# 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?

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
• @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
• I never said that you measure complex numbers (see my other comment below). Neither do you measure the wave itself. What you measure is the intensity and the amplitude, which is derived as the modulus square and is perfectly defined for complex numbers (and a real quantity). – gented Sep 25 '15 at 19:41

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
• 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
• Because the combinations that you obtain considering complex solutions in addition to the real ones are more and different (and physically satisfied) from the ones you could obtain only taking real solutions into account. Again, see quantum mechanics as example. – gented Sep 25 '15 at 19:48
• @Brick I disagree. If you want to make work for yourself and try very hard, then yes, you can always keep real and imaginary parts of dynamical equations separate. Let's step back from QM for a moment and consider the state vector $(x, \,p)^T$ for a body undergoing (classical) simple harmonic motion. Its equation of motion is $\mathrm{d}_t (x, \,p)^T = i\, \mathrm{d}_t (x, \,p)^T$ where $i=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$. Now whether ultimately you explicitly use this or not, the object $i$ fulfilling $i^2+1=0$ and more generally the Lie algebra .... – Selene Routley Sep 26 '15 at 1:01
• @Brick ....$\mathfrak{so}(2)\cong\mathfrak{u}(1)$ of the group of unitary time evolution operators for these dynamics have a Platonic reality every bit as real as the "real" solutions. They exist as "irreducibly" (in the informal sense of needing each other equally, not in "irreducible"'s technical sense) entwined realities. In general, (almost) every Lie group of time evolution operators can be realized as real element matrices, but to ignore their parallel complex description is to ignore most of the structure in the description. In particular, the eigenvalues must in general be complex. – Selene Routley Sep 26 '15 at 1:05