Question

A question about imaginary number :)

Answer 1

"simple"

Answer 2

why is it \[\sqrt[4]{-1} =\sqrt[2]{-1} = i\]
I just can't give an explanation why it is correct :)

Answer 3

How can I explain this?

Answer 4

sqrt(-1) is not a "value" perse

Answer 5

and thats ^1/4 not ^4

Answer 6

let me rewrite that...

Answer 7

root -1 = i ; thts how i is defined

Answer 8

\[\sqrt[4]{-1} \implies \sqrt{\sqrt{-1}} \implies \sqrt i\]
like that? lol

Answer 9

hmm weird...still didnt get i

Answer 10

\[\sqrt[4]{-1}=\sqrt{\sqrt{-1}}=\sqrt{i}\]

Answer 11

@lgbasallote I already tried doing that but it is disturbing me :)

Answer 12

how sure are you the answer is supposed to be i?

Answer 13

http://www.wolframalpha.com/input/?i=sqrt%28i%29-i

they aint equal

Answer 14

because my notebook says \[x=\sqrt[4]{-16}\]
x= +- 2i
??

Answer 15

is it wrong?

Answer 16

that's a different story

Answer 17

wait...why i? o.O

Answer 18

oh wait got it

Answer 19

no on second thought i dont

Answer 20

is it wrong?

Answer 21

yeah....

Answer 22

what should it be?

Answer 23

\[\pm 2\sqrt i\]

Answer 24

what is \[\sqrt[4]{-1}\] ???
in imaginary form?

Answer 25

\[\sqrt i\]

Answer 26

I got it :)
How about \[\sqrt[6]{-1}\] or \[\sqrt[12]{-1}\]

Answer 27

do you need to do it like this?
\[\sqrt[3]{\sqrt{-1}}\]

Answer 28

if you are using the symbol \(\sqrt[4]{-1}\) to mean any solution to \(x^4=-1\) then there are 4 answers

Answer 29

none of them are \(i\) since \(i^4=1\) not \(-1\)

Answer 30

if you want the 4th root of \(-1\) easiest to write in polar or trig form as \(-1=\cos(\pi)+i\sin(\pi)\) then divide the angle by 4 to get
\(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\)

Answer 31

that gives one of them, others are equally spaced about the unit circle

Answer 32

Thanks for the answers :)

Answer 33

could I just stick with \[x = \pm2\sqrt{i}\]
It is one of the solutions for
\[x^{8}+15x^{4}=16\]
??