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1) $\sqrt{-4} = 2\sqrt{-1} = 2i$

2) $\sqrt{5+12i} = \pm (3+2i)$ where $i=\sqrt{-1}$

Why $\sqrt{-4}$ does not have $\pm 2i$ as its two solutions?

Squaring both $\pm 2i$ will lead us to $-4$. Just like in (2), all the complex numbers which give $-4$ on being squared must be a part of the answer. Then why $-2i$ is not taken as one solution?

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"The solution set to $x^2=-4$" is indeed $\pm 2i$, but $\sqrt{-4}$ is conventionally understood to be just $2i$. It is about the definition of $\sqrt{}$, nothing else. The fact that (2) does not choose a particular solution (presumably the one in the first quadrant) while (1) does seems like a flaw in your reference.

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  • $\begingroup$ "It is about the definition of √ , nothing else". Can you please explain this part? $\endgroup$
    – Arishta
    May 29, 2017 at 0:54
  • $\begingroup$ While $\sqrt{a}$ is a solution to $x^2=a$, when $a$ is a real, we view the square root as a single valued function (essentially because we can get away with it in an easy and consistent manner). However, if you try to make the square root function single valued over all the complex numbers, you run into problems with continuity (and you have to take a "branch cut"). $\endgroup$
    – Aaron
    May 29, 2017 at 1:10
  • $\begingroup$ @Cotton The square root of a positive real number $a$ is conventionally taken to be the positive solution to $x^2=a$, and for complex numbers $a$ or negative real numbers $a$, the square root of $a$ is taken to be the solution to $x^2=a$ with imaginary part of $x$ to be greater than $0$. $\endgroup$
    – Simply Beautiful Art
    May 29, 2017 at 1:38
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$\sqrt{x}$ is defined as the nonnegative square root of $x$ when $x\ge 0$. However, as $\mathbb{C}$ is not ordered, we cannot distinguish $i$ and $-i$ by signs. The imaginary unit is preferably defined as the number such that $i^2=-1$. In many situations, like solving polynomial equations with real coefficients, it is perfectly ok to replace $i$ by $-i$. So $\sqrt{-1}=i$ is just a convention, in order to keep the notation single-valued.

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  • $\begingroup$ There are two imaginary numbers $z$ such that $z^2=-1$. The crux of the definition of $i$ is that you choose one of those two number as $i$ (and, as you point out, everything looks the same with either choice). That doesn’t quite come across in your answer. $\endgroup$
    – amd
    May 29, 2017 at 1:24

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