Ok so this was the equation given in my text book $$\implies\sqrt{-a}\sqrt{-a} $$$$= (-1)a $$$$= -a $$

so my question is why can't i solve it this way $$\implies\sqrt{-a}\sqrt{-a}$$$$=\sqrt{(-a)(-a)}$$$$=\sqrt{a^2}$$$$=a$$

so what is wrong with my approach can anyone explain



  • $\begingroup$ See here as possible reference. $\endgroup$ – 2012ssohn Feb 17 '14 at 3:22
  • 2
    $\begingroup$ The question is: is $ \sqrt{a b} = \sqrt{a}\sqrt{b}$? $\endgroup$ – Mhenni Benghorbal Feb 17 '14 at 3:27
  • $\begingroup$ $\sqrt(ab)\neq \sqrt(a)\sqrt(b) $ for complex numbers. $\endgroup$ – user122283 Feb 17 '14 at 3:38
  • $\begingroup$ This should help math.stackexchange.com/questions/44406/… $\endgroup$ – ir7 Feb 17 '14 at 3:43
  • $\begingroup$ What's wrong is when you say that $\sqrt{a^2}=a$. For real numbers $a$, that equation is only true if $a\ge0$. Instead, this is what's true for any real number $a$: $\sqrt{a^2}=|a|$ $\endgroup$ – Steve Kass Feb 17 '14 at 3:58

When dealing with complex numbers, for integer values of n, $\sqrt[n]z$ is not a single number, but rather a set of n numbers, each of which has the property that its n-th power is z. For instance, $\sqrt1=\pm1$, $\sqrt[3]1=\left\{1,\dfrac{-1\pm i\sqrt3}2\right\}$, $\sqrt[4]1=\{\pm1,\pm i\}$, etc. In other words, for complex numbers, the n-th root is a binary relation rather than an actual function. Which is why the property that the n-th root of a product is the same as the product of n-th roots, ultimately no longer holds true anymore.

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