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I assumed that since $a^c \cdot b^c = (ab)^{c}$, then something like $\sqrt{-4} \cdot \sqrt{-9}$ would be $\sqrt{-4 \cdot -9} = \sqrt{36} = \pm 6$ but according to Wolfram Alpha, it's $-6$?

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    $\begingroup$ This is because $a^cb^c=(ab)^c$ is for $a,b>0$. $\endgroup$ – i707107 Aug 11 '16 at 2:32
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    $\begingroup$ It's complicated.. Conventional notation $\sqrt {r}$ for positive real numbers $r$ denote the positive $1/2$-th power of $r$. $\endgroup$ – i707107 Aug 11 '16 at 2:36
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    $\begingroup$ Every positive real number has two square roots, but one of them is so much preferable to the other that we can get away with calling it the square root of the number. When we ask about square roots of negative numbers, there are still two of them, but it’s no longer the case that one is far preferable to the other. Put it this way: though the square root function on nonnegative reals is continuous, there is no square root function that’s continuous on the set of all complex numbers. Deal with it. $\endgroup$ – Lubin Aug 11 '16 at 4:38
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    $\begingroup$ @Lubin I don't agree with your statement- "Every positive real number has two square roots, but one of them is so much preferable to the other that we can get away with calling it the square root of the number." There is nothing like one of the square root is preferable. Its a common mistake to consider the positive square root and forget about the negative square root. $\endgroup$ – Babai Aug 11 '16 at 5:29
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    $\begingroup$ Possible duplicate of Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$? $\endgroup$ – GEdgar Aug 11 '16 at 20:32
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The property $a^c \cdot b^c = (ab)^{c}$ that you mention only holds for integer exponents and nonzero bases. Since $\sqrt{-4} = (-4)^{1/2}$, you cannot use this property here.

Instead, use imaginary numbers to evaluate your expression:

$$ \begin{align*} \sqrt{-4} \cdot \sqrt{-9} &= (2i)(3i) \\ &= 6i^2 \\ &= \boxed{-6} \end{align*} $$

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  • $\begingroup$ It doesn't hold for non-integer exponents? Why not? Even if the bases are integers? Seems to work when I test it on Wolfram Alpha using non-integer exponents $\endgroup$ – Aruka J Aug 11 '16 at 2:35
  • $\begingroup$ @ArukaJ Not all the time, of course. $\endgroup$ – DooplissForce Aug 11 '16 at 2:43
  • $\begingroup$ Those use negative numbers, I (meant to) refer to positive integers with my earlier comment $\endgroup$ – Aruka J Aug 11 '16 at 3:03
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    $\begingroup$ @ArukaJ It can hold for non-integer exponents provided that the bases are non-negative. However, if the bases are negative, then problems may arise, as your question illustrates. $\endgroup$ – wgrenard Aug 11 '16 at 4:01
  • $\begingroup$ Re your remark "not all the time" (meaning not for all arguments), so write the function$f(a,b;c):a^c\cdot b^c\mapsto (ab)^c$, which has a domain $\mathbb{R}^2;\mathbb{R}$ (or $\mathbb{R}^3$), more or less. So what's the subset of $\mathbb{R}^2;\mathbb{R}$ for which $f$ works? For example, I guess we have to remove the lower-left (negative,negative) quadrant of $\mathbb{R}^2$ -- unless the $c\in\mathbb{R}$ exponent happens to be integer. So a somewhat goofy-looking region of $\mathbb{R}^3$. Is there a simple way to characterize it? $\endgroup$ – John Forkosh Aug 11 '16 at 5:51
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For real numbers all numbers are either positive, zero, or negative. And the square of a negative number is positive.

Thus only zero and positive numbers have square roots and positive numbers have two square roots, one positive and one negative, but both equal in magnitude (i.e. absolute value).

NONE of this can be said about complex numbers.

As a result of these observations about real numbers we can make the following assumptions, none of which we can do for the complex:

When we write $\sqrt a$ then by definition $a \ge 0$; for $a >0$ there exist one $q > 0$ such that $q^2=a $ so $\sqrt {a} = \pm q $ and unless we specify in context we may as well arbitrarily define the $\sqrt {a}=q>0$.

And therefore $\sqrt {a}\sqrt {b}=\sqrt {ab} $. Even if we allow square roots to be negative this is true as products of positive and/or negative numbers are positive or negative.

For complex numbers we can not make these assumptions. But we can assume $|ab|=|a||b|$ and so $|\sqrt {a}||\sqrt {b}|=|\sqrt {ab}|$.

So $\sqrt {-4}\sqrt {-9} =\pm 2i * \pm 3i = 6i^2= -6$. But it doesn't equal $\sqrt {-4*-9}=\sqrt {36}=6 $.

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  • $\begingroup$ Unless you choose a particular branch cut, the expression is two-valued and, of course, the values are $6$ and $-6$. $\endgroup$ – egreg Aug 11 '16 at 20:35
  • $\begingroup$ Yeah, I was assuming we would pick a branch cut (do you mean either 2i, 3i or -2,-3i but not 2i, -3i or -2i, 3i?) but I didn't want to get bogged down in minutia. I was hoping no one would notice and I wouldn't have to explain. $\endgroup$ – fleablood Aug 11 '16 at 20:39
  • $\begingroup$ Somebody did. ;-) $\endgroup$ – egreg Aug 11 '16 at 20:43
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Things to understand:

  1. $\sqrt a \times \sqrt b =\sqrt ab$ is only valid when $a,b\geq 0$
  2. The most common mistake I noticed in this thread is like $\sqrt{25} = \pm 5$ or $\sqrt{-4}=\pm 2i$. T. Bongers has mentioned it in his comment as well.

The correct way to think is the equation $x^2=25$ has two solutions $x=\pm\sqrt{25}=\pm 5$ or the equation $t^2=-4$ has two solutions $t=\pm\sqrt{-4}=\pm 2i$.

But nevertheless, $\sqrt{25}$ is just $5$ not $\pm 5$ and $\sqrt{-4}= 2i$ not $\pm 2i$

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  • $\begingroup$ To typeset $\sqrt{ab}$, type \sqrt{ab} when you are in math mode. $\endgroup$ – N. F. Taussig Aug 11 '16 at 18:51

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