Find all three complex cube roots of (2i - 2)

- anonymous

Find all three complex cube roots of (2i - 2)

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- anonymous

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

- ganeshie8

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

- anonymous

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

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## More answers

- ganeshie8

what exactly are we trying to find here ?

- ganeshie8

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

- anonymous

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

- ganeshie8

Yes, what is a root ?

- ganeshie8

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

- 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

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

we can use that trick to find the angles of roots

- anonymous

I think I've seen that before :)

- ganeshie8

whats the angle of 2i-2 ?

- anonymous

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

- ganeshie8

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

- anonymous

sine

- ganeshie8

wat do you mean by sine

- anonymous

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

- anonymous

|dw:1440073112572:dw|

- ganeshie8

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

- anonymous

2nd

- ganeshie8

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

- anonymous

|dw:1440073236919:dw|

- ganeshie8

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

- anonymous

135 degrees

- ganeshie8

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

- ganeshie8

yes 135 is correct

- ganeshie8

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

- anonymous

Alright cool :)

- ganeshie8

lets find the angles first

- ganeshie8

basically you want to find \(\theta\) such that
\[\theta+\theta+\theta = 135 + 360k\]

- ganeshie8

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

- anonymous

So is theta 165?

- ganeshie8

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

- ganeshie8

or
\[\theta = 45 + 120k\]

- 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

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

- ganeshie8

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

- anonymous

Oh ok

- 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

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

- anonymous

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

- 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

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

Hopefully I'll find them fun one day!

- 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

Will do, thanks again for the help! :)

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