A signal may be a complex function of time. The most important complex signal we will investigate is the complex exponential function
\begin{eqnarray}
f(t) = e^{i\omega t}
\end{eqnarray}
which is a periodic signal with frequency $\frac{\omega}{2\pi}$ Hertz.
Theorem:
$$
\sum_{k=-\infty}^{\infty}e^{{2 \pi i} kt}=\lim_{N \rightarrow \infty}\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)}=\sum_{n=-\infty}^{\infty}\delta(t-n)
$$
Proof:
$$\begin{eqnarray}
\sum_{k=-N}^{N}e^{{2 \pi i} kt}&=& e^{-{2 \pi i} Nt}+e^{-{2 \pi i} (N-1)t}+e^{-{2 \pi i} (N-2)t}+\ldots+e^{{2 \pi i} (N-1)t}+e^{{2 \pi i} Nt} \\
&=&e^{-{2 \pi i} Nt} (1+e^{2 \pi i t}+e^{4 \pi i t} + \ldots +e^{2\pi i (2N-1)t}+e^{4 \pi i Nt})
\end{eqnarray}$$
The part within the parantheses is a truncated geomertic series with $r=e^{2\pi i t}$. After summing it, and performing some algebraic manipulations, we arrive at
$$\begin{eqnarray}
\sum_{k=-N}^{N}e^{2 \pi i kt}&=&e^{-2 \pi i Nt}\frac{1-e^{2 \pi i (2N+1)t}}{1-e^{2 \pi i t}}\\
&=&e^{-2 \pi i Nt}\frac{e^{\pi i (2N+1)t}(e^{- \pi i (2N+1)t}-e^{ \pi i (2N+1)t})}{e^{\pi i t}(e^{-\pi i t}-e^{ \pi i t})}\\
&=&\frac{e^{- \pi i (2N+1)t}-e^{ \pi i (2N+1)t}}{e^{-\pi i t}-e^{ \pi i t}}\\
&=&\frac{\frac{e^{\pi i (2N+1)t}-e^{- \pi i (2N+1)t}}{2i}}{\frac{e^{\pi i t}-e^{- \pi i t}}{2i}}\\
&=&\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)}
\end{eqnarray}$$
If we take the limit $N \rightarrow \infty$ of both sides, we reach the expression
$$\begin{eqnarray}
\sum_{k=-\infty}^{\infty}e^{2 \pi i kt}&=&\lim_{N \rightarrow \infty}\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)}
\end{eqnarray}$$
By substituting $2\pi t \rightarrow t$ into the last theorem of the chapter “dirac delta functions”, and using the scaling property of dirac deltas, we arrive at
$$\begin{eqnarray}
\lim_{N \rightarrow \infty}\frac{\sin[(2N+1)\pi t]}{\sin(\pi t)} = 2\pi \sum_{k=-\infty}^{\infty} \delta(2\pi t-2\pi k) = \sum_{k=-\infty}^{\infty} \delta( t- k)
\end{eqnarray}$$
which proves the theorem.