Imaginary Sum

$S_{n} = i^{0}+i^{1}+\cdots+i^{n} \\ -i\cdot S_{n} = -i^{1}-i^{2}-\cdots-i^{n}-i^{n+1}$ $\small\color{#3D99F6}{\text{Adding both equations}}$ $S_{n} \left( 1-i\right)= 1-i^{n+1} \\ 2S_{n} =(1+i)(1-i^{n+1}) \\ S_{n} = \dfrac{1}{2} \left( 1+i-i^{n+1}+i^{n} \right) \\ S_{2009} = \dfrac{1}{2}\left( 1+i + i -(-1)\right) \\ \boxed{S_{2009} = 1+i }$

We note that the powers of $i^k$ are cyclical such that $i^k = \begin{cases} 1 & k \text{ mod 4}=0 \\ i & k \text{ mod 4}=1 \\ -1 & k \text{ mod 4}=2 \\ -i & k \text{ mod 4}=3 \end{cases}$

Similarly, the sum of $i^k$ is also cyclical as follows:

$\begin{aligned} S_0 & = i^0 = 1 \\ S_1 & = 1 + i \\ S_2 & = 1 + i - 1 = i \\ S_3 & = i - i = 0 \\ S_4 & = 1 = S_0 \\ ... & = ... \\ \implies S_k & = \begin{cases} 1 & k \text{ mod 4}=0 \\ 1+i & k \text{ mod 4}=1 \\ i & k \text{ mod 4}=2 \\ 0 & k \text{ mod 4}=3 \end{cases} \end{aligned}$

Since $2009 \text{ mod 4} = 1 \implies S_{2009} = \boxed{1+i}$