Loser66
  • Loser66
Find cube root of i Please, help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zzr0ck3r
  • zzr0ck3r
I am here to learn, I sux at this stuff. Something about a circle :)
anonymous
  • anonymous
|dw:1441159218311:dw|
anonymous
  • anonymous
take one third of that angle to find the first one the other two are evenly spaced around the circle

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
|dw:1441159323165:dw|
anonymous
  • anonymous
\[\frac{\sqrt3}{2}+\frac{1}{2}i\] is that one two more to go
ganeshie8
  • 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
  • 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
  • Loser66
Thanks for the guidance. Question: is there any link among the roots?
freckles
  • 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
  • freckles
\[\sqrt[3]{i}=\pm \frac{\sqrt{3}}{2} + \frac{1}{2} i\]
ganeshie8
  • ganeshie8
Easy to see that these nth roots add up to \(0\) is that the link you talking about ?
ganeshie8
  • ganeshie8
|dw:1441160623986:dw|
ganeshie8
  • 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
  • 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
  • Loser66
My computer is .... crazy!! ok, I got sum of them =-i not 0, how?
ganeshie8
  • 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
  • Loser66
I use algebraic way
freckles
  • 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
  • ganeshie8
^
Loser66
  • Loser66
Like what you said above the root is \(r = e^{i(\pi/6 + 2\pi n/3)}\) , where n = 0,1,2
Loser66
  • Loser66
replace n = 0, I get \(r_1 = e^{i\pi/6}=\sqrt{3}/2 + i/2\)
Loser66
  • Loser66
n =1 , \(r_2 = -\sqrt{3}/2 -i/2\) n =2 \(r_3 = -i\) sum of them =-i
Loser66
  • Loser66
I realize that @freckles has the same answer with mine.
freckles
  • freckles
your r_2 should be...
freckles
  • freckles
\[-\frac{\sqrt{3}}{2}+\frac{i}{2} \text{ \right ? }\]
ganeshie8
  • ganeshie8
yeah the second root is in second quadrant so its imaginary component must be positive
Loser66
  • 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
  • Loser66
oh, yes, I got it =0. Man!! it drove me crazy!! Thanks you guys!!:)
Loser66
  • 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
  • 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
  • Loser66
Yes, it is!! but my prof asked us to try many ways but geometry. hehehe..
ganeshie8
  • ganeshie8
geometry is only to convince you you cannot draw a picture and say "it is a proof"
Loser66
  • Loser66
To this stuff, the first step is geometry to define the arg z , right?
Loser66
  • Loser66
like: find the square root of 7-6i
Loser66
  • Loser66
We need geometry to define where it is and how to get the angle.
ganeshie8
  • 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
  • ganeshie8
geometry helps in visualizing stuff it was not part of definition so strictly speaking you can totally avoid geometry
freckles
  • freckles
like you can say \[\sqrt{7-6i}=x+yi \\ \text{ sqaure both sides and then compare real and imagainry parts }\]
freckles
  • freckles
that is what I did to the previous one
ganeshie8
  • 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
  • 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
  • Loser66
I got you. I have another problem. Let me post a new one.
Loser66
  • Loser66
Ha!! the net doesn't allow me to close this post. hehehe
freckles
  • freckles
the net or openstudy? :p
ganeshie8
  • ganeshie8
i can close it for you, one sec..
Loser66
  • Loser66
Let z = cis (2pi/n) for n > = 2 Show that 1 + z + ...+ z^(n-1) =0
freckles
  • freckles
n is an integer?
ganeshie8
  • ganeshie8
There are like dozens of proofs, il just give you my favorite and simple one
Loser66
  • Loser66
yes
ganeshie8
  • ganeshie8
consider below polynomial of degree \(n\) : \[f(x)=x^n-1\] remember the vieta's formulas for sum of roots ?
Loser66
  • Loser66
No, but go ahead. I will try to follow
ganeshie8
  • 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
  • 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
  • Loser66
Don't get why and how
ganeshie8
  • ganeshie8
given a polynomial you need to know how to find the sum of roots
ganeshie8
  • 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
  • ganeshie8
thats precalculus stuff we can directly use it, don't really need to prove it as part of our current proof
Loser66
  • Loser66
I just remember it is = -b/a where b is the second coefficient of x and a is the first one.
ganeshie8
  • ganeshie8
Exactly!
Loser66
  • Loser66
0
ganeshie8
  • ganeshie8
what can you say about the sum of roots of the polynomial \[f(x) = x^n+0x^{n-1}+\cdots +0x-1\]
ganeshie8
  • ganeshie8
right, and you know that the \(n\) terms in the given sum are the roots of above polynomial that ends the proof
Loser66
  • Loser66
I lost!! how can you link 1 + z +.... + z^ (n-1) =0 with z^n -1
ganeshie8
  • ganeshie8
teamviewer /skype ?
Loser66
  • Loser66
yes
Loser66
  • Loser66
skype
ganeshie8
  • ganeshie8
pm me ur id
ganeshie8
  • ganeshie8
meantime il login to skype, onesec..
Loser66
  • Loser66
oh, I don't have skype for this computer. let me download it. hehhehehe
ganeshie8
  • ganeshie8
wait
ganeshie8
  • ganeshie8
just download teamviewer if u can it will be more fast
ganeshie8
  • ganeshie8
you're on window or mac
Loser66
  • Loser66
window
ganeshie8
  • ganeshie8
here it is http://download.teamviewer.com/download/TeamViewer_Setup_en.exe
Loser66
  • Loser66
what is your meeting id?

Looking for something else?

Not the answer you are looking for? Search for more explanations.