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This is a follow-up of the post here:

using phasors to handle complex numbers

I have decided to create a new post as now I am considering a deeper issue.

Say if we want to compute $\sqrt{-5}$. If I want to find its principal square root then I can use phasor arithmetic as follows: $\sqrt{-5}=\sqrt{5 \angle 180}=\sqrt{5}\angle 90 = \sqrt{5} \; \mathrm{i}$. This agrees with the definition namely (see ):

If $z=r\: \mathrm{e}^{\psi \mathrm{i}}$ with $-\pi < \psi \leq \pi$, then the principal square root of $z$ is defined as $\sqrt{z}=\sqrt{r}\: \mathrm{e}^{\frac{\psi \mathrm{i}}{2}}$.

And we also have the definition that the other square root is simply $-1$ times the principal square root. So for $\sqrt{-5}$, we have the principal square root as $\sqrt{5} \; \mathrm{i}$ and the other root as $-\sqrt{5} \; \mathrm{i}$. This is OK as $-\sqrt{5} \; \mathrm{i} \times -\sqrt{5} \; \mathrm{i} = \sqrt{5}\angle 270 \times \sqrt{5}\angle 270 = 5\angle 540 = 5\angle 180 = -5$.

Now let's consider two intricate cases

  1. $\sqrt{{-1}\times{-1}}=\sqrt{1 \angle 180 \times 1 \angle 180}= \sqrt{1 \angle 360}=\sqrt{1 \angle (360-360)}= 1\angle 0 = 1$ (as the principal square root).

  2. $\sqrt{\frac{1}{-1}}=\sqrt{\frac{1 \angle 0}{1 \angle 180}}=\sqrt{1 \angle -180} = \sqrt{1 \angle (-180+360)}= \sqrt{1 \angle 180}= 1 \angle 90 = \mathrm{i}$ (as the principal square root).

In regard with the above two cases, can we say that another root of $\sqrt{{-1}\times{-1}}$ is $-1$ ( i.e. -ve of the principal square root) and that another root of $\sqrt{\frac{1}{-1}}$ is $-\mathrm{i}$ (i.e -ve of the principal square root)?

If we check:

For, $\sqrt{{-1}\times{-1}}$ having a second root as $-1$, we have on squaring, $1\angle 180 \times 1\angle 180 = 1\angle 360 = 1\angle 0 = 1$. This is equivalent to squaring the principal square root i.e. $1$ to give also $1$.

For $\sqrt{\frac{1}{-1}}$ having a second root as $-\mathrm{i}$ we have on squaring, $-\mathrm{i} \times -\mathrm{i} = 1\angle 270 \times 1\angle 270 = 1\angle 540 = 1\angle 180 = -1$. This is equivalent to squaring the principal square root i.e. $\mathrm{i}$ to give also $-1$.

So $\sqrt{{-1}\times{-1}} = 1 \; \text{(principal root)} \; \mathrm{or} \; -1$ and $\sqrt{\frac{1}{-1}} = \mathrm{i} \; \text{(principal root)} \; \mathrm{or} \; -\mathrm{i}$. Is this correct?

Thanks a lot...

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up vote 4 down vote accepted

It's mathematical semantics. The square root function, which is what $\sqrt{z}$ denotes by convention, only takes on a single value - in which case equations like $\sqrt{-1\times-1}=-1$ are false. However one can refer to "square roots" as solutions to the equations of the form $x^2=a$, in which case statements like $x=1\text{ or }-1$ are meaningful, but the general practice is simple to write $\pm\sqrt{a}$ to refer to either value within a single equation or statement.

Bottom line: there are two "square roots," but symbolically $\sqrt{z}$ only refers to the principal root.

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I agree that a lot of the question is just semantics, but you are missing the crucial point. There is no canonical way of defining the "positive square root" and "the negative square root" of a complex number. There are no "positive complex numbers" or "negative complex numbers". Therefore, $\pm\sqrt{i}$ is a terrible notation, since it is not clear which of the roots you are considering to be $+sqrt{i}$ and which is $-\sqrt{i}$. Instead, to make multivalued functions single valued, one takes branch cuts and carefully specifies which branch one is talking about. – Alex B. Oct 6 '11 at 7:08
@Alex B. I never used the words "positive" or "negative," but I agree that $\pm\sqrt{z}$ should be avoided in certain contexts and that complex analysis requires more care. However, since the square root function by convention is the principal square root, I think it should be clear which sign refers to which root - and in the circumstances you'd use $\pm$ it might not even matter which one is which. – anon Oct 6 '11 at 7:16
@Alex B: the objection you raise against $\pm \sqrt{a}$ is really an objection to the notation $\sqrt{a}$. Whether $f$ is a single-valued or multi-valued expression, $\pm f$ makes exactly as much or as little sense as $f$ does. – zyx Oct 6 '11 at 7:41
Thanks a lot for the discussions. @anon: Why do you say that $\sqrt{-1 \times -1}=-1$ is false when we know that $-1$ on squaring gives $1$, which is same as squaring the principal root of $\sqrt{-1 \times -1}$ which is $1$ to get also $1$. I agree that perhaps we should work only with the principal square root... 1 vote up. – yCalleecharan Oct 6 '11 at 7:41
@yCalleecharan: Yes. The notation $\sqrt{z}$ specifies only one of the roots, not both; you can't set it equal to the nonprincipal one. (A technicality though is that $0$ only has one square root.) – anon Oct 6 '11 at 8:07

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