{8}\) will be the magnitude of roots, it will be same for all three roots

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what is the next step? I'm lost at the moment :(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1what exactly are we trying to find here ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1if you understand the question well, you can answer it easily

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"Find all three complex roots of (2i  2)".

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes, what is a root ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1the root is a number which when you multiply itself three times produces the number 2i2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.1we can use that trick to find the angles of roots

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I've seen that before :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1whats the angle of 2i2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm still trying to find it out. I'm at root8 at the moment

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1in which quadrant does the number (2, 2) lie in ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1wat do you mean by sine

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll just show you what I've been trying to do:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440073112572:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1again, in which quadrant does the number (2, 2) lie in ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1good, can you show it in the graph dw:1440073212144:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440073236919:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1right, what is its angle ? dw:1440073291571:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1when you see the number, you can eyeball the angle, arctan formula and all is not needed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so the magnitude of given complex number is \(\sqrt{8}\) and the angle is \(135\) we're ready to find the roots

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1lets find the angles first

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1basically you want to find \(\theta\) such that \[\theta+\theta+\theta = 135 + 360k\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1which is same as : \[3\theta = 135 + 360k\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1or \[\theta = \dfrac{135}{3} + \dfrac{360k}{3}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1or \[\theta = 45 + 120k\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1letting \(k=0,\) 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
 one year ago
Best ResponseYou've already chosen the best response.1the magnitude is same for all of the roots : \(\large (\sqrt{~8~})^{\frac{1}{3}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1notice that \(\large (\sqrt{~8~})^{\frac{1}{3}}\) simplifies to \(\large \sqrt{~2~}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1simplify them if you want to by plugging in the trig values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm amazed by how much easier it was to understand when I just saw it graphed.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much for helping me to understand this. I've been finding it difficult but I'll keep working on it :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1complex 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
 one year ago
Best ResponseYou've already chosen the best response.0Hopefully I'll find them fun one day!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1sure, you're on right track :) watch this video when you're free http://ocw.mit.edu/resources/res18008calculusrevisitedcomplexvariablesdifferentialequationsandlinearalgebrafall2011/parti/lecture1thecomplexnumbers/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Will do, thanks again for the help! :)
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