That sounds like a simple question but in my head it is not.
Here's the thing. For real numbers, it's been defined for us since elementary school that, for $a,n\in\mathbb{R}$, $a\geq 0$:
$$\sqrt[2]{a} = n\Leftrightarrow a = n^2.$$
We also learn that $\sqrt[2]{a}$ is a non-negative number. So, $\sqrt[2]{4} = 2$, not $\pm 2$ ($2$ and $-2$ are the square roots of $4$). That'd mean that $\sqrt[2]{a}$ doesn't denote the square roots of $a$, only the positive square root.
Then, when we introduce complex numbers, we learn that for a complex number $z = \rho\cdot(\cos\theta + i\sin \theta$), by De Moivre's formula:
$$\sqrt[n]{z} = \sqrt[n]{\rho}\left(\cos\frac{\varphi + 2k\pi}{n} + i\sin\frac{\varphi + 2k\pi}{n}\right)\text{, $k$ varying over the integer values from $0$ to $n − 1$.}$$
Now, here comes the problem to me.
Since $\mathbb{R} \subset \mathbb{C}$, every real number is also a complex number, right? In fact, real numbers are complex numbers with $\varphi = K\pi$, $K\in\mathbb{Z}$. But then, if for example I plug $\rho = 4$ with $\varphi = 0$, we get $z = 4$. Then, according to De Moivre's law (assuming $n = 2$):
\begin{align*} \sqrt[2]{z} &= \sqrt[2]{\rho}\left(\cos\frac{\varphi + 2k\pi}{2} + i\sin\frac{\varphi + 2k\pi} {2}\right)\\ \sqrt[2]{4} &= \sqrt[2]{4}\left(\cos\frac{0 + 2k\pi}{2} + i\sin\frac{0 + 2k\pi} {2}\right)\\ \sqrt[2]{4} &= 2\left(\cos k\pi + i\sin k\pi\right)\\ \end{align*}
Thus, for $k = 0$ we get $\sqrt[2]{4} = 2$ and for $k = 1$ we get $\sqrt[2]{4} = -2$.
But since $4$ is also a real number, shouldn't the definition I've gave at the beginning be also previewing the negative answer? Thing is, the way I see, it doesn't.
So, assuming we're dealing only with real numbers, $\sqrt[2]{4}$ has only one answer. But if we're dealing with complex numbers context, $\sqrt[2]{4}$ has two? So, $\sqrt[2]{x}$ symbol has an ambiguity only solved knowing the set we're dealing with? I'd really appreciate some light here. Thanks in advance!
