# Use of Imaginary Angles in Physics

I am studying higher algebra. I have learnt that as well as for reals, trigonometric functions can also be defined for complex numbers, by means of power series. Such as: $$\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+...(to \infty terms)$$ Here we are taking sine of a complex number $z$, which makes me suspect that there is something called 'imaginary angle'. But as far I know the Subject called 'Mathematics', it is the study of the tools required to describe physics. So, whenever we define a concept in Math, it is supposed to have a particular business with physics. So here, what is the purpose of a complex angle in physics?

• "So, whenever we define a concept in Math, it is supposed to have a particular business with physics." Nope. Commented Apr 8, 2015 at 15:55
• Related: physics.stackexchange.com/q/11396/2451 and links therein. Commented Apr 8, 2015 at 16:28

Hmm. Well, first of all, mathematicians do what mathematicians think is interesting. Often it applies to physics. Sometimes not. I don't want to give any examples of math that I don't think has a physical application though, since I expect someone would be able to find one. Anyway. So, we use complex numbers a lot in physics--it's actually very necessary for quantum mechanics. I'm having a harder time thinking of a particular time when we use a complex argument for the sine function, but here's a shot. Say I want to study the differential equation: $$y'' = -k^2 y$$ Which describes a simple harmonic oscillator, a particle subject to an attractive force like $F(y) = -m k^2 y$. If the boundary conditions are right, and let's say they are, the solution is something like: $$y \propto \cos k t$$ This makes sense--an oscillator has a sinusoidal solution. But what if I find this and then discover that my force wasn't attractive, it was repulsive, $F(y) = m \alpha^2 y$? Well, I can still use my old solution actually, but now I make $k = i \alpha$. Then this sort of "provides its own" minus sign. The cost is that now my frequency is imaginary. But this is fine, because as you may have learned (or soon will), $\cos i z = \cosh z$. Now it's a hyperbolic function, which grows exponentially--my particle doesn't oscillate, it just goes further and further away. By using complex numbers, I've managed to see that these two types of motion aren't actually so separate after all--they solve the same equation in different parts of the complex $k$-plane.

• How does imaginary frequency relate to imaginary angle? Commented Apr 9, 2015 at 9:27
• I couldn't think of a geometrical angle so I took "angle" to mean "argument to a trig function" Commented Apr 9, 2015 at 12:02

I'm not sure how relevant for your question this could be, but you might consider Wick rotations from real to imaginary times, that usually provide a formal way for passing from quantum to statistical mechanics.

For a time-independent Hamiltonian operator $H$, consider the time evolution operator $$U(t) = e^{itH},\qquad t\in\mathbb R.$$ If $\phi$ is any eigenvector of $H$ with eigenvalue $\omega$, the action of $U(t)$ on $\phi$ reduces to $$U(t)\phi = e^{it\omega}\phi.$$ If you now perform a Wick rotation $t\mapsto i\frac\hbar k\beta$ you get a Boltzmann factor $$e^{-\frac{\hbar\omega}k\beta},$$ where $\beta$ is simply the inverse of temperature, i.e. $\beta = 1/T$.

Using Eulers formula you can rewrite every exponential function like this:

$exp(i\varphi)=\cos(\varphi)+i\sin(\varphi)$

Exponential functions occur nearly everywhere in physics, sometimes even with imaginary arguments. A popular example is the tunnel effect or evanescence in general. Here the $\vec{k}$ vector in $exp(i\vec{k}\cdot\vec{r})$ becomes imaginary inside the barrier because $|\vec{k}|=\sqrt{2m(E-V)}$ with the energy of the particle $E$ and the potential $V$. So the argument of the wave-function becomes real and negative. You can easily rewrite this in cos and sin with imaginary arguments if you want.