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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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    That's correct; $\sqrt{(-1)^2}=1$.2011-10-03

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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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    Th$a$nks for making this issue clear to me. And 1 vote up.2011-10-03
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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.

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    @anon : I have a deeper question about the principal square root: http://math.stackexchange.com/questions/70280/prinicipal-square-root-of-a-complex-number2011-10-06