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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?

jstowell
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  • The Wolfram-alpha answer doesn't give you any extra knowledge on the nature of sqrt(-i). – Jacob Wakem May 09 '18 at 18:28
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    https://math.stackexchange.com/q/3315 – Andrew Li May 09 '18 at 18:29
  • @AndrewLi That question is different, it'm asking about the negative square root of i, not just the positive square root of i. – jstowell May 09 '18 at 18:38
  • wolframalpha actually says $\sqrt{-i}=-(-1)^{\frac{3}{4}}$. you must have made a typo – emma May 09 '18 at 18:38
  • Made the edit, it was a typo. – jstowell May 09 '18 at 18:41
  • Every nonzero complex number has two square roots, and four fourth roots. – Angina Seng May 09 '18 at 19:01
  • Be careful with $\sqrt{{}\cdot{}}$ and fractional exponents when the base is negative or complex. They are multivalued, and comparing different expressions with $=$ is precarious. With a calculator like WolframAlpha you never know which of the multiple values it's going to pick. I'd stay clear of it all together, personally, work with integer exponents as much as possible. Instead of $\sqrt i$, use "a solution to $x^2=i$", and so on. – Arthur May 09 '18 at 19:25
  • btw. the typo is still there (the inline equation, not the title) – emma May 10 '18 at 20:40

2 Answers2

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

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

user247327
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