Sqrt (-1) * sqrt (-1)

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  • #1
rishch
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sqrt(A)*sqrt(B)=sqrt(A*B) Therefore,

sqrt(-1)*sqrt(-1)=sqrt(-1*-1)
=sqrt(1)
=1

This doesn't make sense.Did I do something wrong ?
 

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  • #2
Simon Bridge
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In general, [tex]\sqrt{a^2}=\pm a[/tex]... so:[tex]\sqrt{a}^2=\sqrt{a}\sqrt{a}=a=\pm\sqrt{a^2}[/tex]... you have used only the positive square root and a can be positive or negative. But what if a can only be positive? Then:
[tex]\sqrt{-a}^2=\sqrt{-a}\sqrt{-a}=-a=-\sqrt{a^2}[/tex]... because we've excluded one of the possible results from the last term.
 
  • #3
micromass
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In general, [tex]\sqrt{a^2}=\pm a[/tex]

No. The square root only gives out one value: the positive value. You act like there are two solutions, a positive and a negative. There are not.

The problem is that [itex]\sqrt{-1}[/itex] isn't well-defined. The square root operator is only defined for positive real numbers, not for negative numbers.

Furthermore, even if we were to define it, then [itex]\sqrt{ab}=\sqrt{a}\sqrt{b}[/itex] wouldn't hold anymore for negative a and b.
 
  • #4
DivisionByZro
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The confusion between the following is common:

[tex] a = k^{2} [/tex]
[tex] k = \pm \sqrt{a} [/tex]

Here a must be positive (or zero), but k can be anything.
and [tex] \sqrt{s^{2}} = \mid s \mid [/tex]
 
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  • #5
HallsofIvy
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The problem is not well defined because, as micromass says, [itex]\sqrt{-1}[/itex] is not well-defined. You will sometimes see "i" defined as [/itex]\sqrt{-1}[/itex], but that simply is not a very good definition- because we don't know what [itex]\sqrt{-1}[/itex] means until after we have defined i. You will sometimes see i defined as "the complex number such that [itex]i^2= -1[/itex]" but that also is not a good definition. Just as in the real number such an an equation has two solutions but that "definition" doesn't specify which i is. With the real numbers, we can specify that [itex]\sqrt{a}[/itex] is the positive number, x, such that [itex]x^2= a[/itex] but with the complex numbers, we cannot do that- the complex numbers does not form an "ordered field".

What we can do is define the complex numbers as the set of ordered pairs, (a, b) with addition defined by (a, b)+ (c, d)= (a+c, b+d) and multiplication defined by (a, b)*(c, d)= (ac- bd, ad+ bc). Then we can associate the real number a with the pair (a, 0) (that's and "isomorphism" from the ordinary real numbers to the subset of the complex numbers of the form (a, 0)) while defining "i" to be the pair (0, 1): i^2= (0, 1)^2= (0,1)*(0,1)= (0*0- 1*1, 1*0+ 0*1)= (-1, 0)=> the real number -1. Now we can write (a, b)= (a, 0)+ b(0,1)= a+ bi.
 
  • #6
Simon Bridge
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You act like there are two solutions, a positive and a negative.
Aren't there? surely √a2 must be one of +a or -a ... eg, if a=-1 then√(-1)2 = -(-1) or did I need set notation to do this properly?

... it is sloppy notation. You know how it arises though?

a=√a2 is true for a ≥ 0 and -a=√a2 is true for a ≤ 0 isn't it?

;)

What are possible values of a, given b and b=a2?
I would have put: c=√b ... but since |a|=√b, then a=±c [1]

... there are two possible values of a which could give result c.

Thus it seems quite reasonable[2] to write:

a=±√a2

hasn't the surd operation destroyed some of the information about a?
don't we need to reflect that in our notation?

of course, that's how we find a given a2.
the surd returns a single value, but a could be the negative of that value too.
so the notation is trying to say that there are two possible values of a given only a2.

Presumably there is a better way to express this?

---------------------------
[1] or should that be: a [itex]\in[/itex] {-c,+c} ?
[2] though I shall concede that it is better to avoid this sort of construction since a on the LHS is a range of values while the a on the RHS is a single value. Besides, the idea is that we don't have a priori knowledge of a.
 
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  • #7
Deveno
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the trouble lies with the √ symbol.

in the real numbers, we can enforce a definition that turns √ into a function (i.e., single-valued).

but this definition breaks down in the complex numbers. why? because even though the equation:

z2 = a

has 2 solutions (unless a = 0), there's no way to to say which one is "the positive one" for all non-zero complex numbers. that is, on the complex plane √ is "multi-valued" (not a function).

as with the complex logarithm, we have to "choose a branch" before computing a square root (and we have 2 to choose from). so, with complex numbers, it IS NOT TRUE that:

√z√w = √(zw)

in particular, it is NOT TRUE that:

√(-1)√(-1) = √((-1)(-1)) = √1 = 1

(if we choose the branch where √(-1) = i, we get √(-1)√(-1) = i2 = -1, if we choose the branch √(-1) = -i, we get:

√(-1)√(-1) = (-i)2 = √1 = -1,

because THAT branch takes the NEGATIVE square root of a positive real).

as Halls pointed out, we have TWO choices for √(-1), and there isn't any way to KNOW "which one is the "actual" i. complex conjugation is a field automorphism that fixes the reals, so EITHER choice leads to an algebraic entity that "behaves the same".

to make an analogy, in a parallel universe, there might exist an earth that uses the yx-plane, instead of the xy-plane. to our way of looking at things, they do stuff "upside-down" and "backwards", and from their perspective, so do we. who's right?

to make a long story short: the notation √z is ILL-DEFINED, for complex numbers, without further elaboration about what convention you are using. there are similar problems with the so-called "Arg" function, using polar coordinates. you HAVE to specify your intended "angle range".

yes, it's perhaps a bother that the complex exponential is periodic in i. we expect it to be monotonic, like the real exponential, but 'taint so. so logarithms, and roots (the two are intimately bound up, it turns out) don't behave the same way as we expect from our experience with reals. that means certain rules, which we accept as fact in the real numbers, suddenly turn out not to be true in the complex numbers. this is one of those times.
 
  • #8
Char. Limit
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Sigh. One of these again, huh? Maybe my signature needs another line.

The square root operator, √, is defined as the positive square root, i.e. k = √(a) is the positive value such that k^2 = a. The POSITIVE value. Okay, that's one issue solved. One done and I'm on to the next one.

[itex]\sqrt{a} \sqrt{b} = \sqrt{a b}[/itex] only holds if BOTH a and b are POSITIVE. This is not true in your example. Therefore, you've proven nothing except that there's a reason that it only holds when a and b are positive.
 
  • #9
Deveno
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Sigh. One of these again, huh? Maybe my signature needs another line.

The square root operator, √, is defined as the positive square root, i.e. k = √(a) is the positive value such that k^2 = a. The POSITIVE value. Okay, that's one issue solved. One done and I'm on to the next one.

[itex]\sqrt{a} \sqrt{b} = \sqrt{a b}[/itex] only holds if BOTH a and b are POSITIVE. This is not true in your example. Therefore, you've proven nothing except that there's a reason that it only holds when a and b are positive.

with all due respect, √z can be defined on the complex numbers (provided you jump through a few hoops, first....can i hold the hoops? pretty please?), even though no such analogous definition of "positive" exists. but you need to be more specific about what you MEAN by √z (or, to put it another way, "which" angle you're taking "half of", when applying that de moivre business).
 
  • #10
Char. Limit
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with all due respect, √z can be defined on the complex numbers (provided you jump through a few hoops, first....can i hold the hoops? pretty please?), even though no such analogous definition of "positive" exists. but you need to be more specific about what you MEAN by √z (or, to put it another way, "which" angle you're taking "half of", when applying that de moivre business).

All of this is true, but it's hardly relevant to the OP's problem - and I doubt the OP is thinking on that level anyway.
 
  • #11
chiro
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sqrt(A)*sqrt(B)=sqrt(A*B) Therefore,

sqrt(-1)*sqrt(-1)=sqrt(-1*-1)
=sqrt(1)
=1

This doesn't make sense.Did I do something wrong ?

Hey rischch and welcome to the forums.

When you get this kind of thing, it helps to use exponential forms of numbers rather than relying on things that can cause this kind of grief.

The exponential form of a unit length complex number if given by e^(ix) = cos(x) + i sin(x). You can scale the number by multiplying by a positive constant (i.e. r e^(ix)). Then if you want to do things like multiply SQRT(-1) x SQRT(-1) = i * i, if you do this with exponentiation you get e^(i x pi/2) x e^(i x pi/2) = e^(i x pi) = cos(pi) + i sin(pi) = -1. (since sin(pi) = 0 and i x 0 = 0)

This method doesn't fail and its useful to look at because it will help you understand how numbers work when you do all kinds of things geometrically.
 
  • #12
Deveno
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All of this is true, but it's hardly relevant to the OP's problem - and I doubt the OP is thinking on that level anyway.

fair enough. suffice to say, the OP was doomed at line one where he (she?) writes:

sqrt(A)*sqrt(B)=sqrt(A*B)

because this isn't always true.

explaining WHY it isn't true, perhaps is a bit more of a can of worms then the OP expected, although it certainly makes the historical suspicion of complex numbers easier to understand (they're not just "the real numbers, but even better").

Hey rischch and welcome to the forums.

When you get this kind of thing, it helps to use exponential forms of numbers rather than relying on things that can cause this kind of grief.

The exponential form of a unit length complex number if given by e^(ix) = cos(x) + i sin(x). You can scale the number by multiplying by a positive constant (i.e. r e^(ix)). Then if you want to do things like multiply SQRT(-1) x SQRT(-1) = i * i, if you do this with exponentiation you get e^(i x pi/2) x e^(i x pi/2) = e^(i x pi) = cos(pi) + i sin(pi) = -1. (since sin(pi) = 0 and i x 0 = 0)

This method doesn't fail and its useful to look at because it will help you understand how numbers work when you do all kinds of things geometrically.

again, i want to point out that even using "polar form" (which is what you are doing here), still requires a little caution. for example, although it seems obvious that "half of a 0 angle is 0", one needs to remember that an angle of pi is also a possibility (because of the periodicity in i of the complex exponential). you'll get different square roots of cos(5pi/4) + i sin(5pi/4) and cos(-3pi/4) + i sin(-3pi/4) using de moivre's formula, even though both represent the same complex number (the first will be half the angle back clockwise towards the positive x-axis, and the second will be half the angle back counter-clockwise towards the positive x-axis).
 
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  • #13
morphism
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All of this is true, but it's hardly relevant to the OP's problem - and I doubt the OP is thinking on that level anyway.
I think it's extremely relevant to the OP's problem. Your definition of √a as "the positive k such that k^2=a" only makes sense for positive a. Part of the OP's problem was that they naively applied √ to negative numbers. Deveno's very helpful post explained that while you can indeed think about square roots of negative (and, more generally, complex) numbers, you have to be much more cautious, or otherwise you run into problems as the OP has.
 
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  • #14
Char. Limit
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I think it's extremely relevant to the OP's problem. Your definition of √a as "the positive k such that k^2=a" only makes sense for positive a. Part of the OP's problem was that they naively applied √ to negative numbers. Deveno's very helpful post explained that while you can indeed think about square roots of negative (and, more generally, complex), you have to be much more cautious, or otherwise you run into problems as the OP has.

True.
 
  • #15
chiro
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again, i want to point out that even using "polar form" (which is what you are doing here), still requires a little caution. for example, although it seems obvious that "half of a 0 angle is 0", one needs to remember that an angle of pi is also a possibility (because of the periodicity in i of the complex exponential). you'll get different square roots of cos(5pi/4) + i sin(5pi/4) and cos(-3pi/4) + i sin(-3pi/4) using de moivre's formula, even though both represent the same complex number (the first will be half the angle back clockwise towards the positive x-axis, and the second will be half the angle back counter-clockwise towards the positive x-axis).

I should have said this explicitly, but I am talking about doing standard arithmetic with complex numbers which includes addition, subtraction, multiplication and division (if you can do this).

The exponential form will never fail for these operations no matter what numbers you are using (unless you try to divide by a zero modulus complex number).

If you are trying to do exponentiation of some kind (or an operation associated with it) like trying to find roots for example, then that will require the appropriate machinery.

Again I apologize for not making the suggestion explicit, but I still stand by the suggestion that when we are talking about complex numbers in the context of arithmetic, the exponential method will not fail.
 
  • #16
dodo
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I still stand by the suggestion that when we are talking about complex numbers in the context of arithmetic, the exponential method will not fail.

In particular, if you believe that, for [itex]z \in \mathbb{C}[/itex],[tex]\begin{align*}z^a \cdot z^b &= z^{a+b} \qquad \mbox{and}\\ \left(z^a\right)^b &= z^{ab}\end{align*}[/tex]then[tex]z^a \cdot z^a = z^{a+a} = z^{2a} = \left(z^2\right)^a[/tex]No?
 
  • #17
Char. Limit
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In particular, if you believe that, for [itex]z \in \mathbb{C}[/itex],[tex]\begin{align*}z^a \cdot z^b &= z^{a+b} \qquad \mbox{and}\\ \left(z^a\right)^b &= z^{ab}\end{align*}[/tex]then[tex]z^a \cdot z^a = z^{a+a} = z^{2a} = \left(z^2\right)^a[/tex]No?

If you believe that. I don't believe that the things you believe hold for any real a and b and complex z. In particular, I think we've already shown it failing for a = b = 1/2.
 
  • #18
Deveno
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I should have said this explicitly, but I am talking about doing standard arithmetic with complex numbers which includes addition, subtraction, multiplication and division (if you can do this).

The exponential form will never fail for these operations no matter what numbers you are using (unless you try to divide by a zero modulus complex number).

If you are trying to do exponentiation of some kind (or an operation associated with it) like trying to find roots for example, then that will require the appropriate machinery.

Again I apologize for not making the suggestion explicit, but I still stand by the suggestion that when we are talking about complex numbers in the context of arithmetic, the exponential method will not fail.

in a word: no.

writing complex numbers as re just moves the problem from the function at hand to the "arg" function, and you STILL have to choose a branch. why? because:

e = cos θ + i sin θ

and the trigonometric functions are periodic (this means they are NOT 1-1), so "θ" is not uniquely defined. no matter what "range" you pick, somewhere on a circle, arg is going to have a discontinuity.

and it's easy to pick examples where z starts out in your specified range, but z2 (using the "standard exponential formula") is not in your range, and you have to "re-calculate" the argument by adding or subtracting 2π.

here is an explicit example: suppose one (as is typical in engineering) limits the angle range to (-π,π].

then if [itex]z = \cos(2\pi/3) + i\sin(2\pi/3)[/itex], for z2 you DON"T just "double the angle", you double the angle and subtract 2π.

or, if you limit the angle range to [0,2π), then when you square [itex]\cos(3\pi/2) + i\sin(3\pi/2)[/itex] you have a similar situation.

we have to be aware of these subtleties when working with complex numbers, the polar representation is not unique (there's a mod 2π in there somewhere). this rears its ugly head with functions that "undo":

roots (which "undo powers"), related to:
logarithms (which "undo exponentiaion"), related to:
exponentiation (complex exponentiation, like zw, invokes the log function).

it's not correct, and terribly unfair to someone just starting out studying complex numbers, to say something like "you can't go wrong with the exponential method". yes, you can.
 
  • #19
micromass
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Nice answers here.

I would like to get some feedback from the OP on whether we are really answering his question and on whether there are still things unclear in the answers.
 
  • #20
Simon Bridge
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Agreed, it's time to hear from OP!
So far it has been a "make naive statement and sit back and watch the egg-heads argue" type of thing. lulz.
 
  • #21
chiro
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Deveno, the nature of the trigonometric functions is that they are periodic and do the 'modulus' behaviour for you.

Lets go with your example with having an argument of 3pi/2. Specifically,

z = cos(3pi/2) + i sin(3pi/2).

Now z^2 = cos(3pi/2 + 3pi/2) + i sin(3pi/2 + 3pi/2) (Since our number is modulus 1)
= cos(3pi) + i sin(3pi)
= cos(pi) + i sin(pi) [Properties of trigonometric functions]

You have to remember that with arithmetic we don't care about keeping track of the angle: we are only interested in evaluating functions given angles.

If we calculate 3 + 4i multiplied by 2 + i, then with polar we find the principal arguments of both and then use them to evaluate something else that uses the addition of those arguments.

We don't care about keeping memory of those arguments after we have generated a complex number answer to the question, and due to the nature of periodicity of the trigonometric functions, the (mod 2pi) thing is done for us automatically.
 
  • #22
micromass
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Deveno, the nature of the trigonometric functions is that they are periodic and do the 'modulus' behaviour for you.

Lets go with your example with having an argument of 3pi/2. Specifically,

z = cos(3pi/2) + i sin(3pi/2).

Now z^2 = cos(3pi/2 + 3pi/2) + i sin(3pi/2 + 3pi/2) (Since our number is modulus 1)
= cos(3pi) + i sin(3pi)
= cos(pi) + i sin(pi) [Properties of trigonometric functions]

You have to remember that with arithmetic we don't care about keeping track of the angle: we are only interested in evaluating functions given angles.

If we calculate 3 + 4i multiplied by 2 + i, then with polar we find the principal arguments of both and then use them to evaluate something else that uses the addition of those arguments.

We don't care about keeping memory of those arguments after we have generated a complex number answer to the question, and due to the nature of periodicity of the trigonometric functions, the (mod 2pi) thing is done for us automatically.

I think Deveno is trying to say that logarithms are still a problem here. You can get away with exponentiation and multiplying. But the exponential approach does not absolve us from problems with logarithms.
 
  • #23
chiro
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I think Deveno is trying to say that logarithms are still a problem here. You can get away with exponentiation and multiplying. But the exponential approach does not absolve us from problems with logarithms.

I don't doubt that, that is why I said in an earlier post that I clarified that it holds for arithmetic which includes addition, subtraction, multiplication and division (no divide by zero though).

I agree about problems when you are trying to things like finding roots and doing logarithmic operations in general, but I wasn't referring to that (although I had to clarify this specifically as a result of prior posts that pointed this out).
 
  • #24
rishch
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Hey guys,
Sorry I took so long to reply.I couldn't understand most of what you guys said (I'm in 9th grade) and I don't know anything complex numbers and the other things you guys were talking about.One of the posts said that √A√B=√AB only when A and B are positive.So is that where I went wrong?
 
  • #25
Char. Limit
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Hey guys,
I couldn't understand most of what you guys said (I'm in 9th grade) and I don't know anything complex numbers and the other things you guys were talking about.One of the posts said that √A√B=√AB only when A and B are positive.So is that where I went wrong?

Yup. That particular rule only holds when A and B are both positive. If you don't put in that restriction, you get things like... well... what you posted.
 
  • #26
rishch
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Thanks a lot!
So the rule A to the power m * B to the power m =AB to the power m will also be invalid if A = B = -1 and m = 1/2 ?
 
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  • #27
Deveno
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Thanks a lot!
So the rule A to the power m * B to the power m =AB to the power m will also be invalid if A = B = -1 and m = 1/2 ?

exactly so.

Deveno, the nature of the trigonometric functions is that they are periodic and do the 'modulus' behaviour for you.

Lets go with your example with having an argument of 3pi/2. Specifically,

z = cos(3pi/2) + i sin(3pi/2).

Now z^2 = cos(3pi/2 + 3pi/2) + i sin(3pi/2 + 3pi/2) (Since our number is modulus 1)
= cos(3pi) + i sin(3pi)
= cos(pi) + i sin(pi) [Properties of trigonometric functions]

You have to remember that with arithmetic we don't care about keeping track of the angle: we are only interested in evaluating functions given angles.

If we calculate 3 + 4i multiplied by 2 + i, then with polar we find the principal arguments of both and then use them to evaluate something else that uses the addition of those arguments.

We don't care about keeping memory of those arguments after we have generated a complex number answer to the question, and due to the nature of periodicity of the trigonometric functions, the (mod 2pi) thing is done for us automatically.

but we DO care about the angle...usually we just take it as second nature to keep the angle within a specified range (either 0 to 2pi, or -pi to pi, depending on how we're thinking about "angle"). you underscore my point when you say "principal" argument. that (fine) distinction is exactly what we need to keep in mind! remembering that cosines and sines are points on a circle, and that a circle goes around and then repeats itself isn't "difficult" and i will readily agree that that thinking in these terms gives a clear "geometric" idea of what is going on to avoid getting into trouble.

you can easily see that there are TWO points on a unit circle, that when squared, give a desired number (this is the complex analogue of the fact that a positive real number has TWO square roots, positive, and negative. in fact, real numbers are a subfield of the complex numbers, and the positive and negative square roots of a positive real number are in fact, the two COMPLEX square roots of the positive real, so the "real case" is just a "special case" of the "complex case").

it's like this: with a quadratic equation, for example, we often get two roots: and one we can often disregard as "extraneous" (because it yields a negative number, or a number that is too small, or too large for our original problem). but if the discriminant is negative (two complex solutions), it's no longer clear if there is a "preferred" solution (geometrically, there is "symmetry" between a complex number and its conjugate, at least with respect to the reals). this ambiguity is unavoidable, no matter how you choose to represent the complex numbers in question.

i think HallsofIvy pointed out, we really can't tell "i" from "-i". calling one of the two √(-1) is a CHOICE, and one we make arbitrarily (equivalent to choosing an orientation for the plane). it pays to remember this choice, and that it is indeed an aribtrary choice (especially in modelling physical situations, where we may be dealing with the absolute value of an angle ("the angle between"), rather than the angle itself), so that we can at least be consistent.

it may seem obvious that if z2 = w, that z = √w. that's ALMOST true. w is, indeed ONE of the square roots in question. but "which one" (which branch) matters.

IF, √z and √w are on the same branch as √(zw), then it is indeed true that √(zw) = √z√w. in the OP's original "un-proof", the trouble is precisely that choosing the branch where i = √(-1), forces us to the choice √1 = -1, if we wish to invoke √(zw) = √z√w.
 
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