Principal (and secondary) square roots of a complex number

Question

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 http://en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number ):

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...

Answer 1

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.