Why for real number we have just one square root whereas for complex number we have two?

Intuitively, $\sqrt{4}=\pm 2$ but since $x\mapsto \sqrt{x}$ is defined from $\mathbb R^+\to \mathbb R^+$, we have $\sqrt{4}=2$ (if we defined $x\mapsto \sqrt x$ from $\mathbb R^+\to \mathbb R$, the this function wouldn't be well defined since for example $4$ would have $2$ image).

Why such argument doesn't work for complex number ? (because for complex number $z\mapsto \sqrt z$ is defined $\mathbb C\setminus \mathbb R^-\to \mathbb C$ For example, $$\sqrt i=\pm \left(\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}\right).$$

So the complex square root is not well defined ? If yes, why ?

Answer

There is somewhat of a convention on how to choose which complex number is $\sqrt{z}$. Namely, choose $\sqrt{z} = \exp\left[\frac{1}{2}\mathrm{Ln}(x)\right]$, where $\mathrm{Ln}(x)$ is the principal branch of the natural logarithm. This is equivalent to always choosing the root with positive real part, analogous to how with $\mathbb{R}$ the positive root is always chosen. Negative real numbers are then chosen to be the root with positive imaginary part, to preserve $\sqrt{-1} = i$. This is equivalent to writing $z$ as $r e^{i\theta}$, $r \in \mathbb R^+$ and $\theta \in (-\pi,\pi]$, then defining $\sqrt{z} = \sqrt{r}e^{i\theta/2}$. This is not without its downsides, of course. For example, it is no longer true in general that $\sqrt{1/z} = 1/\sqrt{z}$. But there were always going to be compromises when pinning down which root to use.

However, this convention is not as widespread as the universal convention that $\sqrt{x}$ is the positive square root of the positive real number $x$. This is because the positive square root has obvious precedence in $\mathbb R$, whereas in $\mathbb C$ the two are usually on more equal footing. The best solution is probably to just acknowledge that $\sqrt{z}$ is ambiguous in a way that $\sqrt{x}$ isn't, and to be sure to state what convention you're using for $\sqrt{z}$ if that's important in the problem.