Question

Find cube root of i
Please, help

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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 \(\theta\)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

Easy to see that these nth roots add up to \(0\)

is that the link you talking about ?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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 ?

Answer 14

I use algebraic way

Answer 15

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

Answer 16

^

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

your r_2 should be...

Answer 22

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

Answer 23

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

Answer 24

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

\(=(\sqrt3/2+i/2)(-1/2+i\sqrt3/2)\)

\(=-\sqrt3/4+i 3/4 -i/4 -\sqrt 3/4\)

\(= -\sqrt3/2 +i/2\)

Answer 25

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

Answer 26

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. :)

Answer 27

wouldn't it be simpler to just add the exponents

\(r_2 = e^{i(\pi/6 + 2pi/3}= e^{i5\pi/6} = \cos(5\pi/6)+i\sin(5\pi/6)\)

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

like: find the square root of 7-6i

Answer 32

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

Answer 33

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

Answer 34

geometry helps in visualizing stuff
it was not part of definition

so strictly speaking you can totally avoid geometry

Answer 35

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

Answer 36

that is what I did to the previous one

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

the net or openstudy? :p

Answer 42

i can close it for you, one sec..

Answer 43

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

Answer 44

n is an integer?

Answer 45

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

Answer 46

yes