# An example for a calculation where imaginary numbers are used but don't occur in the question or the solution.

In a presentation I will have to give an account of Hilbert's concept of real and ideal mathematics. Hilbert wrote in his treatise "Über das Unendliche" (page 14, second paragraph. Here is an English version - look for the paragraph starting with "Let us remember that we are mathematicians") that this concept can be compared with (some of) the use(s) of imaginary numbers.

He thought probably of a calculation where the setting and the final solution has nothing to do with imaginary numbers but that there is an easy proof using imaginary numbers. I remember once seeing such an example but cannot find one, so:

Does anyone know about a good an easily explicable example of this phenomenon?

("Easily" means that enigneers and biologists can also understand it well.)

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some real integrals are easier to solve using complex analysis – wim Nov 1 '12 at 3:22
I wouldn't call it simple, but the derivation of the spectral theorem uses imaginary numbers even though it's a theorem about real symmetric matrices. – Adam Wuerl Nov 3 '12 at 4:47

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– Dinesh Nov 6 '12 at 20:18

## 16 Answers

Once nice example is the sum

$$\cos x + \cos 2x + \cos 3x + \cdots + \cos nx$$

This can be worked out using trigonometric identities, but it turns out to be surprisingly simple with this neat trick:

$$\sum_{k=1}^n \cos(kx) = \sum_{k=1}^n\mathscr{Re}\{e^{ikx} \} = \mathscr{Re} \sum_{k=1}^n e^{ikx} = \mathscr{Re}\left\{ \frac{e^{i(n+1)x} - e^{ix}}{e^{ix} - 1} \right\}$$

because the sum turns into a geometric series. (computing the real part to get an answer in terms of trigonometric functions is not difficult, but is a little tedious)

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– Gone Oct 31 '12 at 17:00

The canonical example seems to be Cardano's solution of the cubic equation, which requires non-real numbers in some cases even when all the roots are real. The mathematics is not as hard as you might think; and as an added benefit, there is a juicy tale to go with it – as the solution was really due to Scipione del Ferro and Tartaglia.

Here is a writeup, based on some notes I made a year and a half ago:

First, the general cubic equation $x^3+ax^2+bx+c=0$ can be transformed into the form $$x^3-3px+2q=0$$ by a simple substitution of $x-a/3$ for $x$.

We may as well assume $pq\ne0$, since otherwise the equation is trivial to solve.

So we substitute in $$x=u+v$$ and get the equation into the form $$u^3+v^3+3(uv-p)(u+v)+q=0.$$ Now we add the extra equation $$uv=p$$ so that $u^3+v^3+q=0$. Substituting $v=p/u$ in this equation, then multiplying by $u^3$, we arrive at $$u^6+2qu^3+p^3=0,$$ which is a quadratic equation in $u^3$. Noticing that interchanging the two roots of this equation corresponds to interchanging $u$ and $v$, which does not change $x$, we pick one of the two solutions, and get: $$u^3=-q+\sqrt{q^2-p^3},$$ with the resulting solution $$x=u+p/u.$$ The three different cube roots $u$ will of course yield the three solutions $x$ of the original equation.

Real coefficients

In the case when $u^3$ is not real, that is when $q^2<p^3$, we could write instead $$u^3=-q+i\sqrt{p^3-q^2},$$ and we note that in this case $\lvert u\rvert=\sqrt{p}$, so that in fact $x=u+\bar u=2\operatorname{Re} u$. In other words, all the roots are real.

In fact the two extrema of $x^3-3px+2q$ are at $x=\pm\sqrt{p}$, and the values of the polynomial at these two points are $2(q\mp p^{3/2})$. The product of these two values is $4(q^2-p^3)<0$, which is another way to see that there are indeed three real zeros.

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After you substitute $x= u+v$ into the cubic equation, where did  $3u^2v +3uv^2$ go?  More generally, I think I'm just confused about the middle term $3(uv-p)$. – Jason DeVito Oct 31 '12 at 15:47
@JasonDeVito: $3u^2v+3uv^2=3uv(u+v)$ joined up with $-3p(u+v)$ to yield $3(uv-p)(u+v)=0$, since we posited the extra equation $uv=p$. The point being, by setting $x=u+v$ we have gained an extra degree of freedom, which is then used up in the extra equation $uv=p$, so that all these terms cancel. – Harald Hanche-Olsen Oct 31 '12 at 17:31
So, should there be a $(u+v)$ with the $3(uv-p)$ term in the line right after "and get the equation into the form"?  (Not that it matters, because after setting $uv=p$ that term disappears anyway.)  I'm just trying to check my understanding.  (I'm quite embarrassed to admit I've known about the cubic and quartic formulas (and nonexistence of higher degree formulas) for quite some time, but never actually went through the cubic or quartic derivations in any kind of detail.  I had chosen your post to finally do so, so I was just making sure I understood every step.) – Jason DeVito Oct 31 '12 at 17:42
@JasonDeVito: You're right. Now fixed. Thanks. – Harald Hanche-Olsen Oct 31 '12 at 18:12
Yes, this is the example you learn about in history of math where people realized that they should actually pay attention to these complex numbers. – Graphth Nov 6 '12 at 21:46

Well, you can consider this sequence of integers: $$u_0 = 1; u_1 = 1; u_{n+2} = u_{n+1} - u_n$$

This recurring definition is closely linked to the equation: $$x^2 = x - 1 \Leftrightarrow x^2 - x + 1 = 0$$ The solutions in $\mathbb{R}$ are $\frac{1 \pm i\sqrt3}{2}$ (complex numbers), and you can easily prove by recursion that: $$u_n = \left(\frac{1 + i\sqrt3}{2}\right)^n + \left(\frac{1 - i\sqrt3}{2}\right)^n$$

which is an... integer ! So yes, you can have complex numbers that ease calculations of totally non-complex problems.

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The sequence $u_n$ begins 1,1,0,1,-1,2,-3,5,-8,13,-21,34 and is the Fibonacci sequence multiplied by (-1)^n.  The quadratic equation and the formula with n-th powers of 6th roots of 1 is for the recurrence, $u_{n+2} = u_{n+1} - u_n$, which is periodic of order 6, and begins 1,1,0,-1,-1,0,1,1,0,-1,-1,0,1,1,0. – zyx Nov 7 '12 at 23:39

Primes congruent to one mod 4 are sums of two squares. You won't be able to explain the whole argument to an audience of non-mathematicians in a few minutes, but you would be able to show how imaginary numbers give a way of attacking this problem: e.g. you say that 5 is prime so doesn't have a non-trivial factorization in the integers, but you can show them that $(2-i)(2+i) = 5$ so that there is an interesting factorization in the Gaussian integers, and this leads to 5 being the sum of two squares.

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There is a large class of real definite integrals where the evaluation in closed form can be easily (or not so easily) done using the complex method of residues, but for which there is no known "real" method of evaluation.

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One example which will be easily understood (and perhaps appreciated) by many people is the fact that complex numbers allows the complete factorization of a real polynomial.

Suppose we have $p(x)=x^4 + 4$. It may not be immediately obvious how to factor this equation, but in the complex plane, it is easy to see that the equation $x^4 = -4$ has the solutions $$x_1 = 1+i,\ \ x_2 =-1+i,\ \ x_3 =-1-i,\ \ x_4 =1-i$$ This allows the factorization $$p(x) = \left[x - (1+i)\right]\left[x - (1-i)\right]\left[x - (-1+i)\right]\left[x - (-1-i)\right]$$ we can now recombine the conjugate root pairs into quadratic polynomials $$\left[x - (1+i)\right]\left[x - (1-i)\right] = x^2 - 2x + 2$$ $$\left[x - (-1+i)\right]\left[x - (-1-i)\right] = x^2 + 2x + 2$$ for the factorization $$p(x) = (x^2 - 2x + 2)(x^2 + 2x + 2)$$ Techniques like this for example, prove that every polynomial with real coefficients can be reduced into linear and quadratic factors or that every real polynomial has a partial fraction decomposition in it's familiar form.

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I think one of the most useful and simple is obtaining the formulas for $\sin(n\theta)$ and $\cos (n\theta)$ using the binomial theorem on de Moivre's formula:

$$(\cos\theta +i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$

Therefore expanding and identifying real and imaginary parts, one obtains:

$$\sin (n\theta)=\sum_{k=0}^n {n \choose k}\cos^k\theta\sin^{n-k}\theta\sin\left(\frac{n-k}{2}\pi\right)\\ \cos (n\theta)=\sum_{k=0}^n {n \choose k}\cos^k\theta\sin^{n-k}\theta\cos\left(\frac{n-k}{2}\pi\right)$$

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This example is maybe not elementary enough, but it is interesting: The usual proof of the central limit theorem, uses complex analysis, that is, imaginary numbers, via the characteristic functionn of a robabolity distribution. But complex numbers do not appear in the problem formulation.

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The radius of convergence for the power series of $f(x) = \frac{1}{1+e^x}$ about x=0 is $\pi$.

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And how do complex numbers play a role in this case? – Peter Tamaroff Oct 31 '12 at 22:10
@PeterTamaroff: I think it is the same example  as Itay Weiss's: the function has a pole at $x=i\pi$, and all the poles are of the form $x=(2k+1)i\pi$ – tomasz Oct 31 '12 at 22:12
@tomasz Oh! =) ${}{}$ – Peter Tamaroff Oct 31 '12 at 22:12

The fact that an odd degree polynomial $P$ with real coefficients must have at least one real root can be proved very quickly using complex numbers as follows. By the Fundamental Theorem of Algebra $P$ has, counting multiplicities, $n$ complex roots. It is easily verified algebraically that since the coefficients of $P$ are all real, if $z$ is a root then so is $\bar{z}$. Thus we conclude that any non-real solution brings a friend - its conjugate, which is also a root. So the non-real roots come in distinct pairs. The total number of roots is odd which means there must at at least one root with no friend. This root must thus be a real number, hence we found a real root.

Edit: There is also a more analytic proof, via the IVT, of the result (see comments) and there are also proofs of the fundamental theorem of algebra that use the result above. However, there is a very elementary proof of the fundamental theorem of algebra that certainly uses about as little as is required for IVT. It is basically the complex-analytic proof via the minimum modulus argument, but just that for a polynomial the proof of the minimal modulus argument can be given directly without using any sophisticated complex analysis. (see http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Complex-analytic_proofs)

Another example is to show that the radius of convergence of the Maclaurin series for $1/(1+x^2)$ does not exceed $1$. By considering the given function as a complex function it is seen to have a pole at $z=i$ and since the convergence set cannot contain a pole the argument is complete.

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The failure of Maclaurin (or Taylor) series due to non-real singularities was quite an intriguing thing to me back then... – Tobias Kienzler Oct 31 '12 at 13:18

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Your first example is not convincing: For me, every proof of the Fundamental Theorem of Arithmetic uses the fact that a real odd-degree polynomial has a real root. Your second example is a very good one. – GEdgar Oct 31 '12 at 14:03

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@GEdgar: every proof? That seems rather strong. But the standard proof for odd-degree real polynomials is way simpler than any proof of FTA, on that I agree. ($P(x)$ diverges to opposite infinities as $x\to\pm\infty$; now use the intermediate value theorem.) – Harald Hanche-Olsen Oct 31 '12 at 14:23
there is a very elementary proof of FTA that uses nothing but very straightforward properties of complex numbers. That proof is about as complicated as the more analytic proof using IVT. – Ittay Weiss Oct 31 '12 at 20:35

An easy enough example for engineers would be expressing vector rotation matrix as multiplying two complex numbers. $$\left[{x'\atop y'}\right] = \left[{x \cos a - y\sin a\atop y\cos a + x\sin a}\right]$$

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Er, is this using complex numbers? Where is $i$? Or is this just using the same matrix math that you would have used for complex multiplication? – Larry Gritz Nov 5 '12 at 20:05
R*e^iphi * e^ialpha means rotating a complex number x+iy with cos_a +i sin_a, where |x+iy|=R – Aki Suihkonen Nov 5 '12 at 20:09

Contour integration. Integrals of real-valued functions on the real line can sometimes be computed easily by transforming them to integrals over a closed path in the complex plane, using the residue theorem, and letting some part of the path go to some limit.

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When you solve différential equations with the characteristic polynomial and you get complex roots, you combine solutions to get real-valued function.

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You can use complex numbers to solve in $\mathbb{N}^4$ the following system of equations : $\left\{ \begin{array}{l} ac-bd=1 \\ bc+ab=2 \end{array} \right.$

Let $z_1=a+ib$ et $z_2=c+id$. Thus, the system is equivalent to $z_1z_2=1+2i$. So $(a^2+b^2)(c^2+d^2)=|z_1z_2|^2=5$, but $5$ is a prime so either $\left\{ \begin{array}{l} a^2+b^2=1 \\ c^2+d^2 =5 \end{array} \right.$ or $\left\{ \begin{array}{l} a^2+b^2=5 \\ c^2+d^2 =1 \end{array} \right.$. You deduce that $(|a|,|b|,|c|,|d|) \in \{ (0,1,1,2), (0,1,2,1),(1,0,2,1),(1,0,1,2) \}$. Finally, you find that the solutions are $(0, \pm 1, \pm 1, \mp 2)$, $(0, \pm 1,\pm 2,\mp 1)$, $(\pm 1,0, \pm 2, \pm 1)$ and $(\pm 1, 0, \pm 1, \pm 2)$.

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Suppose you have a mass vibrating about equilibrium, and it's subject to a frictional force that's proportional to the velocity. The equation of motion is of the form $d^2x/dt^2+bdx/dt+cx=0$, where $b$ and $c$ are real-valued constants. By far the easiest way to solve this is to take trial solutions of the form $x=e^{rt}$. Typically (if $b$ is not too big) we end up with two possibilities $r_1$ and $r_2$, which are complex conjugates of one another. However, you can match the initial conditions with a linear combination of the corresponding solutions $x_1$ and $x_2$, and the resulting motion is purely real.

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$\log(1+i*x)=\log {\sqrt(1+x^2)*\exp(i*\arctan(x))} = (1/2)*\log(1+x^2)+i*\arctan(x)... 2*\log(1+i*x)=\log(1+x^2)+i*2*\arctan(x)=\log{(1+i*x)^2} =\log{(1-x^2)+2*i*x} =\log(1+x^2)+i*\arctan{2*x/(1-x^2)}$

so

$2*\arctan(x) = \arctan(2*x/(1-x^2)...$ This procedure can be extended to $3*\arctan(x)$...to $n*\arctan(x)$

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You can use LaTeX-commands in the answers. It improves the readability. – AndreasS Nov 10 '12 at 16:05
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