Question

I'm trying to show the cube roots of i

Apparently there are 3, which I have gotten,

(3)^(1/2)/2 + i/2 for 0pi

-(3)^(1/2)/2 + i/2 for 2pi

0 - i for 4pi

when I try to take 6pi though I get

in this case r = 1 so,

z = e^((pi/2 + 6pi)/3)i = e^(13pi/6)i
apparently it is suppose to repeat for 6pi but that doesn't seem the case, this is driving me crazy can anyone show me where I'm making my mistake.

Apparently,

z^(1/3) = r^(1/3) * e^(i*(a/3 + 2*pi)) = r^(1/3) * e^(i*a/3)

where a = pi/2

This algebra just seems wrong

http://www.wolframalpha.com/input/?i=i^%281%2F3%2

Answer 1

@nincompoop

Answer 2

\[e^\frac{\frac{\pi}{2}+ 6\pi}{3}\]

Answer 3

omg I am dizzy already just looking at it

Answer 4

ya it's wrong… wrong link of wolfram alpha

Answer 5

so,

\[e^{\frac{\frac{\pi}{2} + \frac{12\pi}{2}}{\frac{3}{1}}}\]

Answer 6

\[e^{\frac{13\pi}{2(3)}}\]

Answer 7

why would there be cube roots of i? i is what?

Answer 8

cube of i = -√-1

Answer 9

i^(1/3) or \[0 + \sqrt[3]{i}\]

Answer 10

cube of i is -i

Answer 11

I'm looking for the cube root no the cube

Answer 12

not*

Answer 13

can you see a problem in my algebra?

Answer 14

13pi/6 is not one of the roots of the cube root of i

Answer 15

im going to just go to sleep and wait for someone to come on that knows this stuff :\ thanks for trying to help I appreciate it

Answer 16

can you rewrite your expression using the equation toolbar or tex commands ?

Answer 17

\[e^\frac{\frac{\pi}{2} + 6\pi}{3}\]
=
\[e^\frac{\frac{\pi}{2} + \frac{12\pi}{2}}{3}\]
=
\[e^{\frac{\pi}{2(3)} + \frac{12\pi}{2(3)}}\]

=
e^(13pi/6)

Answer 18

which was confirmed by wolfram maybe I'm doing something wrong but I arrived at the other roots just fine for 0pi, 2pi and 4pi

Answer 19

ya I think there's something wrong …
so your e has the exponent (2pi/2+6pi)/3?

Answer 20

stupid equation is so small in my screen, I couldn't tell…

Answer 21

pi/2 not 2pi/2, but otherwise yes

Answer 22

You must be making a mistake... the fact the 13pi/6 is greater than 2pi is a pretty big clue (the angle should be less than 2pi)

this may help: http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx

I'm too tired to check your work now but I might come back to this tomorrow.

The angle of i is pi/2, so your exponents should have angles
\[\Large \frac{ 1 }{ 3 } \times \frac{ \pi }{ 2 }\],
\[\Large \frac{ 1 }{ 3 } \times \frac{ \pi }{ 2 } + \frac{ 2 \pi }{ 3 }\],
\[\Large \frac{ 1 }{ 3 } \times \frac{ \pi }{ 2 } + \frac{ 4 \pi }{ 3 }\],

Answer 23

Since you add 2*k*pi/n, where k =0, 1, 2... and n is the 3 from the cube root.

Answer 24

Wait.. why are you trying to find four roots?

"Apparently there are 3, which I have gotten,

(3)^(1/2)/2 + i/2 for 0pi

-(3)^(1/2)/2 + i/2 for 2pi

0 - i for 4pi

when I try to take 6pi though I get"

That's your mistake right there... there is no 6pi, you found all three roots.

Answer 25

oh so you only go up to 4pi? I thought you continued until it repeated

Answer 26

that doesn't make sense

Answer 27

No... you continue until k = n. You can't find four roots of a cube root...

Answer 28

thank you for explaining that lol, makes sense now ugh ha

Answer 29

It makes perfect sense. A square root has two roots. A cube root has three roots. A fourth root has four roots.

Answer 30

yeah I dont know what I was thinking

Answer 31

Thank you soooo much

Answer 32

haha no prob :)

Answer 33

No... you continue until k = n.

That should say until k=n-1, since k starts from zero.

Answer 34

right :) thanks for being clear