How complex numbers cancel out in complex physics equations

Question

I think I have read somewhere that one of the cool things about complex numbers is that "they just cancel out in the end". Maybe I read this in some physics thing or perhaps a math thing, but I'm wondering what that looks like when they cancel out. So far to me it seems that complex numbers like you get in solutions to simple example polynomial equations aren't what you'd be dealing with in real-world stuff. Wondering generally what it looks like when they cancel out or just disappear at the end. It would be nice to see a practical example if it's not too much.

Answer 1

This can happen in several ways - indeed, in several senses:

• For any complex number $z$, $z$ and its complex conjugate $z^\ast$ have real sum and non-negative real product. This is relevant in the roots of polynomials with real coefficients, including the characteristic polynomials of $2\times 2$ matrices (whose eigenvalues have sum equal to its trace and product equal to its determinant).
• Complex inner products satisfy $\langle x|x\rangle\ge 0$, so naturally $\langle x+y|x+y\rangle-\langle x|x\rangle-\langle y|y\rangle$ is real. But it's $\langle x|y\rangle+\langle y|x\rangle$. Luckily, this is a sum of complex conjugates.
• Complex inner products are important in quantum mechanics because their square moduli are probabilities. These are of the form $zz^\ast$, so as aforesaid are non-negative.
• Quantum-mechanical observables have real eigenvalues. They're (typically infinite-dimensional) complex-valued matrices, but the complexification "cancels" in an unusual way: the identity $X_{ij}=X_{ji}^\ast$ ensures real eigenvalues.
• Raising an imaginary number to an even power makes it real. This arises in such physics as the Klein-Gordon equation.
• Cardano discovered how to solve cubic equations. Even when all roots are real, complex numbers are needed in the technique, but their imaginary parts cancel. The need for complex numbers is called the casus irreducibilis. This was the main historical impetus for taking complex numbers seriously (mathematicians didn't mind considering some quadratics insoluble, as opposed to having non-real roots). A more complicated version of this phenomenon is also found if we solve quartics. Cubics aren't that common in physics, but you'll find them in e.g. the climate of oceans.

Answer 2

One example is to deliberately put a reducible cubic into Cardano. We take polynomial $$ x^3 - 2x + 1 \; , $$ The main root that comes out of Cardano (the others come from multiplying things by cube roots of unity) is $$ \sqrt[3]{ \frac{1}{2} \left( -1 + i \sqrt {\frac{5}{27}} \right)} + \sqrt[3]{ \frac{1}{2} \left( -1 - i \sqrt {\frac{5}{27}} \right)} $$

If you write $\frac{1}{2} \left( -1 + i \sqrt {\frac{5}{27}} \right)$ in polar coordinates $r \approx 0.544331,$ $\theta \approx 156.716^\circ$ and use trigonometry to find the cube root, you wind up with the cube root (in the first quadrant) having $r \approx 0.8164965,$ $\theta \approx 52.2387^\circ.$ Put that back into rectangular, my calculator says we have $\frac{1}{2} + 0.645497 i \; . \;$ Add this complex number to its conjugate, and we get $1.$

As I said, I deliberately chose a reducible cubic. $$ x^2 - 2x+1 = (x-1)(x^2 + x - 1), $$ the roots are $$ 1, \frac{-1 \pm \sqrt 5}{2} $$

Things cancel.