Why is $\sqrt{-i} = -(-1)^{\frac{3}{4}}$ I was doing some research to look for fun on the properties of $i$, the square root of $-1$, and it got me thinking about what the square root of $-i$ was, or the square root of $-1$.  I found on wolfram alpha that it is $\sqrt{-i} = -(-1)^{\frac{3}{4}}$
, but I can't find an explanation anywhere.  Can someone help me?
 A: The standard is to define the principal argument of a complex number as
$$
 - \pi  < {\rm Arg}\left( z \right) \le \pi 
$$
and not from $0$ to $2\pi$.
And all modern CAS respect this convention, and give the results of calculations with complex numbers , square root in particular, as the principal value , i.e. with the argument in that range.
That by default, unless of course proper options are activated.
Therefore, always reducing to the principal value, we have
$$
 - i = e^{\, - i\,\pi /2} 
$$
thus
$$
\sqrt { - i}  = e^{\, - i\,\pi /4} 
$$
and on the other hand
$$
 - \left( { - 1} \right)^{3/4}  =  - \left( {e^{\,\,i\,\pi } } \right)^{3/4}  =  - e^{\,\,i3/4\,\pi }  = e^{\,\,i\,\pi } e^{\,\,i3/4\,\pi }  = e^{\,\,i7/4\,\pi }  \to e^{\, - i\,\pi /4} 
$$
A: If two complex numbers are equal, if a+ bi= c+ di then the real and imaginary parts are equal, a= c and b= d.  That is part of the definition of the "a+ bi" notation.  Her you have $A^2- B^2+ 2ABi= 0+ i$.  You must have the real parts $A^2- B^2= 0$ and imaginary part 2AB= 1.
Personally I would have used the "polar form".  $-i= e^{3\pi/2}$.  The square root of that is $e^{3\pi/4}$.  On the other hand, $-1= e^{i\pi}$ so that $(-1)^{3/4}= e^{3i\pi/4}$.
