## Plasmataco one year ago what is the sqrt of i(imaginary)

1. Plasmataco

$\sqrt[4]{-1}$

2. Plasmataco

ppl... @jim_thompson5910 @dan815

3. Plasmataco

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

4. Plasmataco

tied for @jim_thompson5910

5. Plasmataco

haaaaalp

6. jim_thompson5910

so you're asking why this is true? $\Large \sqrt{i} = \sqrt[4]{-1}$

7. Plasmataco

no, what does it equal.

8. Plasmataco

is there like a different letter for that?

9. jim_thompson5910

I just said it $$\LARGE \sqrt{i}$$ is equal to $$\LARGE \sqrt[4]{-1}$$

10. Plasmataco

but to reduce it without a radical sign.

11. jim_thompson5910

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

12. Plasmataco

Im sry i might be asking the impossible.

13. jim_thompson5910

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

14. Plasmataco

nvm

15. Plasmataco

gtg tho bye

16. anonymous

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

17. Plasmataco

oh. sry afk but thx!

18. IrishBoy123

$$\large \sqrt{i}=\frac{1+i}{\sqrt{2}}$$ 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|

19. anonymous

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.

20. IrishBoy123

thank you