What is a simple, physical situation where complex numbers emerge naturally? [duplicate]

Solution 1:

I don't know

a simple, physical situation where complex numbers emerge naturally

but I can suggest a way to help you

teach middle schoolers about the emergence of complex numbers and I want to motivate this organically.

I did this once as a guest lecturer in a middle school classroom by developing a geometric interpretation of arithmetic on the number line.

Adding a fixed number $r$ is a shift by $r$, to the right if $r > 0$, to the left if $r < 0$. Successive shifts add the shift amounts. Each geometric shift is characterized by the position that $0$ moves to. You illustrate this visually by physically shifting a yardstick along a number line drawn on the board.

The answer to the question "what do you shift by so that doing it twice shifts by $r$?" is clearly $r/2$.

This is looking ahead to square roots, but you don't say that yet. The underlying idea is that the group of shifts is the additive group of the real numbers, but you don't say that ever.

Now that addition is done you go on to multiplication. Multiplying by a fixed positive $r$ rescales the number line. If $r>1$ things stretch, if $r < 1$ they shrink and multiplying by $r=1$ changes nothing. To know what a scaling does all you need to know is the image of $1$.

Successive scalings multiply, just as successive shifts add. What should you do twice to scale by $9$? Half of $9$ doesn't work, but $3$ does. The class will quickly grasp that the geometric way to halve a scaling is to find the square root.

What about multiplication by a negative number? The geometry is clear: it's reflection over $0$ followed by a scaling by the absolute value. Again the transformation is characterized by the image of $1$.

Now you're ready for the denoument. What geometric transformation can you do twice to move $1$ to $-1$ on the number line? Take your yardstick, place it on the line on the board, rotate by a quarter of a circle so that it's vertical, then another quarter and you're there. The image of $1$ is not on the line. It's at position $(0,1)$ in the cartesian coordinate system middle schoolers know about. They will find it cool to think of that point as a new number such that multiplying by it twice turns $r$ into $-r$. Name that number "$i$".

If you have brought the class along this far the rest is easy. They will quickly see the $y$ axis as the real multiples of $i$. Clearly adding $i$ should be a vertical translation by one unit. Vector addition for complex numbers follows quickly. Ask for the square root of $i$ and they will rotate the yardstick $45$ degrees. If they know about isosceles right triangles they will know that the (actually a) square root of $i$ is $(\sqrt{2}/2)(1+i)$, which they can check formally with the distributive law (which they will not ask you to prove).

A caveat. I think this should be pure fun for the class. Make that clear, so if some don't follow they don't worry. I would not try to integrate it into whatever the standard curriculum calls for. It should probably not extend over multiple class periods. Save it for a day near the end of the school year.

Solution 2:

The historical origin of the complex numbers is, I think, the finest approach. Consider the problem of solving cubic equations of the type $x^3+px+q=0$. For this, you have Cardano's formula:$$x=\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}.$$But what do you do if it turns out that $\frac{q^2}{4}+\frac{p^3}{27}<0$? This happens, for instance, in the case of the equation $x^3-15x-4=0$; in this case we have $\frac{q^2}{4}+\frac{p^3}{27}=-121<0$. So, Cardano's formula tells us that a root of the equation is$$\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}.\tag1$$Could this mean that the equation has no solutions? No, since $4$ is clearly a solution. However, if we accept that we can work with square roots of negative numbers, then\begin{multline}\left(2+\sqrt{-1}\right)^3=2+11\sqrt{-1}=2+\sqrt{-121}\text{ and }\\\left(2-\sqrt{-1}\right)^3=2-11\sqrt{-1}=2-\sqrt{-121}.\end{multline}Therefore, it is natural to say that$$(1)=2+\sqrt{-1}+2-\sqrt{-1}=4.$$So, this shows that we can work with complex numbers in order to find real roots of cubic equations with real coefficients. And, in the XIX^th century, Pierre Wantzel proved that, if we wish to have an algebraic formula to do that, it is impossible to avoid complex numbers.