Subsection 5.5.1 Complex Form of Fourier Series
The complex form of Fourier series is obtained by expressing \(\cos nx\) and \(\sin nx\) in the exponential form, i.e.,
\begin{equation*}
f(x) = \frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[a_{n}\cos nx + b_{n}\sin nx\right]
\end{equation*}
\begin{equation*}
=\frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}a_{n}\left[\frac{e^{inx}+e^{-inx}}{2}\right] + \sum\limits_{n=1}^{\infty}b_{n}\left[\frac{e^{inx}-e^{-inx}}{2i}\right]
\end{equation*}
\begin{equation*}
=\frac{a_{o}}{2} + \sum\limits_{n=1}^{\infty}\left[\frac{a_{n}}{2}-\frac{ib_{n}}{2}\right]e^{inx}+\sum\limits_{n=1}^{\infty}\left[\frac{a_{n}}{2}+\frac{ib_{n}}{2}\right]e^{-inx}
\end{equation*}
\begin{equation*}
=c_{0}+\sum\limits_{n=1}^{\infty}c_{n}e^{inx}+\sum\limits_{n=1}^{\infty}c_{-n}e^{-inx}
\end{equation*}
\begin{equation}
\therefore \quad f(x) = \sum\limits_{n=-\infty}^{\infty}c_{n}e^{inx};\tag{5.5.3}
\end{equation}
where,
\begin{equation*}
\begin{cases}
c_{0} = \frac{a_{0}}{2}\\
c_{n} = \frac{1}{2}(a_{n}-ib_{n})\\
c_{-n} = \frac{1}{2}(a_{n}+ib_{n})
\end{cases}
\end{equation*}
Equation (5.5.3) is a complex form of Fourier series. On multiplying both sides of equation (5.5.3) by \(e^{-imx}\) and integrating w.r.t. ’x’, we get -
\begin{equation*}
\int\limits_{c}^{c+2\pi}f(x)e^{-imx} \,dx = \sum\limits_{n=-\infty}^{\infty}c_{n}\int\limits_{c}^{c+2\pi}e^{i(n-mx)} \,dx
\end{equation*}
\begin{equation*}
= \sum\limits_{n=-\infty}^{\infty}c_{n}2\pi \delta_{mn}=2\pi c_{m}
\end{equation*}
\begin{equation*}
\therefore c_{n} =\frac{1}{2\pi}\int\limits_{c}^{c+2\pi}f(x)e^{-inx} \,dx.
\end{equation*}
The function \(f(x)\) is being defined in the interval \([c, c+2\pi]\text{.}\) Hence,
\begin{equation*}
c_{0}=\frac{1}{2\pi}\int\limits_{c}^{c+2\pi}f(x) \,dx = \frac{a_{0}}{2}.
\end{equation*}
