what is the sqrt of i(imaginary)

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what is the sqrt of i(imaginary)

Mathematics
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\[\sqrt[4]{-1}\]
@dan815 ? ur like the smartest guy ever so...

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haaaaalp
so you're asking why this is true? \[\Large \sqrt{i} = \sqrt[4]{-1}\]
no, what does it equal.
is there like a different letter for that?
I just said it \(\LARGE \sqrt{i}\) is equal to \(\LARGE \sqrt[4]{-1}\)
but to reduce it without a radical sign.
\[\Large \sqrt{i} = \sqrt{\sqrt{-1}}\] \[\Large \sqrt{i} = \sqrt{(-1)^{1/2}}\] \[\Large \sqrt{i} = ((-1)^{1/2})^{1/2}\] \[\Large \sqrt{i} = (-1)^{1/2*1/2}\] \[\Large \sqrt{i} = (-1)^{1/4}\] \[\Large \sqrt{i} = \sqrt[4]{-1}\]
Im sry i might be asking the impossible.
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
nvm
gtg tho bye
The answer to your question is \[\frac{1+i}{\sqrt 2}\]
oh. sry afk but thx!
\(\large \sqrt{i}=\frac{1+i}{\sqrt{2}}\) but why not also \(\large -\frac{1+i}{\sqrt{2}}\)?? \(\large\sqrt{e^{i \frac{\pi}{2} + 2n \pi}} = e^{i \frac{\pi}{4} + n \pi} = e^{i \frac{\pi}{4}}, e^{i \frac{5\pi}{4}}\) |dw:1442740104302:dw|
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.
thank you

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