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It is for years that I teach complex numbers following a historical route. I start with the famous problem of Cardano:

Find two numbers whose sum is equal to 10 and whose product is equal to $40$.

After "solving" we try to check the "things" we have found actually satisfies the conditions of the problem. To do so, we need to multiply $\sqrt {-15}$ by $\sqrt {-15}$. "Relying on" and at the same time "extending" the rule of real numbers, we have two choices: take the product to be $-15$ or $15$. We choose the first choice since it gives the solutions of the original problem. I wonder whether there is a better pre-definition explanation of this choice.

Update Please consider that you all know the right answer! Go back in time for a moment and remember that the Great Euler took $\sqrt {-1}$ . $\sqrt {-4} = 2$. See for example here. And remember that I follow a historical route. And I have no $i $, no nothing yet. So later on, I use something like $\sqrt {-4} = \sqrt {4} \sqrt {-1} = 2\sqrt {-1}$. Now from a student's point of view, I guess everything seems like an ad hoc game: we use whatever we need whenever we want!

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    $\begingroup$ $\sqrt{x}$ is defined to be something, let's call it $y$, with the property $y^2 = x$. It is something incidental that when you choose the non-negative square roots of non-negative real numbers, you have $\sqrt{ab} = \sqrt{a}\sqrt{b}$. That "rule" holds only in very constrained situations, and one must take care not to extend it (wrongly) to the complex case. $\endgroup$ Commented Oct 21, 2013 at 20:51
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    $\begingroup$ If $\sqrt{-15}$ is to mean any single number at all, then "by defintion" multiplying that number by itself ought to give $-15$. If not, then there's no justification for notating it as a square root of $-15$. $\endgroup$ Commented Oct 21, 2013 at 20:52
  • $\begingroup$ Regarding complex numbers, there is no “square root function” as for real numbers. Period. $\endgroup$ Commented Oct 21, 2013 at 21:12
  • $\begingroup$ The choice is not random, nor according to whim, but guided by context. $\endgroup$
    – Lucian
    Commented Oct 21, 2013 at 23:23

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Consider the roots of $x^2=-1$. If its roots are presented as $\sqrt{-1}$ and $-\sqrt{-1}$ then we can say that $(\sqrt{-1})^2=-1$.


Look at here to see why the notation of $\sqrt{-1}$ is confusing.

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  • $\begingroup$ In particular, we can say that $(\sqrt {-1})^2=(-\sqrt{-1})^2=-1$. $\endgroup$
    – abiessu
    Commented Oct 21, 2013 at 20:54
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In general, when you "extend" the definition of something to a larger domain, as when real numbers are extended to the complex numbers, some of the rules in the original domain may not apply in the larger domain.

For example, two real numbers can always be compared, but the law of trichotomy (for any reals $x$ and $y$, exactly one of $x < y$, $x = y$, $x > y$ must be true) does not hold for complex numbers.

Similarly, the law in positive reals that $\sqrt{a} \sqrt{b} =\sqrt{ab} $ does not hold generally for complex numbers. It is an interesting exercise to determine for which complex numbers that equation holds.

Other examples:

Matrix multiplication is not commutative.

Quaternions, Hamilton's extension of complex numbers, are not commutative.

Octonions, the extension of quaternions, are not associative.

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In your derivation you do not come along the symbol $\sqrt{-15}$ for no reason (maybe you actually do, because you use some formula for the roots of a quadratic equation, but then you should have the students look at how that comes about), but because you need a numbers whose square is $-15$. That should answer the question what that number multiplied by itself yields ;)

If you are brave enough to investigate this with the students, you can use $\sqrt{-15}\cdot\sqrt{-15}=\sqrt{(-15)^2}=\sqrt{15^2}=15$ to raise the question if maybe it is not possible to consistently work with roots of negative numbers.

And by the way, my favorite way to derive a solution for $x+y=a$, $xy=b$ is $$ (x-y)^2=(x+y)^2-4xy=a^2-4b, $$ hence $x=\frac12((x+y)+(x-y))=\frac12\left(a+\sqrt{a^2-4b}\right)$, $y=\frac12\left(a-\sqrt{a^2-4b}\right)$. Here we are reminded that we need the square of $\sqrt{a^2-4b}$ to be $a^2-4b$. It is also nice, because one can discuss where the apparent asymmetry between $x$ and $y$ suddenly comes from. Of course, if there is a square root of $a^2-4b$, then there is another, namely its negative.

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Well, you have the fact that $i = \sqrt{-1}$, In this case then we have:

$$i = \sqrt{-1}$$ $$i^2 = -1$$ and since $i = \sqrt{-1}$, we end up having something like this:

$$(\sqrt{-1})^2 = -1$$ $$\sqrt{-1} * \sqrt{-1} = -1$$

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    $\begingroup$ Come on, $i\neq\sqrt{-1}$. $\endgroup$ Commented Oct 21, 2013 at 21:08
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    $\begingroup$ As long as your consistent, there's nothing wrong with saying $i = \sqrt{-1}$ and $-i = -\sqrt{-1}$... $\endgroup$
    – user71641
    Commented Oct 21, 2013 at 22:21
  • $\begingroup$ @MichaelHoppe why $i \neq \sqrt -1$? $\endgroup$
    – jimjim
    Commented Oct 21, 2013 at 22:35
  • $\begingroup$ @Arjang If you consider $\mathbb C$ as a $2$-dimensions $\endgroup$ Commented Oct 21, 2013 at 22:41
  • $\begingroup$ @MichaelHoppe do you mean to consider $\mathbb C$ as $\mathbb R^2$ with an algebraic field structure imposed on it? $\endgroup$
    – jimjim
    Commented Oct 21, 2013 at 23:25
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Because $\sqrt{-1}$ can refer to both $i$ and $-i$, whose product is indeed $1$.

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  • $\begingroup$ Yes, but it is a very sensible convention that if in an expression $\sqrt a$ occurs several times, all occurrences denote the same root. $\endgroup$
    – Carsten S
    Commented Oct 21, 2013 at 22:45
  • $\begingroup$ The foundation of that convention is undermined when certain fine details are overlooked (such as the fact that $\sqrt a\sqrt b$ isn't always the same as $\sqrt{ab}$), hence the inherent bewilderment and confusion. As you just pointed out, the convention is indeed very ‘sensible’ :-) $\endgroup$
    – Lucian
    Commented Oct 21, 2013 at 22:56
  • $\begingroup$ Sometimes, when we put things together, we can't always take them later apart in quite the same way as they were before being joined to one another. The same thing happens, for instance, in the case of collisions between subatomic particles, and it happens to hold for imaginary numbers as well. $\endgroup$
    – Lucian
    Commented Oct 21, 2013 at 23:11
  • $\begingroup$ In this case, after $-1$ and $-1$ are being mixed together inside the joined radical sign in order to form $1$, we must keep in mind that the radical of $1$ can be both $+1$ and $-1$, and the choice has to be made within reason, within the given context. $\endgroup$
    – Lucian
    Commented Oct 21, 2013 at 23:18

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