# Homework Help: I^2=1? How can this be?

1. Jan 30, 2006

### DeathKnight

i^2=1??? How can this be?

First all I must tell you people that I have no doubt that i^2=-1 but what am I doing wrong in the following steps?
i^2
=i * i
=sqrt(-1) * sqrt(-1)
=sqrt(-1*-1)
=sqrt(1)=1?
I asked my mathematics teacher but he was unable to find any error in it. Thanks in advance for any help that you can offer.

2. Jan 30, 2006

### TD

Are you sure it was a math teacher?

Simply put: the square root function doesn't "behave" exactly the same way for complex values as for real values, which is why not all the properties you have in the real case will still hold for (general) complex values. En example of such a property which you used here is sqrt(a)*sqrt(b) = sqrt(ab).

Please do correct me if this wasn't the best way to put it in English

3. Jan 30, 2006

### arildno

As a follow-up to TD's post, remember that -1 may be written as:
$$-1=e^{i\pi}$$

But now, if "normal" rules applied, we should have:
$$-1=e^{i\pi}=e^{i\pi\frac{2}{2}}=(e^{i2\pi})^{\frac{1}{2}}=1^{\frac{1}{2}}=1$$
It is the third step here that is faulty..

4. Jan 30, 2006

### TD

Indeed, for complex values we no longer have that $\left( {z^a } \right)^b = z^{ab}$, at least not in general.

Instead, we do have $\left( {z^a } \right)^b = z^{ab} e^{2bk\pi i}$ with k an integer. This follows from the fact that $e^z$ is periodic with period $2\pi i$ and the definition of complex powers as $z^a = e^{a\ln \left( z \right)}$ where ln(z) is the complex natural logarithm. If I recall correctly...

5. Jan 30, 2006

To put in simple terms,

if x^2 = y^2, then it does NOT mean, x = y.
but x = +/- y. and one of it will be true.

So i^2 = +/- 1; which is true (i^2 = -1)

6. Jan 30, 2006

### Muzza

Is this actually relevant?

7. Jan 30, 2006

### matt grime

Oh, look, another question about the square root branches.

I await the next 0.9... is not one question in this endless effing cycle.

Can we PLEASE have the FAQ that I crave so dearly?

8. Jan 30, 2006

### chroot

Staff Emeritus
If I'm not mistaken, universities typically teach the same classes over and over again, too, in the same "effing cycle."

The Dr. Math project (http://mathforum.org/dr.math/) already has a pretty good compendium of FAQs. When someone asks a question that's already present in a FAQ, they just answer with a link to that FAQ. You can easily do the same sort of thing here. Keep a list of threads handy, sorted by topic, and answer repetitive questions with a link.

- Warren

9. Jan 30, 2006

### matt grime

New incoming students at university are typically taught the material. The analogy is meaningless. The students don't typically have the same pointless questions on the same topics of misunderstanding. So, let's have an official policy of NOT replying to these threads, and deleting any replies that there are. Since a moderator invariably checks them anyway, and almost as invariably replies, there is no extra work in having them post, delete, and close these pointless replies to these tedious threads.

Last edited: Jan 30, 2006
10. Jan 30, 2006

### JasonRox

Are you saying the third or the fourth step?

11. Jan 30, 2006

### TD

It's this one: $$e^{i\pi\frac{2}{2}}=(e^{i2\pi})^{\frac{ 1}{2}}$$, see my second post in this topic for the reason.

12. Jan 30, 2006

### JasonRox

Yeah, I understood.

I thought he was refering to adding 2/2 as wrong. So I was confused since I didn't see what was wrong with that.

13. Jan 31, 2006

### arildno

Oh dear, I can't count further than to three I think..

(At least, I recognize the ambiguity of what I wrote.)

14. Jan 31, 2006

### TD

It is often defined by $$i^2 &= -1$$, precisely because of the subtilities and possible problems with using the square root in the complex case.
Furthermore, I don't see how this explains the fallacy in the 'proof', it uses your definition $$i = \sqrt{-1}$$ but 'abuses' the properties of the square root in the real case by applying them to the complex case and that's where it goes wrong.

Last edited: Jan 31, 2006
15. Jan 31, 2006

### HallsofIvy

Actually, even defining i by "i2= -1" leads to problems since there are TWO complex numbers having that property. And since the complex numbers are not an ordered field, there is no immediate way to distinguish between them (you can't say "the positive number such that.. ). In the real numbers, we define $\sqrt{a}$ to be the positive number, x, such that x2= a. In the complex numbers we can't do that so, strictly speaking, [itex]\sqrt{a}[/tex] is not defined as a single number in the complex numbers.

A more rigorous definition of the complex numbers is a pairs of real numbers, (a,b) with addition defined by (a,b)+ (c,d)= (a+b,c+d) and multiplication defined by (a,b)*(c,d)= (ac- bd,bc+ad). We get immediately that (a,0)+ (c,0)= (a+c,0) and (a,0)*(c,0)= (ac,0) so the real numbers can be thought of as the complex numbers of the form (a,0). Obviously (0, 1)*(0, 1)= (-1, 0) and (0,-1)*(0,-1)= (-1, 0). We define i to be (0, 1).

16. Jan 31, 2006

### TheDestroyer

LOL, I Can't Read all this !! but the answer is very simple,

Just use eulers law in taking roots of a complex number,

A=r e^(i t)
A^(1/b)=(r)^(1/b) e^((i (t +2pi)/b)

It's simple right :P you'll find here 2 roots for 1, is 1 and -1...

17. Jan 31, 2006

### JasonRox

An elementary or high school mathematics teacher isn't expected to much outside of the school curriculum.

This is why when you read history books the author expresses in a surprised fashion that some great mathematicians spent their whole lives as a high school teacher. It's just highly unlikely a high school teacher knows anything beyond they are needed to know.

18. Feb 23, 2006

### heaven eye

In complex numbers field there are many preferences :-
1)brackets
2)exponents and roots
3)multiplication and division

In your Problem you multiplied before you deal with exponent

If two numbers has the same base and they have an exponent then in their multiplication we add their exponents for example:-
A^x *A^y = A^(x+y)

So :-
If i = Sqrt[-1]
Then i^2 = i * i = Sqrt[-1]* Sqrt[-1]=Sqrt[(-1)*(-1)] = Sqrt[((-1)^1)*((-1)^(1))] = Sqrt[(-1)^(1+1)]=(-1)^((1+1)/2)= (-1)^(2/2)= (-1)^(1)= -1

Your mistake was you multiplied two negative quantities in a root and that’s a mistake because first you should add the exponent in the bases was the same
So Sqrt[(-1)*(-1)]=/=Sqrt[1]

Because if a>0 , b>0 then :-
Sqrt[-a]*Sqrt[-b]=Sqrt[(-a)*(-b)] =/=Sqrt[a * b ]
=/= means does not equal

But Sqrt[(-a)*(-a)]= -a =/= a

Am I right????

My regards

19. Feb 23, 2006

### symplectic_manifold

I think in order to stress the pecularities of complex numbers and their special behaviour, which is in many ways different from that of real numbers, we should put in some geometry here.

Since C is a group with respect to multiplication, the following holds: 1*i=i. Now let's look at the complex plane and think what it means. It means that the vector (1,0) is the unit element and when we "multiply" it with the vector (0,1) we simply make a rotation (the actual group operation when we multiply complex numbers) 90° anticlockwise. So i^2=i*i=(0,1)*(0,1) means that we rotate the vector (0,1) 90° anticlockwise and so we get the vector (-1,0), that is i^2=-1.

20. Feb 25, 2006

### krab

I agree with Warren. Deathknight is in highschool. I remember having this sort of question at that stage as well. I also remember my math teacher being unable to answer. AFAIK, Deathknight has not asked this question more than once, so he deserves an explanation, or a polite re-direct to a FAQ. Simply locking a thread because i'm tired of the question is very self-serving.