# What is the need of complex functions in wave analysis?

It is commonly known that waves can be express in terms of sine or cosine function.
But when I study further, I seen that for analyising the waves, it is common to use complex functions in the form
$$y=y_{_0}e^{i(kx-\omega{t})}$$
where $y_{_0}$ is the amplitude, $k$ is the wave number, $\omega$ is the angular velocity and $x$ & $t$ are position an time respectively. Ofcourse, I know that the function $e^{ix}$ can be written in the form $\cos{x}+i\sin{x}$ and so it is a periodic function with period $2\pi$ but my question is for what purpose we define it in terms of complex numbers? It seem to be more convenient to use real functions for real variables such as amplitude, electric and magnetic field of an electro magnetic wave, and also in quantum mechanics. What actually this interpretation means or what is the advantage of such functions?

## 2 Answers

The use of complex numbers is just a mathematical convenience. It makes calculation of derivatives especially easy, it has nice properties when you do Fourier transforms, etc. You're correct that you can do it all using real numbers, so that's not wrong. It's just - in most people's view - more cumbersome.

EDIT In light of the back and forth in the comments, let me provide more detail.

First, starting with classical mechanics: Let $f$ be a (potentially) complex solution to the wave equation. The physically relevant (i.e. measurable) quantity here is the amplitude as a function of space and time. Any complex function can be rewritten in terms of two real-valued functions $g$ and $h$ such that $$f = g + ih$$ The amplitude of $f$ is $\| f \| = (g^2 + h^2)^{(1/2)}$. We basically have two free functions here where we only need one to meet this constraint, so we're free to choose $h=0$, which means that $f$ is actually real-valued. You could choose some other values for $g$ and $h$ that have the same amplitude, but you don't need the complex part. (Note that I'm not dealing with plane wave solutions here, although you could build up your solution from them. I'm dealing with general solutions to the wave equation.)

For quantum mechanics, we have the Schroedinger equation: $$i\hbar \partial_t \Psi = -\frac{\hbar^2}{2m} \nabla^2 \Psi$$ (where I set $V=0$ because it's not going to figure in the rest of the point). This is typically written with complex numbers, as shown above, but this is again a short-hand only. We could instead write the solution in terms of two real-valued functions: $$\Psi = f + ig$$ and then, doing a little simplification, get two, coupled, real-valued PDEs: $$\hbar \partial_t f = -\frac{\hbar^2}{2m} \nabla^2 g$$ $$\hbar \partial_t g = +\frac{\hbar^2}{2m} \nabla^2 f$$ So, again, we can avoid complex numbers in the formulation. The price here is that we now have coupled PDEs for real functions instead of a single PDE over complex values. It turns out for practical reasons, that working with the single, complex-valued formulation is easier.

• That's just not why complex functions are introduced. They emerge as solutions of the equations the functions are subject to and you cannot throw them away (unless specific conditions hold). – gented Sep 25 '15 at 17:34
• @GennaroTedesco The boundary conditions and context in a real physics problem are going to require a real solution, not the most general solution. You could reformulate everything in terms of sine and cosine transforms instead of the complex Fourier transform. (In fact, I believe, that was Fourier's original formulation for what it's worth.) – Brick Sep 25 '15 at 17:50
• Solution in the field of the real numbers $\mathbb{R}$ does not mean that the solution is real, as for a real thing. I see no difference in the meaning between real and complex (unless you calculate modulus square, i. e. positive numbers). And no, you cannot, in general, reformulate everything in terms of sine and cosine. – gented Sep 25 '15 at 19:02
• @GennaroTedesco Of course you can reformulate everything in terms of sine and cosine. Just use $e^{ix} = \cos(x) + i \sin(x)$ and then write two coupled real valued equations: one for the real part and one for the imaginary part. – DanielSank Sep 25 '15 at 19:03
• I get the different uses of the word "real" here. At the same time, there is no way to take a physical measurement that observes a complex number. The complex numbers are useful mathematical solutions, but there's no physical theory that will accept a complex answer as physically meaningful for something you measure in a lab. If you solve in terms of complex-valued solutions, you always end up taking a combination of such solutions that results in a real-valued answer for observables. That is part of the theory and constrains the solution. See also: physics.stackexchange.com/q/11396/2451 – Brick Sep 25 '15 at 19:20

A function $f(x,t)$ is said to obey a wave equation if it holds that $$\Box f(x,t) = 0.$$ The most general solution of the above equation can be usually expressed in Fourier transform as $$f(x,t) = \int \textrm{d}k\,\textrm{d}\omega\,c(k,\omega)\,\textrm{e}^{-i(kx-\omega t)}$$ plus some boundary conditions. Except in special cases, the solution is usually a complex function.

• This answer begs the question. This is one way to write a general solution to the wave equation, but there's no a-priori reason to believe that the most general solutions have physical meaning. – Brick Sep 25 '15 at 17:47
• Likewise, there is no a priori reason to believe that the real solution has more physical meaning than a complex solution (real numbers does not mean real as in the language). – gented Sep 25 '15 at 19:04
• If you can find an accepted physical theory that says you can measure a complex number [e.g. (1+i) Volts], then I will yield. I'm sure that number is not on my voltmeter. Maybe some other device? I can measure amplitude and phase, but those are both real. The fact (on which we agree, I think) that it's mathematically convenient for calculation to combine them into a complex number does not change the fact that the measurable quantities are real. Since you know you need real numbers at the end, you could solve everything using real numbers from the start. – Brick Sep 25 '15 at 19:24
• I did not say that you measure complex numbers. What you measure are amplitudes, namely the modules square of the solution, which is the same whether the function is complex or real. If you exclude complex solution, you exclude a big variety of physically acceptable solutions for the observables to be measured (see quantum mechanics, for example). – gented Sep 25 '15 at 19:32
• Your last comment again begs the question: If you're going to take combinations (such as complex norms) that give you real numbers at the end, why not use real numbers all the way through? – Brick Sep 25 '15 at 19:45