What is $\sqrt{-4}\sqrt{-9}$? 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 $-6$?
 A: Things to understand: 


*

*$\sqrt a \times \sqrt b =\sqrt ab$ is only valid when $a,b\geq 0$

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