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Can you enter the rabbit hole recursively?

If the $ \sqrt{-1} = i $ then, what does $ \sqrt{-i} $ equal?

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It is not a nice practice to write $\sqrt a$ whenever $a$ is either a negative real number, or a complex number. –  Pedro Tamaroff Mar 26 '13 at 1:36

3 Answers 3

I'll add an answer simply because I think it would be helpful or interesting to the original poster to eliminate the exponential. Drawing from either answer by Marvis or Andyk, we have $$e^{i3\pi/4} = \frac{\sqrt{2}}{2}(-1+i), ~ e^{i7\pi/4} = \frac{\sqrt{2}}{2}(1-i) \quad\Longrightarrow\quad \sqrt{-i} = \pm\tfrac{\sqrt{2}}{2}(1-i).$$ Note that unlike with real numbers, you cannot select a principal square root; that is, neither choice is "better" than the other. Working backwards to prove this is correct: $$\left(\pm\frac{\sqrt{2}}{2}(1-i)\right)^2 = \frac{1}{2}(1-i)^2= (1-2i+i^2)/2 = -2i/2 = -i.$$ EDITED TO ADD: In a comment above, Peter Tamaroff says "It is not a nice practice to write $\sqrt{a}$ whenever a is either a negative real number, or a complex number." This is because $\sqrt{\cdot}$ is used to refer to the principal square root of a number, which for positive numbers refers to the nonnegative root. That is, if $x^2=9$, then we know that $x$ could be $+3$ or $-3$; but the principal square root is $\sqrt{9}=3$. There is no corresponding convention to define the principal square root of a negative or complex number---though to be frank, for negative numbers I've seen $\sqrt{-a} \triangleq \sqrt{a}\cdot i$ adopted in more casual discussion.

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We have $$i = e^{2k \pi i+ i \pi/2}$$ Hence, $$-i = e^{2k \pi i - i \pi/2} \implies (-i)^{1/2} = \left(e^{2k \pi i - i \pi/2}\right)^{1/2} = e^{k \pi i - i \pi/4} = e^{i 3 \pi/4}, e^{i 7 \pi/4}$$ We have two possible values, just like we have two possible values for $x^2 = -1$.

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Thank you for the explanation; this is a new area of mathematics for me. I really thought that I had found an infinite recursion. The fact that this sews everything together has left me happy. –  Dave Kirkby Mar 26 '13 at 1:50
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@DaveKirkby: Dear Dave, You may want to learn about the fundamental theorem of algebra. This guarantees that there is no rabbit hole: the problem of solving algebraic equations is solved once and for all by the complex numbers. Regards, –  Matt E Mar 26 '13 at 2:16

We have $i^2=-1$, thus $\sqrt{-i}=\sqrt{i^3}=i^{3/2}$

Also $i=(-1)^{1/2}$, so $i^{3/2}=\pm(-1)^{3/4}$. But $-(-1)^{3/4}=(-1)^{4/4}(-1)^{3/4}=(-1)^{7/4}$.

$$\sqrt{-i}=i^{3/2} = \left\{ \begin{array}{rl} (-1)^{3/4}\\ (-1)^{7/4} \end{array} \right.$$

And because $-1=e^{i\pi}$, you can write this as $e^{i3\pi/4}$ or $e^{i7\pi/4}$, exactly what Marvis has found.

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