Where and when is $\sqrt{0}$ defined?

Question

This may appear a dumb question, but I was thinking about this in a deeper way.

If we are concerning about the set of real numbers, and hence we are in $\mathbb{R}^n$ (or $\mathbb{R}$ for simplicity [or even better in $\mathbb{R}^+$]), then $$\sqrt{0} = 0$$ and well defined.

Actually, this already start to have some flaw when we try to write the square root as: $$\sqrt{x} = \frac{x}{\sqrt{x}}$$ and this function is defined in $\mathbb{R}^+$ except at $x = 0$.

But besides strange ways to write the square root, it shall be "everything ok".

Now, let's pass into $\mathbb{C}$, and here we may read $$\sqrt{z} = \large e^{\frac{1}{2}\ln(z)}$$

Now, this is the fact: $\ln(0)$ is not defined at $0$. The logarithmic function is, in addition to that, a multivalued function.

But $\ln(0)$ is not even a multivalued function, and also $0$ is an essential pole for that function.

That infinity that arises in this way, is even worse, because it's a complex infinity, hence it means it can come from any direction and forming any direction in the complex plane.

This brings me to say that $\sqrt{z}$ is not defined at all in $\mathbb{C}$.

What I am now going to ask you is:

• Am I right, am I wrong?
• Did I miss something? May you add some other way to think?

Thank you in avance, I hope this isn't really a dumb question. It's, at least for me, not.

Answer 1

In $\Bbb R$, $\sqrt x$ is defined to be the non-negative number that, when squared, produces $x$. With this, $\sqrt 0=0$. The fact that the otherwise valid equality $\sqrt x=\frac x{\sqrt x}$ has one side undefined when $x=0$, should not bother you. In the same manner, we have $1+x=\frac{x^2-1}{x-1}$ whenever $x\ne 1$, but that does not suggest that $1+x$ has problems when $x=1$, does it?

In the complex case, $\sqrt z$ can be defined (i.e., a branch picked) nicely for all simply connected domains that do not include $0$. Even then, if $A$ and $B$ are two such domains, it may not be possible to do this in a way that agrees on $A\cap B$. Still, all this does not stand against the fact that $\sqrt 0=0$ (in fact, this is the only point where there is no doubt about the complex square root).

Answer 2

Just because two expressions are the same by algebraic manipulation, it doesn't mean that they're equal as functions.

For two functions being equal means that they must have the same domain in the first place. Therefore, $f(x)=\sqrt{x}$ and $g(x)=\frac{x}{\sqrt{x}}$ are different functions.

In the case of the complex plane, you can define $\sqrt{z}$ when $z$ is approaching $0$ on the real line by a limit process. It won't be problematic if you restrict its domain only to the real line. Again, defining $\sqrt{z}$ by exponentiation doesn't necessarily give the same function because the domains can be different. It depends on your way of defining it though.