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In another question here, about roots of equations being imaginary, the accepted answer said something interesting about "imaginary" (as a technical word in math) not meaning "not real". I understand that imaginary numbers are "manipulable and usable to do all kinds of things". And thus, "they exist" at least, in the realm of mathematics.

So my question is this. I was playing around in wxMaxima, and solving polynomials, to get their roots. In maxima, $\sqrt{-1}$ is represented as %i.
Editor's note: all "%i"s have been replaced with $i$.

Here's an example of a polynomial, say $3x^6 + 4x^4$:

solve($3x^6 + 4x^4$,$x$);

Maxima's results: [$x=\displaystyle -\frac{2i}{\sqrt{3}}$,$x=\displaystyle \frac{2i}{\sqrt{3}}$,$x=0$]

My exact question:

  1. Is it common to have the same imaginary result twice with only the sign different (both q and -q are imaginary roots)? (At first I thought it had reported the same root twice as I overlooked the negative sign.)

  2. One can easily plot a real root and see how it intercepts the x axis. How does one "visualize" or think of an imaginary root -- does it correspond to a local minimum that does not occur at the x axis?

  3. What uses do imaginary roots of polynomials have in math?

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For your first question, notice that it doesn't give you the same root twice but in fact the complex conjugate. It is a theorem that if a polynomial has an imaginary root then it also has its complex conjugate as a root: in other words, they come in pairs. To be clearer, we can write imaginary numbers as $a + bi$ where $i=\sqrt{-1}$. The conjugate of $a+bi$ is simply $a - bi$. So if $a+bi$ is a root, so is $a-bi$. In your case, there is no real part, so you simply have $\frac{2}{\sqrt{3}}i$ and its conjugate $-\frac{2}{\sqrt{3}}i$. – james May 22 '11 at 2:41
up vote 19 down vote accepted

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.

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I really liked the "class of maps from the plane to itself" answer in the linked question. That made my brain do an interesting "rethink" about imaginary numbers. – Warren P May 22 '11 at 3:06
@Warren: I hope that you up-voted that answer, then! – Arturo Magidin May 22 '11 at 3:07
Would it be fair to say that the fundamental reason why conjugation works this way is because $i$ and $-i$ are simply indistinguishable? In the sense that there are clearly two square roots of $-1$, but there is no a priori reason to pick one over the other, so if I called one of them $i$ and the other $-i$ and you did it the opposite way, it would not make any difference at all to the mathematics? Or have I got the implication the wrong way around in my first sentence --- is it only once we have proved conjugation works this way that we can say switching $i$ and $-i$ makes no difference? – Rahul May 22 '11 at 6:19
Playing with wxMaxima has made me want to go back and re-learn all the cool algebra that I've forgotten since college, which was, well, a few years ago. It's a shame that when math is so cool, so many kids just learn how to hate it in school. I loved it, even when I had a mediocre teacher, but 20 years of not using stuff, you lose it. – Warren P May 22 '11 at 20:15
@Warren: Alas, it takes a lot less than 20 years. – Arturo Magidin May 22 '11 at 21:08

You might want to take a look at this as well: Do complex numbers really exist?.

Additionally, you can feel proud of yourself for inquiring along the same lines (with the aid of a computer) that Cardano & other mathematicians did 500 years ago, which led to the acceptance and elaboration of %i. Your observation about $\pm i$ is perspicacious and important in the theory of differential equations (dynamical systems), as well as other places. complex plane from mathworld

complex plane from a tutorial

We think about $\sqrt{-1}$ as part of a larger entity called the complex plane (google is your friend). $\sqrt{-1}$ is 90 degrees around the complex plane, whereas $-1$ is 180 degrees around the complex plane -- a fully turned-around $+1$. The 180-degree turn is the number line you learned in school; going only 90 degrees enters the complex plane.

The unifying idea is that a pair of imaginary and "real" numbers $x + i \cdot y$ are equivalent to a radius and an angle. Simple enough if you're familiar with rectangular and polar coordinates. But the key observation by de Moivre which unlocked a deep and mysterious world is that this can all be done with the natural base $e$.

  • $\exp (i \cdot x) = \cos x + i \sin x$.

Therefore $x+\sqrt{-1} \cdot y = \sqrt{x^2 + y^2} \cdot \exp (\sqrt{-1} \cdot \arctan {y \over x})$, which we think of as

  • $\mathrm{radius} \cdot e^{i \cdot \mathrm{angle}}$.

Here are some concrete examples of writing %i numbers in polar ($\exp$) form:

  • $3 + 4i = 5 \cdot e^{\pi \over 4}$
  • $-1 = e^{ \sqrt{-1} \cdot \pi }$
  • $\sqrt{-1} = e^{\sqrt{-1} \cdot {\pi \over 2}}$
  • $\sqrt{-25} = 5 \cdot e^{\sqrt{-1} \cdot {\pi \over 2}}$

As far as uses within mathematics, %i:

  • makes any polynomial equation soluble
  • matrices with %i model certain groups (google hermitian matrix)
  • complex eigenvalues tell you about the kind of solution to a (system of) differential equation(s)
  • helps model fluid flow
  • %i reveals more about the derivative (google amplitwist or Cauchy-Riemann equations)
  • %i reveals a fundamental to unity to the Moebius transformations (watch this)
  • you need %i for Fourier decomposition
  • the "angle" allows us to model the phase of a signal or quantum state as well as its magnitude
  • some previously non-integrable functions become integrable when you make use of the complex plane (google Laurent series or calculus of residues)
  • twistors and spinors in mathematical physics use %i
  • you can factor any polynomial all the way down using %i
  • the characteristic function of a probability distribution uses %i

If you've got the time, look at Tristan Needham's Visual Complex Analysis or Roger Penrose's The Road to Reality. Two of the best maths books that have been written, period. And they both show really cool things you can do with %i.

You're asking all the right questions to begin a magnificent journey of discovery.

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+1 Great information. – Warren P May 25 '11 at 0:33

I will answer the first question, and perhaps the others if no one else does. It is not a double root, there is a minus sign in front of one of the answers. The fact that the exponents are all even here forces any root to be accompanied by its negative.

The root $x=0$ is a quadruple root, but Maxima seems to mention it only once.

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Silly me. I overlooked the negative sign beside the fraction-line. When I put it in plain text here, I finally noticed the minus. I'm going to have to pay more careful attention in the future! – Warren P May 22 '11 at 2:59
@Andere Nicolas : Maxima has a function : multiplicities which gives a list of the multiplicities of the individual solutions returned by solve, so (%i1) solve(3*x^6+4*x^4,x); (%o1) [x=−(2*%i)/sqrt(3),x=(2*%i)/sqrt(3),x=0] (%i2) multiplicities; (%o2) [1,1,4] – Adam Jan 1 '14 at 12:55

See this question for parts 2 and 3.

As for part 1, if you have a polynomial with real coefficients then for each complex root $x + iy$, its conjugate $x - iy$ is also a root.

The reason is that conjugation "commutes" with all the arithmetic operation, i.e. $$ \overline{z+w} = \overline{z} + \overline{w}, \quad \overline{zw} = \overline{z} \overline{w}. $$ Here the line refers to conjugation. Since all coefficients are real, if you plug in your root and conjugate the entire expression, you end up with $\overline{0} = 0$ again, with the same polynomial, but with the conjugate root. So these roots come in pairs.

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For Q2, You can imagine the absolute value of a polynomial in one complex variable, for a given point $z=x+iy$, as the height of a surface.

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