I've been following a series on understanding about how imaginary numbers came to be, and in the series, it mentions that imaginary numbers mostly follows the algebra rules for real numbers, such as adding or multiplying by real numbers. However, it specifically mentions this "inconsistency(?)" about multiplying square roots of imaginary numbers do not follow the rule for multiplying square roots of real numbers, namely $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$. For example, why can't I do this: $\sqrt{-2} \times \sqrt{-3} = \sqrt{(-2)(-3)} = \sqrt{6}$, which I know is the wrong answer. In the past, I've just memorized to factor out the imaginary parts first, and that $i^2 = -1.$ However, can anyone show me an explanation for this, or tell me where I could learn more about it?
[Math] Why doesn’t multiplying square roots of imaginary numbers follow $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
complex numberssquare-numbers
Related Solutions
For 1: It's not only common, it's always the case: if you have a polynomial with real coefficients, then the "complex roots" come in what are called "conjugate pairs: if $a+bi$ is a root, with $a,b$ real numbers and $b\neq 0$, then $a-bi$ is also a root.
The explanation has to do with something called "complex conjugation". Complex conjugation is an operation on complex numbers that sends the complex number $a+bi$ to the complex number $a-bi$. This is denoted with a line over the number, that is, $\overline{z}$ denotes the complex conjugate of $z$: $$\overline{a+bi} = a-bi,\qquad a\text{ and }b\text{ real numbers.}$$ Conjugation respects sums and products: $$\overline{z+w} = \overline{z}+\overline{w}\text{ and }\overline{z\times w}=\overline{z}\times\overline{w}\text{ for all complex numbers }z\text{ and }w.$$
Also, a number $a+bi$ is real (that is, has $b=0$) if and only if $\overline{a+bi} = a+bi$.
Now suppose you have a polynomial with real coefficients, $$p(x) = \alpha_nx^n + \cdots + \alpha_1x + \alpha_0$$ and that there are real numbers $a$ and $b$ such that $a+bi$ is a root of this polynomial. Plugging it in, we get $0$: $$p(a+bi) = \alpha_n(a+bi)^n + \cdots + \alpha_1(a+bi) + \alpha_0 = 0.$$ Taking complex conjugates on both sides, and applying the properties mentioned above (complex conjugate of the sum is the sum of the complex conjugates; complex conjugate of the product is the product of the complex conjugate; complex conjugate of a real number like $\alpha_i$ is itself) we have: $$\begin{align*} 0 &= \overline{0}\\ &= \overline{\alpha_n(a+bi)^n + \cdots + \alpha_1(a+bi) + \alpha_0}\\ &= \overline{\alpha_n}\overline{(a+bi)}^n + \cdots + \overline{\alpha_1}\overline{(a+bi)} + \overline{\alpha_0}\\ &= \alpha_n(a-bi)^n + \cdots + \alpha_1(a-bi) + \alpha_0\\ &= p(a-bi). \end{align*}$$ So if $p(a+bi)=0$, then $p(a-bi)=0$ as well.
Complex roots don't correspond in any reasonable way to "maxima and minima" of the polynomials. They correspond to irreducible quadratic factors. It's hard to visualize, from a graph of the polynomial in the real axis, where (or even whether) it has complex roots.
The polynomial you tried, $3x^6+4x^4$, can be written as $$3x^6 + 4x^4 = x^4(3x^2 + 4).$$ The product is equal to $0$ if and only if one of the factors is $0$, so either $x=0$ or $3x^2+4 = 0$; for the latter, you would need $x^2 = -\frac{4}{3}$, which is impossible with real numbers; that is, $3x^2 + 4$ is an irreducible quadratic, which is where the complex roots (in this case, purely imaginary) come from. Since you want a (complex) number whose square is $-\frac{4}{3}$, one possibility is to take a real number whose square is $\frac{4}{3}$, namely $\frac{2}{\sqrt{3}}$, and then multiply it by a complex number whose square is $-1$. There are two such numbers, $i$ and $-i$, so the two complex roots are $\frac{2}{\sqrt{3}}i$ and $\frac{2}{\sqrt{3}}(-i) = -\frac{2}{\sqrt{3}}i$.
Each distinct irreducible quadratic factor will give you a pair of conjugate complex roots.
As a consequence of the Fundamental Theorem of Algebra, every polynomial with real coefficients can be written as a product of polynomials that are either degree $1$, or irreducible quadratics.
Might as well mention one application here: one reason people were interested in knowing that the Fundamental Theorem of Algebra was true (that every real polynomial could be written as a product of linear and irreducible quadratic polynomials) was to ensure that the method of partial fractions would always be available to solve an integral of a rational function (a function given as the quotient of a polynomial by another polynomial).
Other uses are too many to mention, but as Yuval mentions, this question and its answers may get you started.
Every complex and real number except $0$ have two square roots. If $r$ is one of the square roots $-r $ is the other.
So it is not true that "$\sqrt i $ has two values". It's that, "$i$ has two values as square roots. And it's not true that positive real numbers have one square root. They have two.
So what does the symbol $\sqrt {} $ and what does "the square root" mean? Well, nothing really. It's convenience to have a single value "square root function" so we arbitrarily chose that the positive value of square roots of positive real would be "the" square root. And the negative square root was the "other".
We could do the same for complex square roots. We could arbitrarily decide the one with a no negative real component was "the" square root and the one with a negative component would be the "other". But what would be the point?
The main reason we do this for the reals is because the real numbers is convenience really. And none of that convenience is useful in the complex numbers. The complex numbers don't have an greater/less than ordering. You are going to learn that exponentiation is cylic and there are multiple logarithms of each number (don't worry about that; you'll learn it later).
So basically we say $\sqrt z$ to mean the set of the two complex numbers, $w $ and $-w $, so that $w^2=(-w )^2=z$. Or we write $\sqrt z =\pm w$ to mean that the number to be consider a square root of $z$ could be either of $w$ or $-w $.
Best Answer
$\sqrt\cdot$ is well defined on $\Bbb R_+$ because $\Bbb R$ has a total order, which makes us able to make a sensible choice between the square roots of $b$ (ie $\sqrt b$ and $-\sqrt b$), that is, take the positive one. However in $\Bbb C$, you no longer can compare numbers ($3i \le 5$ makes no sense, for instance).
Therefore in your example $i\sqrt 2 \times i\sqrt 3 = -\sqrt 6$ is in fact a square root of $6$, but it is not the square root of $6$ as we like to call it in $\Bbb R$. In $\Bbb R$ since the product of two positive numbers is positive, we do not have this issue.