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I wanted to figure out the square root of $-i$.

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

$\sqrt{-i}$ should equal $i\sqrt {i}$, however, WolframAlpha said it was false.

However, if I do say that $\sqrt{-i} = i\sqrt{i}$, I can replace $\sqrt{i}$ with $\dfrac {1+i} {\sqrt{2}}$, leaving me with $\sqrt{-i} = \dfrac {i(1+i)} {\sqrt{2}}$

Now I distribute, giving me $\sqrt{-i} = \dfrac {i-1} {\sqrt{2}}$

Then I square both sides, giving me $-i = \dfrac {(i-1)(i-1)} {2}$

Then I FOIL, giving me $-i = \dfrac {-2i} {2}$

Then the $2$'s cancel out, leaving $-i = -i$, making $\sqrt {-i} = i\sqrt{i}$ a true statement.

ZTqvhI5vpo
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    Recommend reading http://math.stackexchange.com/questions/438/why-sqrt-1-times-1-neq-sqrt-12?rq=1 In particular, $\sqrt{a}\cdot \sqrt{b} = \sqrt{a\cdot b}$ need not be true for nonpositive numbers (including nonreal and negative) – JMoravitz Jun 14 '15 at 04:34

3 Answers3

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Remember that every number (except $0$) has two complex square roots -- and if you choose appropriate square roots then your relation does indeed hold.

In particular, if we take $\sqrt i = \frac{1+i}{\sqrt 2}$ and $\sqrt{-i}=\frac{-1+i}{\sqrt 2}$ (and these are indeed possible square roots of $i$ and $-i$), then you do get $\sqrt{-i}=i\sqrt{i}$.

Wolfram Alpha probably chooses to set $\sqrt{-i}=\frac{1-i}{\sqrt 2}$, in which case you have $\sqrt{-i}=-i\sqrt{i}$ instead.

hmakholm left over Monica
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  • Yes, it looks to me like WA chooses the "principal" square-root as the one with its argument in the domain of the arctangent function. – colormegone Jun 14 '15 at 05:22
  • @Henning Makholm I understand your argument about the multivalued nature of the square root function. But that can be fixed by a branch cut right? But how it is that $\sqrt{1}=\sqrt{-1 \times -1}=\sqrt{-1}\sqrt{-1}=ii=-1$? What is the going on? – Miz Jun 14 '15 at 05:44
  • @AloysiusGodinho: You can use a branch cut to (more or less arbitrarily) choose one of the values of the square root in each case. But then you lose the property that $\sqrt a \sqrt b = \sqrt{ab}$, except when $a$ and $b$ are both in a restricted range of arguments that depends on where your branch cut is. – hmakholm left over Monica Jun 14 '15 at 11:31
  • @ Henning Makholm Yes you are right. I have thought about it and I get how that will work out. So apparently with a branch cut in place you are insured that the relation $\sqrt{ab}=\sqrt{a}\sqrt{b}$ will hold properly. – Miz Jun 14 '15 at 11:48
  • @AloysiusGodinho: No, no, no -- that is precisely what a branch cut does not guarantee. When you have a branch cut, there will be some combinations of $a$ and $b$ where you get $\sqrt{ab} = \sqrt a\sqrt b$ and some where you get $\sqrt{ab} = -\sqrt a\sqrt b$. – hmakholm left over Monica Jun 14 '15 at 12:06
  • @ Henning Makholm You are right. I didn't see that coming. I guess I need to study this out. Thanks a lot. Is there some rule to avoid this? or is $(z_1z_2)^z=z_1^zz_2^z$ not valid in $\mathbb{C}$? – Miz Jun 14 '15 at 13:52
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    @AloysiusGodinho: In general $(z_1z_2)^z=z_1^zz_2^z$ is not valid in $\mathbb C$. It holds when $z$ is an integer, as well as when $z_1$ and $z_2$ are both positive reals. Beyond those cases it requires so tight controls on which branches for the multi-valued operation you choose for $z_1$, $z_2$ and $z_1z_2$ that it is easiest not to rely on it at all. – hmakholm left over Monica Jun 14 '15 at 16:10
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Your assumption that $\sqrt{-x} = i\sqrt{x}$ is incorrect. Suppose $x=-4.$ Then $\sqrt{-x}=\sqrt{4}=2\neq i\sqrt{-4}=i\cdot2i=-2$

Sujaan Kunalan
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You are running up against a complication that arises with roots of complex numbers: they are not single-valued, but multi-valued. If you express the question "what is $ \ \sqrt{-i} \ $ ?" as "what are the roots of the equation $ \ x^2 \ - \ (-i) \ = \ 0 \ ? $ " , there are two solutions. DeMoivre's Theorem tells us that the square-roots of $ \ -i \ = \ \cos(\frac{3 \pi}{2}) \ + \ i \ \sin(\frac{3 \pi}{2}) \ $ are

$$ \cos(\frac{3 \pi}{4}) \ + \ i \ \sin(\frac{3 \pi}{4}) \ = \ \frac{-1 + i}{\sqrt{2}} \ \ \text{and} \ \ \cos(\frac{7 \pi}{4}) \ + \ i \ \sin(\frac{7 \pi}{4}) \ = \ \frac{1 - i}{\sqrt{2}} \ \ . $$

The square-roots of $ \ i \ = \ \cos(\frac{ \pi}{2}) \ + \ i \ \sin(\frac{ \pi}{2}) \ $ are

$$ \cos(\frac{ \pi}{4}) \ + \ i \ \sin(\frac{ \pi}{4}) \ = \ \frac{1 + i}{\sqrt{2}} \ \ \text{and} \ \ \cos(\frac{5 \pi}{4}) \ + \ i \ \sin(\frac{5 \pi}{4}) \ = \ \frac{-1 - i}{\sqrt{2}} \ \ . $$

We can see that if we multiply each of the square-roots of $ \ -i \ $ by $ \ i \ $ , we will get one of the square-roots of $ \ i \ $ , but the correspondence is not as simple as is suggested by the putative equation $ \ \sqrt{-i} \ = \ i \ \sqrt{i} \ $ .

What WolframAlpha appears to do is take as the principal values for these square-roots the ones where the angle (argument) in the complex plane is in the domain of the arctangent function, $ \ -\frac{\pi}{2} \ < \ \theta \ < \ \frac{\pi}{2} \ $ . When you just look at that, the equation doesn't work. But further down the pages, when you look at all the square-roots, you see that the locations of the two square-roots of $ \ i \ $ get rotated by $ \ \frac{\pi}{2} \ $ (upon multiplication by $ \ i \ $ ) to give the locations of the two square-roots of $ \ -i \ $ . So your equation is "wrong" in the sense that the complex square-root function has to be described more carefully than the one defined only for real numbers.

colormegone
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  • I understand your argument about the multivalued nature of the square root function. But that can be fixed by a branch cut right? But how it is that $\sqrt{1}=\sqrt{-1 \times -1}=\sqrt{-1}\sqrt{-1}=ii=-1$? What is the going on? – Miz Jun 14 '15 at 05:46
  • The issue is the same for your example: if $ \ z \ = \ \sqrt{1} \ $ is interpreted as the roots of $ \ z^2 \ - \ 1 \ = \ 0 \ $ , the solutions are $ \ cis 0 \ = \ 1 \ $ and $ \ cis \pi \ = \ -1 \ $ ; similarly, the two square-roots of $ \ -1 \ $ are $ \ cis \frac{\pi}{2} \ = \ i \ $ and $ \ cis \frac{3 \pi}{2} \ = \ -i \ $ . So by this definition, $ \ \sqrt{1} \ = \ i \ (\pm i) \ = \ \pm 1 \ $ . If we stick to "principal" values, then $ \ \sqrt{1} \ $ is just 1 . We have to be more careful with definitions when discussing $ \ n^{th} \ $ roots of complex numbers. – colormegone Jun 14 '15 at 06:16