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Normandy6644
2005-May-04, 06:09 PM
Okay, this seems way too basic to be confusing, but I haven't come up with a decent answer yet.

What is sqrt(-1)*sqrt(-1)?

There seem to be two ways of looking at it:

sqrt(-1)*sqrt(-1)=sqrt[(-1)*(-1)]=1

or

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

I'm sure this is an order of operations thing, so we have:

Parenthesis
Exponents
Multiplication
Division
Subtraction

(Sally, you are excused)

This seems to imply that the exponents would come first, and you would have

(-1)^1/2*(-1)^1/2=(-1)^1=-1.

Maple seems to agree with me, but it does seem slightly tricky!

WaxRubiks
2005-May-04, 06:12 PM
but what is the square root of -i?

Swift
2005-May-04, 06:15 PM
sqrt(-1)*sqrt(-1)=i*i=-1

Is correct.

ToSeek
2005-May-04, 06:21 PM
Or

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

Remember that every number has two square roots.

01101001
2005-May-04, 06:37 PM
What is sqrt(-1)*sqrt(-1)?
Possibly more than a single value.

When you evaluate square, or even, roots, you get two quantities: the positive and negative root.

i ^ 2 is -1
(-i) ^ 2 is also -1

So sqrt(-1)*sqrt(-1) has two results, either: 1 or -1.

Just like sqrt(1)*sqrt(1) is either 1 or -1.

A Thousand Pardons
2005-May-04, 06:42 PM
but what is the square root of -i?
sqrt(2) * (1 - i) /2 :)

Remember that every number has two square roots.
But not two sqrt( )s :)

Disinfo Agent
2005-May-04, 06:51 PM
Okay, this seems way too basic to be confusing, but I haven't come up with a decent answer yet.

What is sqrt(-1)*sqrt(-1)?

There seem to be two ways of looking at it:

sqrt(-1)*sqrt(-1)=sqrt[(-1)*(-1)]=1 [-X

or

sqrt(-1)*sqrt(-1)=i*i=-1
The property that sqrt(a) * sqrt(b) = sqrt (a*b) fails for negative numbers.

...Unless you regard the square root as a multivalued "function", and think in terms of sets, as 01101001 did above (http://www.badastronomy.com/phpBB/viewtopic.php?p=465775#465775).

peter eldergill
2005-May-04, 07:14 PM
i is defined to be the square root of -1, and the square root function is defined to be the positive value

The solutions to the equation x^2+1=0 are +i or -i

The value of the expression sqrt(-1) is +i

I could go on about complex numbers all day

Has anyone here actually ever proven deMoivre's Theorem for all real numbers? Any book I've ever seen (including my universtiy books) say "The proof of this is beyond the scope of this book"

I don't remember if we ever proved it in my 4th year complex analysis course or not..

Hope this helps

Pete

Disinfo Agent
2005-May-04, 07:22 PM
i is defined to be the square root of -1, and the square root function is defined to be the positive value
Actually, if you're really rigorous, you first construct i, and then define the square root(s) of negative numbers.
Also, pure imaginary numbers aren't usually called "negative" or "positive", although I can see what you mean. (But how would you go about defining the square root of a general complex number in that fashion? Note: it can be done, but it's not trivial.)

Has anyone here actually ever proven deMoivre's Theorem for all real numbers? Any book I've ever seen (including my universtiy books) say "The proof of this is beyond the scope of this book"

I don't remember if we ever proved it in my 4th year complex analysis course or not..
What theorem do you mean? What does it state?

peter eldergill
2005-May-04, 07:41 PM
In polar form a+bi = r(cosA + isinA), where A is the principal angle (between 0 and 360) between the positive real axis and the complex number.

demoivre's theorem say that
(r(cosA + isinA))^n = r^n(cos(nA) + isin(nA))

Very simple to prove using mathematical induction but really hard to prove for all real numbers

I know what I said before is not perfectly rigorous, but the square root of a complex number can be found using demoivre's theorem where n = 1/2

In this case, for the imaginary number i, we have r = 1 and A =90 deg

So (cos90 + isin90)^(1/2) = cos(45) + isin45 = (1+i)/sqrt(2), exactly as stated by Mr. pardons earlier

Note that there is another root of i, by choosing A = 90 + 360

I think it is (-1-i)/sqrt(2)

I certainly disagree that

So sqrt(-1)*sqrt(-1) has two results, either: 1 or -1.

Using the above theorem, this is (i)^2 = (cos90 + isin90)^2
= cos180 + isin180 = -1 + 0 = -1

Definately not +1

Later

Pete

Disinfo Agent
2005-May-04, 07:47 PM
Notice that you said, yourself, that i had two square roots. If i can have two square roots, why not -1?

jfribrg
2005-May-04, 08:24 PM
We're dealing with the field of complex numbers here. There is one and only one answer to the question of what a* b is, where a and b are both complex numbers. In this case a and b are both = i. The correct answer is that i * i = -1. The fact (proven by Gauss) that there are two solutions to the equation x^2 + 1 = 0 is irrelevant.

Now back to the original question:

Given complex numbers a,b, and recalling middle school math, (a^b) * (a^b) = a^(2b), not a^(b^2).

let a= -1 and b = 1/2, you get sqrt(-1) * sqrt(-1) = -1^1 = -1.

Consider the (incorrect) assertion that sqrt(-1) * sqrt(-1) = sqrt(-1 * -1). In terms of bases and exponents, this assertion is equivalent to :
-1^(1/2) * -1^(1/2) = (-1 *-1) ^ (1/2). This is true only if a^b * a^b = a^(b*b). Almost any choice of numbers for a,b, except a=1 will give you a counterexample. As as specific example, let a= 2, b = 3
a^b * a^b = 64
a^(b*b) = 512

This counterexample is in the real numbers, but since every real number is a complex number (with imaginary part = 0), this also serves as a counterexample in the Complex field.

So, in summary, it is an order of operations issue, like Normandy suspected.

Now does anyone care to discuss whether (sqrt(0.9999....) * sqrt(0.999999....) = -1? :) (runs and hides before anyone can throw a tomato at him)

Disinfo Agent
2005-May-04, 08:38 PM
We're dealing with the field of complex numbers here. There is one and only one answer to the question of what a* b is, where a and b are both complex numbers. In this case a and b are both = i.
It depends. In this case, a and b can each be i or -i, according to some definitions.

peter eldergill
2005-May-04, 08:40 PM
Disinfo, I understand now the confusion. It took me a while

Yes, by the way I defined it, -1 will have 2 roots +i and -i, again using deMoivres theorem

These roots, however, come from different angles in the complex plane
The principle angle (called the argument) comes from A = 180 deg, corresponding to the number -1 being on the negative real axis

The next rotation gives 180 + 360 = 540

So (-1)^1/2 = (cos540 +isin540)^(1/2) = cos270 +isin270 = -i

Any other multiple of 360 will yield either + or - i

I wonder if the nth root of a complex number is defined to be given by the Principal Argument. Does anyone know? It certainly is for real numbers, that is square roots are positive.

I show this to my students all the time:

sqrt(4) = +2, however
x^2 = 4 has two solutions: +or- sqrt(4)

Does this help? This could go on forever, just like .99999 :lol:

Later

Pete

A Thousand Pardons
2005-May-04, 11:21 PM
Note that there is another root of i, by choosing A = 90 + 360

I think it is (-1-i)/sqrt(2)
No, it is the negative of the one I gave (http://www.badastronomy.com/phpBB/viewtopic.php?p=465780&amp;sid=9ab5c266e69efac05194d6f bed749181#465780).

Now does anyone care to discuss whether (sqrt(0.9999....) * sqrt(0.999999....) = -1? :) (runs and hides before anyone can throw a tomato at him)
It does not.

crosscountry
2005-May-05, 12:39 AM
[(-1)^.5]^2= -1

any number squared is positive except (i)

meaning 5i^2=5^2 * i^2

Disinfo Agent
2005-May-05, 10:52 AM
I wonder if the nth root of a complex number is defined to be given by the Principal Argument. Does anyone know? It certainly is for real numbers, that is square roots are positive.
It can work for square roots, but not for n-th roots in general, because, since each complex number z has n n-th roots, all lying on the same circle with centre at the origin and radius sqrt(|z|), you end up with more than one root in any given quadrant.

peter eldergill
2005-May-05, 12:46 PM
peter eldergill wrote:

Note that there is another root of i, by choosing A = 90 + 360

I think it is (-1-i)/sqrt(2)

A thousand pardons wrote:

No, it is the negative of the one I gave.

What I have is one of the roots of i and the one you wrote is a root of -i

Pete

A Thousand Pardons
2005-May-05, 03:03 PM
What I have is one of the roots of i and the one you wrote is a root of -i
Arggh (http://www.badastronomy.com/phpBB/viewtopic.php?p=1124#1124), you are right, I was wrong.

I got confused when you said "exactly as stated by Mr. pardons earlier" but I think it's mostly because of the medicine I was taking. :)