$\sqrt {-i}$ 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 ?

Oh, so we have $$(-i)^{1/2}=\left(\cos(-π/2+2πk)+i\sin(-π/2 + 2πk)\right)^{1/2}.$$ When you consider the periodicity of the trigonometric functions, it should be apparent why sometimes some powers of some complex numbers are not unique.
So also in this case. By the way, it may be less confusing to apply the symbol $\sqrt{}$ only to nonnegative real arguments, and use powers for other arguments.
