SpaceDude said that he would die if I write in somemore. Well, here is pay back for killing me too many time in magicka
The common confusion with sqrt is more about notation. sqrt(4) is strictly 2 but 4 have two sqrt -sqrt(4) and sqrt(4).
Sqrt(ab) doesn't equal sqrt(a)sqrt(b) when both number are negative
Short answer. NoOOO0oO00oOO0o000
Pseudosemiquasilong answer: 2 weeks ago...
"Oh a baddy in there, i is not the square root of -1 which is not defined, or else we may have big problems with our logarithms.
i is the principal square root of -1, or as some would call it the square root of -1.
Then -1 = i * i = sqrt(-1) * sqrt(-1) = sqrt((-1) * (-1)) = sqrt(1) = 1. Somewhere you have to cross out one of the equal signs, and it is usually done at the second one. So no, i is not the square root of -1, but i * i is -1."
Well, the issue comes from two different places. Some people say that defining i to be sqrt(−1) is bad since there are two complex numbers which are square roots of −1. So how do you pick which one to use? From an algebraic perspective, it doesn't matter, since the algebraic structures are the same, regardless of which one you pick. (So if you pick one of the square roots of −1 to call i and I pick the other, any algebraic manipulations we perform on them will yield the exact same notation - algebra can't tell the difference between them). But if you think of them as points on a plane, the one you choose to represent as i does matter a little bit. That being said, what I said about picking the "principal" square root (which people normally just call the square root) completely removes this issue. It specifies which of the two square roots you want to call i. I mean, this problem could also come up with sqrt(1). There are technically two square roots of 1, both 1 and −1, but people don't seem to have a problem with calling 1 the square root of 1, since we use the principal square root.
That being said, the place where most people would put a "not equals to" sign in that string of equations would not be at the second equality, but rather at the third equality. Let's explore why this is.
The heart of the problem stems from the fact that the square root operation is not a function. It is what's called a multi-valued function. It gives multiple outputs for each input, but if we keep track of an extra detail, then we can consider the square root operation to behave like a function.
So, every complex number can be described in polar form, and this polar form is actually really enlightening in terms of multiplication and exponentiation (and therefore in terms of taking roots and logarithms too). Every point in the complex plane can be described by its distance from the origin and angle it makes with the positive real axis. Of course, this angle is not unique. So a complex number with distance r from the origin and angle θ with the positive real axis can be described as (r, θ). If you have two complex numbers (r, θ) and (ρ, φ), then their product is (rρ, θ+φ). In other words when you multiply two complex numbers, you multiply their lengths and add their angles. You can also think of complex multiplication as a dynamic transformation in the complex plane. Namely, if you take (r, θ) and multiply by (ρ, φ), then it's like taking the vector (r, θ), scaling its length by a factor of ρ, and rotating it by an angle of φ. From this, it follows that the nth power of a complex number (r, θ) is (r^n, nθ). Of course, it seems reasonable then that the nth root of (r, θ) should be (r^(1/n), θ/n) (where r^(1/n) is the positive real nth root of r). But here's the issue: the result depends on which θ you start with.
Let's look at the square roots of 1. You can have all sorts of angles for the complex number 1. The standard choice is 0. But you could also choose 2π, or −2π, or any integer multiple of 2π. If you choose to start with the angle 0, then halving this angle again yields the angle 0. So with this choice of angle, thinking of 1 as (1, 0), we remain at 1 when we take the square root (since (1^(1/2), 0/2) = (1, 0)). But if we choose the angle 2π, then halving this angle, we get 2π/2 = π, and the square root is −1 (since (1^(1/2), 2π/2) = (1, π)). The square root depends on what our choice of angle is.
Now, there are two ways to make the square root multi-valued function actually behave like a function.
One: We could change the domain and range from instead of being complex numbers to being pairs of real numbers (r, θ) where r > 0 and θ is any real number, and have the square root function be defined as sqrt(r, θ) = (r^(1/2), θ/2). Of course, this isn't quite what we want since there is no longer a one-to-one correspondence with complex numbers: the complex number 1, for example, is associated with (1, 0), (1, 2π), (1, −2π), (1, 4π), (1, −4π), etc.
Two: We could fix an "argument" θ for each complex number and put it in polar form (r, θ), and then define the square root of the complex number to be the complex number with a distance of r^(1/2) from the origin and an angle of θ/2 from the origin.
Generally, we go with option 2, but option 1 is deeply connected to this, and we shouldn't forget it. In option 2, picking the "argument" for each complex number is called "choosing a branch" of the root function. The standard branch to pick is to have −π < θ ≤ π. With this standard (principal) branch of the square root function, −1 can be represented in polar form as (1, π), so sqrt(−1) is the complex number whose polar form is (1^(1/2), π/2) = (1/ π/2), which is i. Of course, if we chose our branch differently, such as −π ≤ θ < π, the sqrt(−1) could be different. In this other branch, −1 is represented in polar form as (1, −π), so then sqrt(−1) is the complex number whose has a polar form (1^(1/2), −π/2) = (1, −π/2), which is −i. But as I already mentioned, the standard (principal) way to define the square root function is to have argument in −π < θ ≤ π.
Something else to point out here: after choosing a branch for the square root function, we have a very important curve in the complex plane. For our standard choice of −π < θ ≤ π, the curve in the complex plane is the negative real axis. This curve is called the "branch cut." It's basically the set of points in the plane where the argument function is discontinuous since the argument jumps when you cross over the branch cut (just "above" the cut the argument is close to π, and just "below" the branch cut, the argument is close to −π).
You don't need to use so many letters, I know these arguments too, but you are wrong if you think most people would void so easily the rules to calculate with logarithms. You take quite an effort to try to discuss the complex plane, but really, I do know it. Still, it is not necessary to do so, and in my experience, only few people would the needlessly void the rules to calculate with logarithms, which you just do if you want to define i (or, for people from the signal processing field, j) as the square root of -1. And honestly speaking, they are usually engineers who just do not care.
Using very simple math, you can see the problem with your "equalities".
For added simplicity, let's say:
sqrt(1) * sqrt(1) = sqrt((1) * (1))
Is this true? No, since the left side of the equation can be -1 when the right side is 1 and vice versa, depending on which square root you choose in each case (positive or negative).
This doesn't magically change when you replace the 1's with -1's.
What I was interested in was you showing me a statement in the language of logarithms. And it's quite difficult to convert "sqrt(−1) · sqrt(−1) = sqrt(−1 · −1)" into the language of logarithms. There are some major problems with doing so. First, you would probably have to be using logs with bases −1 and 1, which are terrible bases to use for logarithms in the first place. And second, you're relying on a more general statement about positive bases which states that a^n · b^n = (ab)^n. Since this switches bases throughout the entire statement, it's very tricky to write this in terms of logarithms.
Of course, if you want to ban the use of non-integer exponents, then you won't have to worry about a^n · b^n no longer being equal to (ab)^n when we allow a and b to be complex numbers. And if you think that's a better approach for dealing with engineering and science students, then so be it. But that doesn't mean that it's wrong to say that i is the square root of −1. There is a standard meaning which makes complete sense. And I would hope that engineering students would be able to grasp the concept of some properties no longer holding when we expand our domain... and quite frankly, the ones who make it far along the program can. Engineering students, once they reach that level, have certainly demonstrated that they could handle things like that no longer holding.
No, the square root is a function and it is not the inverse of the square in the whole number plane. And yes, the formula above is correct. Yes, it is the usual argument. And no, the square root is not defined on negative values.
There is a consistent, well-understood way to define the square root of any complex number. It's true that some of the properties we enjoy for the principal square root function defined on the nonnegative real numbers no longer hold for the principal square root function defined on all complex numbers. Nevertheless, why those properties no longer hold and how to address those properties no longer holding is well-understood and standard material in complex analysis. Similar to my long explanation above, a multitude of references (such as Wikipedia, Math stack exchange, and any introductory textbook on complex analysis) explain this.
If you want to ignore these explanations and leave the square root function undefined on the complex plane for your own purposes, you can do that; however, you cannot by your own proclamation invalidate the well-understood, consistent way to define the principal square root of -1 that most mathematicians use.
You are free to view engineering students as intellectually lazy, but this does not invalidate consistent, well-understood mathematics.