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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?

user801923
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    You could ask yourself if taking for granted that $\sqrt{-2}$, for example, is $\mathrm i\sqrt2$ rather than $-\mathrm i\sqrt2$ is really a good idea. This might lead you to avoid altogether using the notation $\sqrt{x}$ when $x$ is something else than a nonnegative real number (a move I support wholeheartedly). – Did Feb 25 '12 at 07:56
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    There is no such thing as the square root function on the complex numbers. There is on the non-negative reals, but only by convention. – André Nicolas Feb 25 '12 at 07:56
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    see http://math.stackexchange.com/questions/49169/i2-why-is-it-1-when-you-can-show-it-is-1/49198 or http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake or ... – anon Feb 25 '12 at 07:57
  • nice question,please see this from wolframalpha http://www.wolframalpha.com/input/?i=sqrt%28-1%29*sqrt%28-1%29 – dato datuashvili Feb 25 '12 at 08:01
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    Can your human do complex analysis? – Simon Feb 25 '12 at 08:40
  • @Simon did you ask me? – dato datuashvili Feb 25 '12 at 08:46
  • @dato: My previous comment was just a poor attempt at humour - not addressed to anyone in particular. "Can you computer do complex analysis?" is a bit of a classic paper that discusses these things in the context of computer algebra. It also discusses the useful "unwinding number" approach to thinking about such problems. – Simon Feb 25 '12 at 08:56
  • @Simon i have one ask for you,please if you can upvote my answer for make it zero,because i have seen that mostly guys did not downvote it,just someone who did not understand my answer,thanks beforehand – dato datuashvili Feb 25 '12 at 09:08
  • @dato: I don't think anyone misunderstood your answer, but rather thought that it did not add anything to the discussion. You presented a simple calculation with no attempt at addressing the underlying issue of multivaluedness and branch-cut conventions. You also didn't address the 2nd part of the question (cf the $1=-1$ that Alex provided). Finally, upvoting to balance a downvote is not a good reason to vote. I don't think your answer was worth a downvote, but I also won't upvote it. – Simon Feb 25 '12 at 09:23
  • 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:) – dato datuashvili Feb 25 '12 at 09:28

3 Answers3

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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.

Alex Becker
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    I think that "$\sqrt{a^2}=a$ is not necessarily true." is not quite accurate. Probably, we should say "is not the only solution in general", but it is always true in general because it is one of the solutions. – NoChance Feb 25 '12 at 09:07
  • @EmmadKareem When I see the $\sqrt$ symbol used, it is used to denote a function, specifically the principle square root function. – Alex Becker Feb 25 '12 at 18:54
  • Thanks for the explanation. – NoChance Feb 25 '12 at 20:06
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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.

André Nicolas
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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.

Michael Hardy
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fretty
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