Why is $\sqrt{-i} \neq i\sqrt{i}$? I wanted to figure out the square root of $-i$.  
Since $\sqrt{-x} = i\sqrt{x}$,  
$\sqrt{-i}$ should equal $i\sqrt {i}$, however, WolframAlpha said it was false.  
However, if I do say that $\sqrt{-i} = i\sqrt{i}$, I can replace $\sqrt{i}$ with $\dfrac {1+i} {\sqrt{2}}$, leaving me with $\sqrt{-i} = \dfrac {i(1+i)} {\sqrt{2}}$  
Now I distribute, giving me $\sqrt{-i} = \dfrac {i-1} {\sqrt{2}}$  
Then I square both sides, giving me $-i = \dfrac {(i-1)(i-1)} {2}$  
Then I FOIL, giving me $-i = \dfrac {-2i} {2}$  
Then the $2$'s cancel out, leaving $-i = -i$, making $\sqrt {-i} = i\sqrt{i}$ a true statement.
 A: Remember that every number (except $0$) has two complex square roots -- and if you choose appropriate square roots then your relation does indeed hold.
In particular, if we take $\sqrt i = \frac{1+i}{\sqrt 2}$ and $\sqrt{-i}=\frac{-1+i}{\sqrt 2}$ (and these are indeed possible square roots of $i$ and $-i$), then you do get $\sqrt{-i}=i\sqrt{i}$.
Wolfram Alpha probably chooses to set $\sqrt{-i}=\frac{1-i}{\sqrt 2}$, in which case you have $\sqrt{-i}=-i\sqrt{i}$ instead.
A: Your assumption that $\sqrt{-x} = i\sqrt{x}$ is incorrect. Suppose $x=-4.$ Then $\sqrt{-x}=\sqrt{4}=2\neq i\sqrt{-4}=i\cdot2i=-2$
A: You are running up against a complication that arises with roots of complex numbers: they are not single-valued, but multi-valued.  If you express the question "what is $ \ \sqrt{-i} \ $ ?" as "what are the roots of the equation $ \ x^2 \ - \ (-i) \ = \ 0 \ ? $ " , there are two solutions. DeMoivre's Theorem tells us that the square-roots of $ \ -i \ = \ \cos(\frac{3 \pi}{2}) \ + \ i \ \sin(\frac{3 \pi}{2}) \ $ are
$$  \cos(\frac{3 \pi}{4}) \ + \ i \  \sin(\frac{3 \pi}{4}) \ = \ \frac{-1 + i}{\sqrt{2}} \ \ \text{and} \ \ \cos(\frac{7 \pi}{4}) \ + \ i \  \sin(\frac{7 \pi}{4}) \ = \ \frac{1 - i}{\sqrt{2}} \ \ . $$ 
The square-roots of $ \ i \ = \ \cos(\frac{ \pi}{2}) \ + \ i \ \sin(\frac{ \pi}{2}) \ $  are 
$$  \cos(\frac{ \pi}{4}) \ + \ i \  \sin(\frac{ \pi}{4}) \ = \ \frac{1 + i}{\sqrt{2}} \ \ \text{and} \ \ \cos(\frac{5 \pi}{4}) \ + \ i \  \sin(\frac{5 \pi}{4}) \ = \ \frac{-1 - i}{\sqrt{2}} \ \ . $$ 
We can see that if we multiply each of the square-roots of $ \ -i \ $ by $ \ i \ $ , we will get one of the square-roots of $ \ i \ $ , but the correspondence is not as simple as is suggested by the putative equation $ \ \sqrt{-i} \ = \ i \ \sqrt{i} \ $ .
What WolframAlpha appears to do is take as the principal values for these square-roots the ones where the angle (argument) in the complex plane is in the domain of the arctangent function, $ \ -\frac{\pi}{2} \ < \ \theta \ < \ \frac{\pi}{2} \ $ .  When you just look at that, the equation doesn't work.  But further down the pages, when you look at all the square-roots, you see that the locations of the two square-roots of $ \ i \ $ get rotated by $ \ \frac{\pi}{2} \ $ (upon multiplication by $ \ i \ $ ) to give the locations of the two square-roots of $ \ -i \ $ .  So your equation is "wrong" in the sense that the complex square-root function has to be described more carefully than the one defined only for real numbers.
