The complex Fourier Series is more compact and easier to manipulate.
Euler's formula is
$\begin{array}{rcl} e^{it} & = & \cos t + i \sin t\\ \text{or}\ \ \ e^{-it} & = & \cos t - i \sin t \end{array}$
(See some proofs here.)
By adding those equations, we can get an expressions for $\cos t$:
$\begin{array}{rcl} \cos t & = & \frac{1}{2} ( e^{-it} + e^{it} ) \\ \end{array}$
By subtracting those equations, we can get an expression $\sin t$:
$\begin{array}{rcl} e^{it} - e^{-it} &=& (\cos t + i \sin t) - (\cos t - i \sin t) \\ e^{it} - e^{-it} &=& 2 i \sin t \\ \sin t & = & {1 \over 2i} (e^{it} - e^{-it}) \\ \sin t & = & {i^2 \over 2i} (e^{-it} - e^{it}) \\ \sin t & = & \frac{i}{2} ( e^{-it} - e^{it} )\ \ \ \ \ \ \ \ \text{(Note the $i$ in $\frac{i}{2}$)} \end{array}$
Substituting these expressions into the $k^{th}$ term of the Fourier Series yeilds:
$\begin{array}{rcl} a_k \cos kt + b_k \sin kt & = & a_k \frac{1}{2} ( e^{-ikt} + e^{ikt} ) + b_k \frac{i}{2} ( e^{-ikt} - e^{ikt} ) \\ & = & \frac12 ( a_k - i b_k ) e^{ikt} + \frac12 ( a_k + i b_k ) e^{-ikt} \end{array}$
So the Complex Fourier Series is
$f(t) = \sum_{k=-\infty}^\infty\limits c_k\ e^{ikt}$
where $c_k = \left\{ \begin{array}{cl} \frac12 ( a_k - i b_k ) & \text{for}\ k > 0 \\ a_0 & \text{for}\ k = 0 \\ \frac12 ( a_{-k} + i b_{-k} ) & \text{for}\ k < 0 \end{array} \right.$
with the $a_k$ and $b_k$ coming from the Real Fourier Series.
Given a $c_k$ (with $k>0$) from the complex series, we can extract the $a_k$ and $b_k$ for the corresponding real series:
$\begin{array}{cl} a_k & = 2\ \text{real}( c_k ) \\ b_k & = -2\ \text{imaginary}( c_k ) \end{array}$
Which means that $c_k$ represents a shifted, modulated sinusoid.