Is sqrt(-1) x sqrt(-1) = -1 Algebraically Valid?

[TheAlkemist]

Question:

Is the following mathematical argument algebraically valid:

sqrt(-1) x sqrt(-1) = [sqrt(-1)]² = -1

I know sqrt(-1) is a complex number, I just want to know if the argument above is valid.

Thanks.

 

[jbunniii]

Yes, it's valid.

There is one subtlety: there are two numbers whose square is -1, namely $i$ and $-i$.

Your argument works fine if you choose sqrt(-1) = $i$:

$$\sqrt{-1} \times \sqrt{-1} = i \times i = i^2 = -1$$

and equally well if you choose sqrt(-1) = $-i$:

$$\sqrt{-1} \times \sqrt{-1} = (-i) \times (-i) = (-i)^2 = -1$$

But what happens if I choose sqrt(-1) = $i$ for one of the square roots, and sqrt(-1) = $-i$ for the other one? Then I get

$$\sqrt{-1} \times \sqrt{-1} = (i) \times (-i) = -(i^2) = 1$$

Uh oh!

 

[TheAlkemist]

Yes, it's valid.

There is one subtlety: there are two numbers whose square is -1, namely $i$ and $-i$.

Your argument works fine if you choose sqrt(-1) = $i$:

$$\sqrt{-1} \times \sqrt{-1} = i \times i = i^2 = -1$$

and equally well if you choose sqrt(-1) = $-i$:

$$\sqrt{-1} \times \sqrt{-1} = (-i) \times (-i) = (-i)^2 = -1$$

But what happens if I choose sqrt(-1) = $i$ for one of the square roots, and sqrt(-1) = $-i$ for the other one? Then I get

$$\sqrt{-1} \times \sqrt{-1} = (i) \times (-i) = -(i^2) = 1$$

Uh oh!

LOL @ "uh oh!"

Thanks. I actually just searched through the threads and see that this question has been asked several times in different ways. But it make sense now. Thanks again.

 

[robert Ihnot]

That so I was told, is the reason for i. Then the matter becomes $$\sqrt(-1)x\sqrt(-1) = i^2 = -1.$$

 

[Gerenuk]

The point is that you have to define sqrt() to be a function, i.e. return only one value. I think this is called principal value. Then your calculations works. You just have to remember that $sqrt(x^2)\neq x$ in general.

But I'm not completely sure about this Riemann surface business and what it's for :)