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Why is $\sqrt{-2} \cdot \sqrt{-3} \neq \sqrt{6}$? Are there other examples where regular arithmetic goes wrong for complex numbers?

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    i have added comment at the end of my answer,but ok it does not matter ,for me important is to study something then manipulating marks:)2012-02-25

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You have to be careful with square roots, because for any $y$ we have $2$ values of $x$ satisfying $x^2=y$, while the square root function gives you only one of these. In your case, even though $(\sqrt{-2} \cdot \sqrt{-3})^2 = \sqrt{6}^2$, we have that $\sqrt{-2} \cdot \sqrt{-3} = -\sqrt{6}$, the other possible solution to $x^2=6$. So you must keep in mind that $\sqrt{a^2}=a$ is not necessarily true. A related erroneous assumption often made is $\sqrt{ab}=\sqrt{a}\sqrt{b}$. While $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true for some, it does not hold for all $a,b$. What if it did? This would make the following "proof" valid: $1=\sqrt{1^2}=\sqrt{(-1)^2}=\sqrt{-1}\sqrt{-1}=-1$ thus it clearly cannot be.

Edit: Note that many people do not even define $\sqrt{y}$ for $y<0$, for the reasons described in Didier's comment. In this post I let $\sqrt{y}$ be defined as the solution to $x^2=y$ with positive imaginary component, as seems to be done in the original post.

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    Thanks for the explanation.2012-02-25
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The argument of a complex number is only determined upto addition/subtraction of multiples of $2\pi$.

Now when multiplying complex numbers $z_1, z_2$ we can visualise the answer as having modulus $|z_1||z_2|$ and argument in the set $\{\text{arg}(z_1) + \text{arg}(z_2) + 2k\pi\mid k\in\mathbb{Z}\}$.

So taking the square root should essentially (positive) square root the modulus and half all possible arguments.

Let's see how this works with $\sqrt{-4}$. As a complex number $-4$ has modulus $4$ and possible arguments $\{\ldots, -\pi, \pi,\ldots\}$.

If we consider the argument to be $\pi$ then the square root should be $2i$ (since this is of modulus $2$ and principal argument $\frac{\pi}{2}$).

However, if we consider the argument to be $-\pi$ then the square root should be $-2i$ (since this is of modulus $2$ and principal argument $-\frac{\pi}{2}$).

All other possible arguments give one of these two complex numbers.

You see how we get a choice of square root, and it is not really favourable to choose one over the other. The complex numbers don't have a nice ordering like the real numbers do.

Compare this with square roots of positive real numbers. Considering the number as having argument $0$ we get the positive square root but considering the number as having argument $2\pi$ we get the negative square root. We usually choose the positive square root over the negative one but both are equally important.

So really something like $\sqrt{-2}\sqrt{-3}$ has $4$ different values depending on which square root you take for each term.

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Interestingly, the great Euler, in his Algebra, writes explicitly that $\sqrt{-2}\sqrt{-3}=\sqrt{6}$. (This is in Article $148$, p. $43$.) Earlier, he writes that $\sqrt{9}=3$, so he is not treating square root as a multiple-valued function. Your question reminded me that I saw this many years ago, and wondered why he did it. Very nice book, better than current stuff that does the basics of algebra.

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    From http://math.stackexchange.com/questions/49169/i2-why-is-it-1-when-you-can-show-it-is-1/49224#49224: It is said that even Euler got confused with $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$. Or did he? See [Euler's "mistake''? The radical product rule in historical perspective](https://webspace.utexas.edu/aam829/1/m/Euler_files/EulerMonthly.pdf) (Amer. Math. Monthly 114 (2007), no. 4, 273–285).2012-02-27