fourier transform - What is the advantage of using exponential function over trigonometric function in analyzing waves?

A.P.French in his book Vibrations and Waves writes:

. . . Why should the exponential function be such an important contribution to the analysis of vibrations? The prime reason is the special property of the exponential function. . .its reappearance after every operation of differentiation or integration.

Now,what is the advantage of using exponential function over trigonometric function? They are directly linked by De-Moiver's theorem.

Answer

I'm taking a bit of a gamble here from your age on your user page: we do have a 15 year old string theorist on this site (who is also from your homeland), so at the risk of seeming belittling, here is something I found really satisfying in relation to your question when I was about your age.

Differentiation of trigonometric functions is fiddly. When you differentiate a $\sin$ it becomes a $\cos$, when you differentiate a $\cos$ it becomes a $-\sin$. So you have to differentiate twice to get a trigonometric function back to its original form. The second derivative $\mathrm{d}_t^2 f(\omega\,x)$ is equivalent to multiplying by $-\omega^2$ (where $f$ is any linear combination of $\sin\,\omega\,t$ and $\cos\,\omega\,t$), but the first derivative in general changes the function to something linearly independent from it.

In contrast, differentiation of $\exp(i\,\omega t)$ is equivalent to a simple scaling, which means it can greatly simplify operations containing derivatives of all orders, not just even ones, as is the case with $\sin$ or $\cos$.

This is all equivalent to saying that $\sin$ and $\cos$ fulfill second but not first order differential equations, whereas $e^{\pm i\,t}$, which are special linear combinations of $\sin$ and $\cos$ fulfill first order DEs. Here is a wonderful way of thinking that for me unified $e^{i\,t}$, $\sin t$, $\cos t$ and justified to me the complex number field when I was 17. We begin by thinking of the kinematics of something moving on a path around the unit circle. A position vector on this circle is $\vec{r} = \left(\begin{array}{c}x\\y\end{array}\right)$ such that $\left<\vec{r},\,\vec{r}\right>=x^2+y^2=1$. We now find the equation of motion of a point $\vec{r}(t) = \left(\begin{array}{c}x(t)\\y(t)\end{array}\right)$ constrained to the circle is defined by ${\rm d}_t \left<\vec{r},\,\vec{r}\right> = 0$, whence $\left<{\rm d}_t \vec{r}(t),\,\vec{r}(t)\right> = 0$, whence (with a little fiddling):

$${\rm d}_t \left(\begin{array}{c}x(t)\\y(t)\end{array}\right) = v(t) \left(\begin{array}{cc}0&-1\\1&0\end{array}\right) \left(\begin{array}{c}x(t)\\y(t)\end{array}\right)$$

where $v(t) = \sqrt{\dot{x}^2+\dot{y}^2}$. We take, for simplicity, $v(t) = 1$ so that we immediately have, by the universal convergence of the matrix exponential, when supposing that the the path begins at the point $x=1,y=0$ at time $t=0$:

$$\vec{r}(t) = \exp\left[\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\,t\right]\left(\begin{array}{c}1\\0\end{array}\right)=\left(\mathrm{id} + i\,t + \frac{i^2\,t}{2!} + \frac{i^3\,t}{3!} + \cdots\right)\left(\begin{array}{c}1\\0\end{array}\right)$$

where I have defined

$$i= \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$$

You can play around with this one and check that this $i$ has all the properties that the "everyday" $i$ has; in particular $i^2=-1$. Indeed, you can go a little further and prove that "numbers" of the form:

$$a \,\mathrm{id} + b\,i = \left(\begin{array}{cc}a&-b\\b&a\end{array}\right)$$

add, subtract, multiply and divide exactly like the everyday complex numbers. The field of matrices of the form above is isomorphic to the complex number field. Mathematically it is therefore indistinguishable from the complex number field. Then we separate real parts (multipliers of the identity matrix) and imaginary parts (multipliers of the $i$ matrix to define, in this field:

$$e^{i\,t} = \cos(t) + \sin{t}\,i$$

and this entity has the useful property that its derivative is a simple scale factor times the original function and we get:

$$\cos(t) = \mathrm{id} - \frac{t^2}{2!}\mathrm{id}+ \frac{t^4}{4!}\mathrm{id} + \cdots$$ $$\sin(t) = \mathrm{id}\, t - \frac{t^3}{3!}\mathrm{id}+ \frac{t^5}{5!}\mathrm{id} + \cdots$$

How do these match up with everyday $\sin$ and $\cos$? Well, work out the co-ordinates of the point beginning at position $(1,0)$ and you will see that they are indeed given by the same $\sin$ and $\cos$ as defined above.

You may also enjoy the lecture by Richard Feynman:

"Algebra"; Lecture 22, Volume 1, The Feynman Lectures on Physics