<s>[INST] $\sqrt {-i}$ by De Moivre's Theorem $\sqrt{-i} = \left(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2})\right)^{\frac{1}{2}}$
according to De Moivre's Theorem
 $\sqrt{-i} = \left(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2})\right)^{\frac{1}{2}}$
$\sqrt{-i} = \left(\cos(\frac{-\pi}{4})+i\sin(\frac{-\pi}{4})\right)$
$\sqrt{-i} = \frac{1-i}{\sqrt{2}}$
similarly ,
$\sqrt{-i} = \left(\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}
{2})\right)^{\frac{1}{2}}$
$\sqrt{-i} = \left(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4})\right)$
$\sqrt{-i} = \frac{i-1}{\sqrt{2}}$
why there is two different answer ?
 [/INST] You get two answers because each complex number (except $0$) has two complex square roots. Note that one is the negative of the other. Thus squaring either of them gives the same result. Which one of them deserves to go under the name $\sqrt{-i}$, as which is forced to make do with $-\sqrt{-i}$ is more or less up to you.
That being said, I advise you to get away from using $\sqrt{\phantom 5}$ when dealing with complex numbers as soon as possible. The utility of the root symbol is not worth the confusions it can lead to.
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