$ Let\\ c=-1\ and\ c+2l=1\\[2ex]\therefore{}-1+2l=1\\[2ex] \therefore{}2l=2\\[2ex]\therefore{}l=1$

$C_n=\frac{1}{2l}\int_c^{c+2l}f(x)e^{\frac{-in\pi{}x}{l}}dx\\[2ex] \displaystyle=\frac{1}{2}\int_{-1}^1{e^{ax}e}^{\frac{-in\pi{}x}{l}}dx\\[2ex]=\frac{1}{2}\int_{-1}^1e^{-\left(-a+in\pi{}\right)x}dx\\[2ex] \displaystyle=\frac{1}{2}\int_{-1}^1e^{-\left(-a+in\pi{}\right)x}dx\\[2ex]\displaystyle=\frac{1}{2}{\left[\frac{e^{-\left(-a+in\pi{}\right)x}}{-\left(-a+in\pi{}\right)}\right]}_{-1}^1=\frac{-1}{2\left(-a+in\pi{}\right)}[e^{-\left(-a+in\pi{}\right)}-e^{\left(-a+in\pi{}\right)}] $

$\displaystyle \therefore{}\ Cn=-\frac{1}{2\left(-a+in\pi{}\right)}\ \left[-2\sinh{\left(-a+in\pi{}\right)}\right] \\[2ex] \displaystyle C_n=\ \frac{\sinh{\left(-a+in\pi{}\right)}}{-a+in\pi{}} \\[2ex] \displaystyle In\ complex\ Fourier\ series, \\[2ex] \displaystyle f(x)=\ \sum_{n=-\infty{}}^{\infty{}}Cne^{\frac{in\pi{}x}{l}} \\[2ex] \displaystyle {\therefore{}e}^{ax}=\sum_{n=-\infty{}}^{\infty{}}\frac{\sinh{\left(-a+in\pi{}\right)}}{-a+in\pi{}}e^{in\pi{}x} \\[2ex] $