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I have been reading the post:

Simple Complex Number Problem: $1 = -1$

and other similar posts based on the manipulation of complex numbers.

If I want to simplify $\sqrt{(-1) \times (-1)}$, I would, using phasor algebra, do

$$\sqrt{(-1) \times (-1)} = \sqrt{-1} \times \sqrt{-1} = 1 \angle 90 \times 1 \angle 90 = 1 \angle 180 = -1$$

I believe the above way is neater and that I have the right answer.

Please do let me know what you think.

PS: What I'm doing is to avoid simplifying when under the square-root sign.

Thanks a lot...

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yCalleecharan
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    The error you made was before you translated the numbers into phasor algebra. You can't split square roots up willy-nilly; there are conditions that, if not satisfied, prevent it from being a valid maneuver. That should be made clear in this question. – anon Oct 03 '11 at 07:58
  • Thanks. I understand that we cannot have [;\sqrt{-1} \times \sqrt{-1} = \sqrt{-1 \times -1};] but I was thinking that [;\sqrt{-1 \times -1};] can be decomposed as [;\sqrt{-1} \times \sqrt{-1};] – yCalleecharan Oct 03 '11 at 08:03
  • Don't use [; and ]; because they don't work. On M.SE we put LaTeX markup inside dollar signs, e.g. $\LaTeX$ gives $\LaTeX$. You should be able to edit the original question and see how I've revised the original markup. And no, you can't do that decomposition as it's only valid for nonnegative numbers. ($-1$ is obviously negative.) – anon Oct 03 '11 at 08:08
  • If I cannot split the square root (which requires both arguments to be positive), then $\sqrt{-1 \times -1}$ = $ \sqrt{1\angle 360} = 1.$ – yCalleecharan Oct 03 '11 at 08:12
  • Ok for the markup remark. – yCalleecharan Oct 03 '11 at 08:13
  • That's correct; $\sqrt{(-1)^2}=1$. – anon Oct 03 '11 at 08:18

2 Answers2

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Using phasors to multiply complex numbers is fine; it's equivalent to the usual way of working in the polar form of complex numbers $z=re^{i\phi}$, where $\phi$ is the phase angle in radians and $r$ is the modulus. But there are symbolic manipulations of complex numbers that only obtain under certain conditions, mainly due to branch cuts of elementery functions like the square root. For the case of the square root function applied to numbers on the real axis, the answers at this question pretty much cover the issue of $\sqrt{ab}=\sqrt{a}\sqrt{b}$.

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anon
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If we use the definition $\sqrt{-1}=\sqrt{1 \angle 180}= 1\angle 90 = + \mathrm{i}$,

then we have

$\sqrt{{-1}\times{-1}}=\sqrt{1 \angle 180 \times 1 \angle 180}= \sqrt{1 \angle 360}=1\angle 180 = -1$

If however we use the definition $\sqrt{-1} = \sqrt{1 \angle -180}$ then we get $\sqrt{-1} = 1\angle -90 = -\mathrm{i}$ which is wrong. Though $-1 = 1 \angle 180$ or $1 \angle -180$, we use only the former definition when under the square root sign.

yCalleecharan
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  • This is incorrect; $\sqrt{1\angle360}\ne1\angle180$. You have to reduce 360 to 0 first, then obtain $\sqrt{1\angle0}=1\angle0$. Look at the graphic at Wikipedia of the complex square root function. To represent the evaluation of the square root in phasor form, you put the angle inside the range $(-180^\circ,180^\circ]$ before halving it. So any angle in $[0^\circ,180^\circ]$ can be divided by 2 as-is, while any angle $\phi$ in $(-180^\circ,0]$ must be converted to $\phi-360^\circ$ and then halved. – anon Oct 05 '11 at 06:46
  • Sorry, that second interval should be $(-180^\circ,0^\circ)$ (it doesn't include $0$). – anon Oct 05 '11 at 06:52
  • Thanks. That is very interesting. I guess that you mean the second interval is $-180 < \theta <0 $. Am I right? So if we have $\theta = -90$, then we do $\frac{-90 -360}{2}=-225$ which is still outside $-180 < \theta \leq 180$. What am I doing wrong? – yCalleecharan Oct 05 '11 at 07:37
  • Sorry, I really goofed. That second interval (where you should subtract 360 before halving) should be $(180^\circ,360^\circ)$. The angle $\theta=-90^\circ$ is already in the range $(-180^\circ,180^\circ]$ so nothing needs to be done before halfing; just take $\sqrt{1\angle-90}=1\angle-45$. Ultimately, it all boils down to this: whatever angle you have, before halving it you have to add or subtract multiples of $360^\circ$ so that it's in the necessary range. – anon Oct 05 '11 at 07:47
  • If I understand well, then the principle range is $-180 < \theta \leq 180$. And anything that is outside that range should be adjusted by adding or subtracting 360 (and multiples if necessary) so that the final angle $\theta$ lies in our principal range. Is this correct? – yCalleecharan Oct 05 '11 at 08:00
  • Yes. That way halving it will correctly represent the complex square root function. – anon Oct 05 '11 at 08:04
  • Thanks for taking time to clear out my misunderstandings. I am re-accepting your answer :). – yCalleecharan Oct 05 '11 at 08:08
  • @anon : I have a deeper question about the principal square root: http://math.stackexchange.com/questions/70280/prinicipal-square-root-of-a-complex-number – yCalleecharan Oct 06 '11 at 05:58