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I've been wondering for quite a while, what the result of $\sqrt{-i}$ might be. After some research, I've found what the square root of $\sqrt{i}$ is.

In the link above, they are assuming (or applying a rule), that $a^2-b^2=0$ in the calculation

$\displaystyle i=z^2=(a+bi)^2=a^2+2abi+-b^2=(a^2-b^2)+2abi$

So far so good.

Then, by using $a^2-b^2=0$, he reduces the equation to $i=2abi\Longrightarrow 2ab=1$. By inserting $a=\frac{1}{2b}$ into $a^2-b^2=0$, he finally gets $a=b=\pm\frac{1}{\sqrt{2}}$.

My question is: If this is allowed, why is this allowed? Can I apply the same rule for $\sqrt{-i}$?

Thanks in advance.

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thruun
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  • In the solution $a,b$ are real numbers. The rule comes from the observation that if $x+iy=u+iv$ with $x,y,u,v$ all real, then we must have $x=u$ and $y=v$, and go from there. This is an instance of linear independence. The complex numbers $1$ and $i$ are linearly independent over the reals. Have you heard of this concept in linear algbera? In the present case the equation was $0+1\cdot i= (a^2-b^2)+2ab i$. Thus the solvers could deduce that $a^2-b^2=0$ and $2ab=1$. Anyway, yes, you can use the same technique to find square roots of $-i$. – Jyrki Lahtonen Nov 02 '14 at 09:31
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    Probably the answer of http://math.stackexchange.com/users/354/zar to the question in the link would be more useful for you. – Przemysław Scherwentke Nov 02 '14 at 09:32
  • $\sqrt{-i}=i\cdot \sqrt{i}$. If you want to proceed formally, Sami already has given the answer. – Swapnil Tripathi Nov 02 '14 at 09:35
  • @JyrkiLahtonen thank you for your detailed explanation, this helped a lot :) – thruun Nov 02 '14 at 09:42
  • @PrzemysławScherwentke yes, I've already found the solution by using polar coordinates, but it is about finding the solution by using the method above. – thruun Nov 02 '14 at 09:43
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    Possible duplicate of What is $\sqrt{i}$? – amWhy Jul 23 '17 at 18:37

2 Answers2

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We can evaluate the $\sqrt{-i}$ using the complex plane. Since $-i$ has a magnitude of $1$ and an angle of $-90^{\circ}$, we just need a number with a magnitude of $1$ and an angle of $-45^{\circ}$, so that when we square it, we get $-i$. Let's call this number $x$.

$$\sqrt{-i} = x$$

$$-i = i^3 = x^2$$

If $x$ has a magnitude of $1$ and an angle of $-45^{\circ}$, then if we square this number, we will get $-i$ with a magnitude of $1$ and an angle of $-90^{\circ}$. So all we have to do is superimpose the unit circle onto the complex plane and we see that:

$$x = \pm \frac{\sqrt 2}{2} \mp \frac{\sqrt 2}{2}i$$

$\therefore \sqrt {-i} = \pm \frac{\sqrt 2}{2} \mp \frac{\sqrt {-2}}{2}$ which is just another ordinary complex number.

This explanation was brought to you from 'Welch Labs' in their "Imaginary Numbers Are Real" YouTube Series. Check it out, it has ten videos and it will change the way you view imaginary numbers! It blew my mind when I first watched it, and it gets better and better the further into the series you watch.

I got this information from Part 9 (Video 9) of the series from $4$:$26$ to $5$:$08$. The link is down below:

https://m.youtube.com/watch?v=dLn5H69lS0w

George N. Missailidis
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We solve the equation

$$x^2=-i=e^{-i\pi/2}\implies x=e^{-i\pi/4+ik\pi},\; k=0,1$$ and you can write if you want the two solutions on the algebraic form.