Why is $\sqrt{-i} = -(-1)^{\frac{3}{4}}$

Question

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?

Answer 1

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 $$ - \left( { - 1} \right)^{3/4} = - \left( {e^{\,\,i\,\pi } } \right)^{3/4} = - e^{\,\,i3/4\,\pi } = e^{\,\,i\,\pi } e^{\,\,i3/4\,\pi } = e^{\,\,i7/4\,\pi } \to e^{\, - i\,\pi /4} $$

Answer 2

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 $(-1)^{3/4}= e^{3i\pi/4}$.