We have Trigonometric fourier series,
\[\begin{array}{l}
f(t) = {a_0} + {a_1}\cos {\omega}t + {a_2}\cos 2{\omega}t + .... + {a_n}\cos n{\omega}t + {b_1}\sin {\omega}t + {b_2}\sin 2{\omega}t + .... + {b_n}\sin n{\omega}t\\
= {a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos n{\omega}t + {b_n}\sin n{\omega}t} \right)}
\end{array}\]
We know that,
[tex]\sin n\omega t = \frac{{{e^{jn\omega t}} - {e^{ - jn\omega t}}}}{{2j}}[/tex]
and [tex]\cos n\omega t = \frac{{{e^{jn\omega t}} + {e^{ - jn\omega t}}}}{2}[/tex]
Thus,
\[\begin{array}{l}
f(t) = {a_0} + \sum\limits_{n = 1}^\infty {\left[ {{a_n}\frac{{({e^{jn\omega t}} + {e^{ - jn\omega t}})}}{2} + {b_n}\frac{{({e^{jn\omega t}} - {e^{ - jn\omega t}})}}{{2j}}} \right]} \\
= {a_0} + \sum\limits_{n = 1}^\infty {} \left[ {(\frac{{{a_n} - j{b_n}}}{2}){e^{jn\omega t}} + (\frac{{{a_n} + j{b_n}}}{2}){e^{ - jn\omega t}}} \right]
\end{array}\]
Let, \[\begin{array}{l}
{C_0} = {a_0}\\
{C_n} = \frac{{{a_n} - j{b_n}}}{2}\\
{C_{ - n}} = \frac{{{a_n} + j{b_n}}}{2}
\end{array}\]
And the series becomes,
[tex]f(t) = {C_0} + \sum\limits_{n = 1}^\infty {\left[ {{C_n}{e^{jn\omega t}} + {C_{ - n}}{e^{ - jn\omega t}}} \right]} [/tex]
This is the exponential form of the Fourier series.
Now ,
\[\begin{array}{l}
{C_n} = \frac{{{a_n} - j{b_n}}}{2} = \frac{1}{T}\int\limits_0^T {f(t)(\cos n\omega t - j\sin n\omega t)dt} \\
= \frac{1}{T}\int\limits_0^T {f(t)} {e^{ - jn\omega t}}dt
\end{array}\]
This equation is valid for both positive, negative and zero values of n.
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