Fourier Series on the Complex Plane

Let us now show that the usual formulas giving the Fourier coefficients, in terms of integrals involving the corresponding real functions, follow as consequences of the analytic properties of certain complex functions within the open unit disk. In order to do this, let us consider a pair of FC trigonometric series that have the rather weak property that there is at least one value of $\theta$ for which both elements of the pair are convergent. Note that this constitutes an indirect restriction on the coefficients of the series. It follows at once that the power series $S_{z}$ converges at the point on the unit circle that corresponds to that value of $\theta$. Consequently, it follows from the basic convergence theorem that the power series is strongly convergent at least on the open unit disk. Furthermore, it converges to a complex function that is analytic at least on the open unit disk, which we will denote by $w(z)$, and therefore we may now write

$$w(z) = \sum_{k=1}^{\infty} a_{k}z^{k}.$$

Note that, since the coefficients are real, the function $w(z)$ reduces to a purely real function on the open interval $(-1,1)$ of the real axis. It is therefore the analytic continuation of a real analytic function defined on that interval. Apart from this fact, from the fact that $w(0)=0$, and from the fact that it is analytic on the open unit disk, it is an otherwise arbitrary analytic function. In addition to all this, we have that $S_{z}$ is the Taylor series of $w(z)$ around $z=0$. We will call an analytic function that has these properties an inner analytic function. Let us list the defining properties. An inner analytic function is one that:

• is analytic at least on the open unit disk;

• is the analytic continuation of a real function defined in $(-1,1)$;

• assumes the value zero at $z=0$.

Let us now examine another property of $w(z)$ implied by the fact that it is the analytic continuation of a real function. If we use polar coordinates and write $z=\rho\exp(\mbox{\boldmath$\imath$}\theta)$, with $\theta\in[-\pi,\pi]$, then we may write out the Taylor series of $w(z)$ as

$$w(z) = \sum_{k=1}^{\infty} a_{k}\rho^{k} \left[ \cos(k\theta) + \mbox{\boldmath$\imath$} \sin(k\theta) \right].$$

Since the coefficients are real, we have at once that

$$w(z) = \left[ \, \sum_{k=1}^{\infty} a_{k}\rho^{k} \c... ..., \sum_{k=1}^{\infty} a_{k}\rho^{k} \sin(k\theta) \right],$$

where the expressions within square brackets are real. If we write $w(z)$ in terms of its real and imaginary parts,

$$w(z) = f_{\rm c}(\rho,\theta) + \mbox{\boldmath$\imath$} f_{\rm s}(\rho,\theta),$$

then the real part $f_{\rm c}(\rho,\theta)$ must be even on $\theta$, because it is the function that the cosine series contained in $S_{z}$ converges to,

$$f_{\rm c}(\rho,\theta) = \sum_{k=1}^{\infty} a_{k}\rho^{k} \cos(k\theta).$$

Similarly, the imaginary part $f_{\rm s}(\rho,\theta)$ must be odd on $\theta$, because it is the function that the sine series contained in $S_{z}$ converges to,

$$f_{\rm s}(\rho,\theta) = \sum_{k=1}^{\infty} a_{k}\rho^{k} \sin(k\theta).$$

With these preliminaries established, we may now proceed towards our objective here, which consists of the inversion of the relations above, so that we may write $a_{k}$ in terms of $f_{\rm c}(\rho,\theta)$, or in terms of $f_{\rm s}(\rho,\theta)$, by means of the use of the analytic structure within the open unit disk. Consider then the Cauchy integral formulas for the function $w(z)$ and its derivatives, written around $z=0$ for the $k^{\rm th}$ derivative,

$$w^{k\prime}(0) = \frac{k!}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz\, \frac{w(z)}{z^{k+1}},$$

where $C$ is the circle centered at $z=0$ with radius $\rho\in(0,1)$. The coefficients of the Taylor series of $w(z)$ may be written in terms of these integrals, so that we have for $a_{k}$

$$\begin{eqnarray*} a_{k} & = & \frac{w^{k\prime}(0)}{k!} \\ & = & \frac{1}... ...i\mbox{\boldmath$\imath$}} \oint_{C}dz\, \frac{w(z)}{z^{k+1}}. \end{eqnarray*}$$

It is very important to note that since $w(z)$ is analytic in the open unit disk, by the Cauchy-Goursat theorem the integral is independent of $\rho$ within that disk, and therefore so are the coefficients $a_{k}$. We now write the integral explicitly, using the integration variable $\theta$ on the circle of radius $\rho$. We have $dz=\mbox{\boldmath$\imath$}\rho\exp(\mbox{\boldmath$\imath$}\theta)d\theta$, and therefore get

$$\begin{eqnarray*} a_{k} & = & \frac{1}{2\pi\mbox{\boldmath$\imath$}} \int_{-... ...ho,\theta) \sin(k\theta) \right] \rule{0em}{2.5ex} \right\}. \end{eqnarray*}$$

Since we know that $a_{k}$ are real, we may at once conclude that the imaginary part of this last integral is zero. But we can state more than just that, because all the functions appearing in all these integrals have definite parities on $\theta$, and hence we see that the integrands that appear in the imaginary part are odd, while the integrals are over symmetric intervals. We therefore conclude that the following two integrals are separately zero,

$$\begin{eqnarray*} \int_{-\pi}^{\pi}d\theta\, f_{\rm s}(\rho,\theta) \cos(k\th... ...{\pi}d\theta\, f_{\rm c}(\rho,\theta) \sin(k\theta) & = & 0, \end{eqnarray*}$$

for all $k$. We are therefore left with the following expression for $a_{k}$,

$$a_{k} = \frac{\rho^{-k}}{2\pi} \int_{-\pi}^{\pi}d\theta\... ...s(k\theta) + f_{\rm s}(\rho,\theta) \sin(k\theta) \right].$$ (2)

In order to continue the analysis of the coefficients $a_{k}$ we consider now the following integral on the same circuit $C$,

$$\oint_{C}dz\, w(z)z^{k-1} = 0,$$

with $k>0$. The integral is zero by the Cauchy-Goursat theorem, since for $k>0$ the integrand is analytic on the open unit disk. As before we write the integral on the circle of radius $\rho$ using the integration variable $\theta$, to get

$$\begin{eqnarray*} 0 & = & \int_{-\pi}^{\pi}d\theta\, \mbox{\boldmath$\imath$... ...ho,\theta) \sin(k\theta) \right] \rule{0em}{2.5ex} \right\}. \end{eqnarray*}$$

Once again the integrals that appear in the imaginary part of this last expression are zero by parity arguments, and since $\rho\neq 0$ we are left with

$$\int_{-\pi}^{\pi}d\theta\, \left[ f_{\rm c}(\rho,\theta) ... ...eta) - f_{\rm s}(\rho,\theta) \sin(k\theta) \right] = 0,$$

which is valid for all $k>0$. We conclude therefore that the two integrals shown are equal,

$$\int_{-\pi}^{\pi}d\theta\, f_{\rm c}(\rho,\theta) \cos(k\... ..._{-\pi}^{\pi}d\theta\, f_{\rm s}(\rho,\theta) \sin(k\theta),$$

for all $k>0$. If we now go back to the expression in Equation (2) for $a_{k}$ we see that the two integrals appearing in that expression are equal to each other. We may therefore write for the coefficients

$$\begin{eqnarray*} a_{k} & = & \frac{\rho^{-k}}{\pi} \int_{-\pi}^{\pi}d\theta... ...nt_{-\pi}^{\pi}d\theta\, f_{\rm s}(\rho,\theta) \sin(k\theta). \end{eqnarray*}$$

We observe now that these formulas for the coefficients $a_{k}$ are simple extensions of the usual formulas for the Fourier coefficients of the even function $f_{\rm c}(\rho,\theta)$ and the odd function $f_{\rm s}(\rho,\theta)$, and therefore are related in a simple way to the Fourier coefficients for the real function of $\theta$

$$f(\rho,\theta) = f_{\rm c}(\rho,\theta) + f_{\rm s}(\rho,\theta),$$

with $\rho$ interpreted as an extra parameter. In fact, these formulas become the usual ones in the $\rho\to 1$ limit, thus completing the construction of a pair of FC Fourier series on the unit circle.

Whether or not we may now take the limit $\rho\to 1$ in these formulas depends on whether or not the coefficients, and hence the integrals that define them, are continuous functions of $\rho$ at the unit circle, for limits coming from within the unit disk. We saw before that the coefficients are constant with $\rho$, and therefore are continuous functions of $\rho$ within the open unit disk. We therefore know that their $\rho\to 1$ limits exist. Furthermore, by construction these are the coefficients of the FC pair of DP trigonometric series we started with, on the unit circle. Therefore the coefficients assume at $\rho=1$ the values given by their limits when $\rho\to 1$.

Consequently, the coefficients $a_{k}$ and the expressions giving them within the open unit disk are continuous from within at the unit circle, as functions of $\rho$, and so are the integrals appearing in those expressions. We may now take the $\rho\to 1$ limit and therefore get the usual formulas for the Fourier coefficients,

$$\begin{eqnarray*} a_{k} & = & \alpha_{k} \\ & = & \beta_{k}, \\ \alpha... ... \int_{-\pi}^{\pi}d\theta\, f_{\rm s}(1,\theta) \sin(k\theta), \end{eqnarray*}$$

where

$$\begin{eqnarray*} f_{\rm c}(1,\theta) & = & \lim_{\rho\to 1} f_{\rm c}(\rho,... ...m s}(1,\theta) & = & \lim_{\rho\to 1} f_{\rm s}(\rho,\theta). \end{eqnarray*}$$

We see therefore that the two trigonometric series of the pair of FC series we started with, under the very weak hypothesis that they both converge at one common point, are in fact the DP Fourier series of the DP functions $f_{\rm c}(1,\theta)$ and $f_{\rm s}(1,\theta)$ which are obtained as the $\rho\to 1$ limits of the real part $f_{\rm c}(\rho,\theta)$ and of the imaginary part $f_{\rm s}(\rho,\theta)$ of the inner analytic function $w(z)$.

It is important to note that $w(z)$ might not be analytic at some points on the unit circle. Also, so far we cannot state that the Taylor series $S_{z}$ converges anywhere on the unit circle, besides that single point at which we assumed the convergence of the pair of FC trigonometric series. For it to be possible to define the real integrals over the unit circle, the $\rho\to 1$ limits of the functions $f_{\rm c}(\rho,\theta)$ and $f_{\rm s}(\rho,\theta)$ must exist at least almost everywhere on the unit circle parametrized by $\theta$. They may fail to exist at points where $w(z)$ has isolated singularities on that circle. Therefore, for the moment the definition of the trigonometric series as Fourier series on the unit circle must remain conditioned to the existence of these limits almost everywhere.

Note that, if the limits to the unit circle result in isolated singularities in $f_{\rm c}(1,\theta)$ or $f_{\rm s}(1,\theta)$, then these must be integrable ones along the unit circle, since the $a_{k}$ coefficients are all finite.