Question

what is the sqrt of i(imaginary)

Answer 1

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

Answer 2

ppl... @jim_thompson5910 @dan815

Answer 3

@dan815 ? ur like the smartest guy ever so...

Answer 4

tied for @jim_thompson5910

Answer 5

haaaaalp

Answer 6

so you're asking why this is true?

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

Answer 7

no, what does it equal.

Answer 8

is there like a different letter for that?

Answer 9

I just said it

\(\LARGE \sqrt{i}\) is equal to \(\LARGE \sqrt[4]{-1}\)

Answer 10

but to reduce it without a radical sign.

Answer 11

\[\Large \sqrt{i} = \sqrt{\sqrt{-1}}\]
\[\Large \sqrt{i} = \sqrt{(-1)^{1/2}}\]
\[\Large \sqrt{i} = ((-1)^{1/2})^{1/2}\]
\[\Large \sqrt{i} = (-1)^{1/2*1/2}\]
\[\Large \sqrt{i} = (-1)^{1/4}\]
\[\Large \sqrt{i} = \sqrt[4]{-1}\]

Answer 12

Im sry i might be asking the impossible.

Answer 13

You can either write it with a fractional exponent, or as a radical. I don't think it's possible to do it any other way

Answer 14

nvm

Answer 15

gtg tho bye

Answer 16

The answer to your question is \[\frac{1+i}{\sqrt 2}\]

Answer 17

oh. sry afk but thx!

Answer 18

\(\large \sqrt{i}=\frac{1+i}{\sqrt{2}}\)

but why not also \(\large -\frac{1+i}{\sqrt{2}}\)??

\(\large\sqrt{e^{i \frac{\pi}{2} + 2n \pi}} = e^{i \frac{\pi}{4} + n \pi} = e^{i \frac{\pi}{4}}, e^{i \frac{5\pi}{4}}\)

Answer 19

That's also true, of course. I was answering the question in the same way that one might say "The square root of nine is three".

If we're being strict about it, the radical sign should not be used in this context because it only applies to positive, real numbers. It should be written something like
\[ i^{1/2} =\left\{ \frac{(1+i)}{\sqrt{2}} , -\frac{1+i}{\sqrt{2}} \right\}\]
\(\sqrt{i},\sqrt{-1}\), and other things like that don't actually make sense.

Answer 20

thank you