Find cube root of i
Please, help

- Loser66

Find cube root of i
Please, help

- Stacey Warren - Expert brainly.com

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- jamiebookeater

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- zzr0ck3r

I am here to learn, I sux at this stuff. Something about a circle :)

- anonymous

|dw:1441159218311:dw|

- anonymous

take one third of that angle to find the first one
the other two are evenly spaced around the circle

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## More answers

- anonymous

|dw:1441159323165:dw|

- anonymous

\[\frac{\sqrt3}{2}+\frac{1}{2}i\] is that one
two more to go

- ganeshie8

geometry is awesome!
you want to find a number which when cubed produces \(i\) :
\[\large {(e^{i\theta})^3 = e^{i\frac{\pi}{2}}\\\implies 3\theta = \frac{\pi}{2}+2k\pi}\]
you can solve \(\theta\)

- ganeshie8

notice the trick with exponents
when we rise a complex number to a power, the argument(angle) simply gets multiplied by the power

- Loser66

Thanks for the guidance.
Question: is there any link among the roots?

- freckles

assume
\[\sqrt[3]{i}=x+yi \\ \text{ cube both sides } i=(x+yi)^3 \\ i=x^3+(yi)^3+3x^2(yi)+3x(yi)^2 \\ i=x^3-3xy^2+3x^2yi-y^3i \\ \text{ so we have } x^3-3xy^2=0 \text{ and } 3x^2y-y^3=1 \\ \text{ assume } x \neq 0 \text{ so } x^2=3y^2 \text{ so the second equation can be written as } \\ 3(3y^2)y-y^3=1 \\ 9y^3-y^3=1 \\ 8y^3=1 \\ y=\frac{1}{2} \\ x^2=3(\frac{1}{2})^2 \\ x^2=\frac{3}{4} \\ \]
\[x=\pm \frac{\sqrt{3}}{2}\]

- freckles

\[\sqrt[3]{i}=\pm \frac{\sqrt{3}}{2} + \frac{1}{2} i\]

- ganeshie8

Easy to see that these nth roots add up to \(0\)
is that the link you talking about ?

- ganeshie8

|dw:1441160623986:dw|

- ganeshie8

as you might know we can treat these literally same as vectors
look at the geometry, they have symmetry, they must add up to 0

- freckles

oh and I guess x=0 can be a solution to that one equation
which makes \[-y^3=1 \\ \text{ so } y^3=-1 \\ \text{ so } y=-1 \text{ and so the last solution would be } x+yi=0-i =-i \]

- Loser66

My computer is .... crazy!!
ok, I got sum of them =-i not 0, how?

- ganeshie8

i wouldn't know what you did to get -i haha
must be some arithmetic error somewhere.. just double check :)
but isn't the geometry so convincing ?

- Loser66

I use algebraic way

- freckles

\[(\frac{\sqrt{3}}{2}+\frac{1}{2}i)+(\frac{-\sqrt{3}}{2}+\frac{1}{2}i)+(0-i) \\ (\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+0)+(\frac{1}{2}+\frac{1}{2}-1)i \]

- ganeshie8

^

- Loser66

Like what you said above
the root is \(r = e^{i(\pi/6 + 2\pi n/3)}\) , where n = 0,1,2

- Loser66

replace n = 0, I get \(r_1 = e^{i\pi/6}=\sqrt{3}/2 + i/2\)

- Loser66

n =1 , \(r_2 = -\sqrt{3}/2 -i/2\)
n =2 \(r_3 = -i\)
sum of them =-i

- Loser66

I realize that @freckles has the same answer with mine.

- freckles

your r_2 should be...

- freckles

\[-\frac{\sqrt{3}}{2}+\frac{i}{2} \text{ \right ? }\]

- ganeshie8

yeah the second root is in second quadrant
so its imaginary component must be positive

- Loser66

\(r_2 = e^{i(\pi/6 + 2pi/3}= e^{i\pi/6}*e^{i2\pi/3} = (\sqrt {3}/2+i/2)*(cos 2pi/3 + isin2pi/3)\)
\(=(\sqrt3/2+i/2)(-1/2+i\sqrt3/2)\)
\(=-\sqrt3/4+i 3/4 -i/4 -\sqrt 3/4\)
\(= -\sqrt3/2 +i/2\)

- Loser66

oh, yes, I got it =0. Man!! it drove me crazy!!
Thanks you guys!!:)

- Loser66

Question: no matter what the result is, to test whether I can get it right or not, the sum of roots =0, right? just make sure. :)

- ganeshie8

wouldn't it be simpler to just add the exponents
\(r_2 = e^{i(\pi/6 + 2pi/3}= e^{i5\pi/6} = \cos(5\pi/6)+i\sin(5\pi/6)\)

- Loser66

Yes, it is!! but my prof asked us to try many ways but geometry. hehehe..

- ganeshie8

geometry is only to convince you
you cannot draw a picture and say "it is a proof"

- Loser66

To this stuff, the first step is geometry to define the arg z , right?

- Loser66

like: find the square root of 7-6i

- Loser66

We need geometry to define where it is and how to get the angle.

- ganeshie8

not really
complex numbers form a field, we define complex number as an ordered pair \((a,b)\) and give it structure by defining other necessary operations

- ganeshie8

geometry helps in visualizing stuff
it was not part of definition
so strictly speaking you can totally avoid geometry

- freckles

like you can say
\[\sqrt{7-6i}=x+yi \\ \text{ sqaure both sides and then compare real and imagainry parts }\]

- freckles

that is what I did to the previous one

- ganeshie8

above equation,\(\sqrt{~z~}=w\) is equivalent to
\[7 = x^2-y^2\\-6 = 2xy\]
so any equation in complex variable can be split as a system of equations of real variables

- ganeshie8

finding \(n\)th roots of a complex number is essentially same as solving a system of polynomial equations of degree \(n\)
unless you take advantage of the geometry, finding nth roots is very difficult by hand as solving \(n\)th degree polynomial equations is a major pain

- Loser66

I got you. I have another problem. Let me post a new one.

- Loser66

Ha!! the net doesn't allow me to close this post. hehehe

- freckles

the net or openstudy? :p

- ganeshie8

i can close it for you, one sec..

- Loser66

Let z = cis (2pi/n) for n > = 2
Show that 1 + z + ...+ z^(n-1) =0

- freckles

n is an integer?

- ganeshie8

There are like dozens of proofs, il just give you my favorite and simple one

- Loser66

yes

- ganeshie8

consider below polynomial of degree \(n\) : \[f(x)=x^n-1\]
remember the vieta's formulas for sum of roots ?

- Loser66

No, but go ahead. I will try to follow

- ganeshie8

we're done!
by vieta's formulas, the sum of roots = 0
and its easy to see that the terms in given expression are precicely the roots of above polynomial

- ganeshie8

\[\sum\limits_{n=0}^{n-1}cis(2\pi/n) \]
each term in above series is a distinct root of \(f(x) = x^n-1\)

- Loser66

Don't get why and how

- ganeshie8

given a polynomial
you need to know how to find the sum of roots

- ganeshie8

consider a generic polynomial
\(f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots + x+1\)
do you remember how to find the sum of its roots ?

- ganeshie8

thats precalculus stuff
we can directly use it, don't really need to prove it as part of our current proof

- Loser66

I just remember it is = -b/a where b is the second coefficient of x and a is the first one.

- ganeshie8

Exactly!

- Loser66

0

- ganeshie8

what can you say about the sum of roots of the polynomial
\[f(x) = x^n+0x^{n-1}+\cdots +0x-1\]

- ganeshie8

right, and you know that the \(n\) terms in the given sum are the roots of above polynomial
that ends the proof

- Loser66

I lost!! how can you link 1 + z +.... + z^ (n-1) =0 with z^n -1

- ganeshie8

teamviewer /skype ?

- Loser66

yes

- Loser66

skype

- ganeshie8

pm me ur id

- ganeshie8

meantime il login to skype, onesec..

- Loser66

oh, I don't have skype for this computer. let me download it. hehhehehe

- ganeshie8

wait

- ganeshie8

just download teamviewer if u can
it will be more fast

- ganeshie8

you're on window or mac

- Loser66

window

- ganeshie8

here it is
http://download.teamviewer.com/download/TeamViewer_Setup_en.exe

- Loser66

what is your meeting id?

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