$\sqrt {-i}$ by De Moivre's Theorem

Question

$\sqrt{-i} = \left(\cos(\frac{-\pi}{2})+i\sin(\frac{-\pi}{2})\right)^{\frac{1}{2}}$

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 ?

Answer 1

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.

Answer 2

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.