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If $\sqrt{64}$ is equal to $\pm{}8$, is $-64$ equal to $\pm{}8i$, or just $8i$?

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  • $\begingroup$ How do you define $\sqrt{\cdot}$? Do note that there is a difference between $\sqrt{a}$ and a number being a square root of $a$. Generally, $\sqrt{\cdot}$ is used to denoted the non-negative root function root, in which case $\sqrt{-64}$ meaningless. $\endgroup$ – Brian May 11 at 22:45
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$\sqrt{a}$, for real $a$, is almost always defined to be the positive solution of the equation $x^2 - a = 0$, so $\sqrt{64}$ is $8$ and not $\pm 8$. The reason the square root only takes on one value is because it is a function and so each element in the domain can be mapped to at most one element in the codomain.

As for your second question, $i$ is defined to as a number satisfying the equation $i^2 + 1 = 0$ and so we can say that $\sqrt{-1} = i$.

Following from this, we have $\sqrt{-64} = \sqrt{-1} \sqrt{64} = i\sqrt{64} = 8i$.

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  • $\begingroup$ Is $i$ "positive"? $\endgroup$ – Deepak May 11 at 22:47
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    $\begingroup$ This answer is slightly misleading since $i$ is not defined to be equal to $\sqrt{-1}$ (which is meaningless), but rather to be a number such that $i^2 = -1$. $\endgroup$ – Brian May 11 at 22:50
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    $\begingroup$ @Brian You’re right. I’ve tried to improve my answer. $\endgroup$ – 雨が好きな人 May 11 at 22:52
  • $\begingroup$ Actually, there are two numbers whose square is $-1$; neither of them is "positive" or "negative", but one of them is called $i$ (or $+i$) and the other is called $-i$. Every nonzero (real or complex) number has two square roots. Just as $x^2-64=(x-8)(x+8)$, also $x^2-(-64)=(x-8i)(x+8i)$. $\endgroup$ – bof May 11 at 23:27
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Well, in theory, the square root of any number should return both its negative and positive root. Meaning $\sqrt{x^2}=\pm x$. But if you think about the geometric meaning of the square root, it’s finding the side length which makes a square of area $x^2$. So some people think that the square root should only be a positive answer since length cannot be negative. In the case where we do want both the negative and positive root, here’s what you can do. $\sqrt{-16}=\sqrt{-1} \cdot \sqrt{16}$ $\sqrt{-1}=i$, $\sqrt{16}=\pm 4$, which gives us $i \cdot \pm 4=\pm 4i$

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