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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 ?
sqrt(-1)*sqrt(-1)=sqrt(-1*-1)
=sqrt(1)
=1
This doesn't make sense.Did I do something wrong ?
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.In general, [tex]\sqrt{a^2}=\pm a[/tex]
Aren't there? surely √a^{2} 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?You act like there are two solutions, a positive and a negative.
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).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.
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.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).
Hey rischch and welcome to the forums.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 ?
fair enough. suffice to say, the OP was doomed at line one where he (she?) writes: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.
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).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.
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.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.
True.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.
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).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).
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?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.
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.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?
in a word: no.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.
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.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 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 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.
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.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?