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Loser66
 one year ago
Find cube root of i
Please, help
Loser66
 one year ago
Find cube root of i Please, help

This Question is Closed

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I am here to learn, I sux at this stuff. Something about a circle :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441159218311:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0take one third of that angle to find the first one the other two are evenly spaced around the circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441159323165:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\sqrt3}{2}+\frac{1}{2}i\] is that one two more to go

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3geometry 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
 one year ago
Best ResponseYou've already chosen the best response.3notice the trick with exponents when we rise a complex number to a power, the argument(angle) simply gets multiplied by the power

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the guidance. Question: is there any link among the roots?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1assume \[\sqrt[3]{i}=x+yi \\ \text{ cube both sides } i=(x+yi)^3 \\ i=x^3+(yi)^3+3x^2(yi)+3x(yi)^2 \\ i=x^33xy^2+3x^2yiy^3i \\ \text{ so we have } x^33xy^2=0 \text{ and } 3x^2yy^3=1 \\ \text{ assume } x \neq 0 \text{ so } x^2=3y^2 \text{ so the second equation can be written as } \\ 3(3y^2)yy^3=1 \\ 9y^3y^3=1 \\ 8y^3=1 \\ y=\frac{1}{2} \\ x^2=3(\frac{1}{2})^2 \\ x^2=\frac{3}{4} \\ \] \[x=\pm \frac{\sqrt{3}}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt[3]{i}=\pm \frac{\sqrt{3}}{2} + \frac{1}{2} i\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Easy to see that these nth roots add up to \(0\) is that the link you talking about ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1441160623986:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3as 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
 one year ago
Best ResponseYou've already chosen the best response.1oh and I guess x=0 can be a solution to that one equation which makes \[y^3=1 \\ \text{ so } y^3=1 \\ \text{ so } y=1 \text{ and so the last solution would be } x+yi=0i =i \]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0My computer is .... crazy!! ok, I got sum of them =i not 0, how?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i 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 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[(\frac{\sqrt{3}}{2}+\frac{1}{2}i)+(\frac{\sqrt{3}}{2}+\frac{1}{2}i)+(0i) \\ (\frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2}+0)+(\frac{1}{2}+\frac{1}{2}1)i \]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Like what you said above the root is \(r = e^{i(\pi/6 + 2\pi n/3)}\) , where n = 0,1,2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0replace n = 0, I get \(r_1 = e^{i\pi/6}=\sqrt{3}/2 + i/2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0n =1 , \(r_2 = \sqrt{3}/2 i/2\) n =2 \(r_3 = i\) sum of them =i

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I realize that @freckles has the same answer with mine.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1your r_2 should be...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\sqrt{3}}{2}+\frac{i}{2} \text{ \right ? }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah the second root is in second quadrant so its imaginary component must be positive

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(r_2 = e^{i(\pi/6 + 2pi/3}= e^{i\pi/6}*e^{i2\pi/3} = (\sqrt {3}/2+i/2)*(cos 2pi/3 + isin2pi/3)\) \(=(\sqrt3/2+i/2)(1/2+i\sqrt3/2)\) \(=\sqrt3/4+i 3/4 i/4 \sqrt 3/4\) \(= \sqrt3/2 +i/2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, yes, I got it =0. Man!! it drove me crazy!! Thanks you guys!!:)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Question: 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
 one year ago
Best ResponseYou've already chosen the best response.3wouldn'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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it is!! but my prof asked us to try many ways but geometry. hehehe..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3geometry is only to convince you you cannot draw a picture and say "it is a proof"

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0To this stuff, the first step is geometry to define the arg z , right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0like: find the square root of 76i

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0We need geometry to define where it is and how to get the angle.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3not 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
 one year ago
Best ResponseYou've already chosen the best response.3geometry helps in visualizing stuff it was not part of definition so strictly speaking you can totally avoid geometry

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like you can say \[\sqrt{76i}=x+yi \\ \text{ sqaure both sides and then compare real and imagainry parts }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1that is what I did to the previous one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3above equation,\(\sqrt{~z~}=w\) is equivalent to \[7 = x^2y^2\\6 = 2xy\] so any equation in complex variable can be split as a system of equations of real variables

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3finding \(n\)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
 one year ago
Best ResponseYou've already chosen the best response.0I got you. I have another problem. Let me post a new one.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Ha!! the net doesn't allow me to close this post. hehehe

freckles
 one year ago
Best ResponseYou've already chosen the best response.1the net or openstudy? :p

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i can close it for you, one sec..

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Let z = cis (2pi/n) for n > = 2 Show that 1 + z + ...+ z^(n1) =0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3There are like dozens of proofs, il just give you my favorite and simple one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3consider below polynomial of degree \(n\) : \[f(x)=x^n1\] remember the vieta's formulas for sum of roots ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0No, but go ahead. I will try to follow

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3we'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
 one year ago
Best ResponseYou've already chosen the best response.3\[\sum\limits_{n=0}^{n1}cis(2\pi/n) \] each term in above series is a distinct root of \(f(x) = x^n1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3given a polynomial you need to know how to find the sum of roots

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3consider a generic polynomial \(f(x) = a_nx^n+a_{n1}x^{n1}+\cdots + x+1\) do you remember how to find the sum of its roots ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3thats precalculus stuff we can directly use it, don't really need to prove it as part of our current proof

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I just remember it is = b/a where b is the second coefficient of x and a is the first one.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3what can you say about the sum of roots of the polynomial \[f(x) = x^n+0x^{n1}+\cdots +0x1\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3right, and you know that the \(n\) terms in the given sum are the roots of above polynomial that ends the proof

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I lost!! how can you link 1 + z +.... + z^ (n1) =0 with z^n 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3meantime il login to skype, onesec..

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, I don't have skype for this computer. let me download it. hehhehehe

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3just download teamviewer if u can it will be more fast

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you're on window or mac

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3here it is http://download.teamviewer.com/download/TeamViewer_Setup_en.exe

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0what is your meeting id?
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