# What's the correct way of defining the use of square root symbol?

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!

• You might want to read this question and answer, and also this one Commented May 5, 2020 at 21:32
• The idea that $\sqrt{n^2}=|n|$ for reals is a custom, not a mathematical truth. Square root is intrinsically a two valued function. Commented May 5, 2020 at 21:36
• Hmmm that didn't help me fight this ambiguity (probably for lack of links for me). Commented May 5, 2020 at 22:07
• So, did anyone find a way of explaining this? Commented May 9, 2020 at 2:43

So, $$\sqrt4=\pm2$$, however, we like $$\sqrt4=2$$ better (mostly for the purpose of $$f(x)=\sqrt{x}$$ to be a function) hence, we refer to it as the Principal root of 4. After that little by little we drop the word Principle and that is why $$\sqrt4$$ is considered to be $$+2$$. If we enter in the WolframAlpha we can see both positive and Negative roots are the outcomes.