Theorem. Suppose that a function $f$ is analytic throughout a disk $|z-z_0|<R_0$ centered at $z_0$ and with radius $R_0$. Then $f(z)$ has the power series representation
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\ (|z-z_0|<R_0),$$
where
$$a_n=\frac{f^{(n)}(z_0)}{n!}\ (n=0,1,2,\cdots).$$

Proof. First consider the case $z_0=0$ and show that
$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n\ (|z|<R_0).$$
This particular Taylor series is called a Maclaurin series. Let us write $|z|=r$ and let $C_0$ denote any positively oriented circle centered at the origin and with radius $r_0$, where $r<r_0<R_0$.

Since $f$ is analytic inside and on the circle $C_0$ and since $z$ is interior to $C_0$, by the Cauchy Integral Formula, we have
$$f(z)=\frac{1}{2\pi i}\oint_{C_0}\frac{f(s)}{s-z}ds.$$
Note that $\frac{1}{s-z}$ may be written as
\begin{align*}
\frac{1}{s-z}&=\frac{1}{s}\frac{1}{1-\frac{z}{s}}\\
&=\sum_{n=0}^{N-1}\frac{z^n}{s^{n+1}}+z^N\frac{1}{(s-z)s^N}.
\end{align*}
Using this, $f(z)$ may be written as
\begin{align*}
f(z)&=\frac{1}{2\pi i}\oint_{C_0}\frac{f(s)}{s-z}ds\\
&=\sum_{n=0}^{N-1}\frac{1}{2\pi i}\oint_{C_0}\frac{f(s)}{s^{n+1}}ds z^n+\frac{z^N}{2\pi i}\oint_{C_0}\frac{f(s)}{(s-z)s^N}ds\\
&=\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{n!}z^n+\rho_N(z),
\end{align*}
where
$$\rho_N(z)=\frac{z^N}{2\pi i}\oint_{C_0}\frac{f(s)}{(s-z)s^N}ds.$$
We are done if we can show that $\displaystyle\lim_{N\to\infty}\rho_N=0$. By triangle inequality, $|s-z|\geq ||s|-|z||=r_0-r$. So, we obatin
\begin{align*}
|\rho_N(z)|&\leq \frac{r^N}{2\pi}\frac{M}{(r_0-r)r_0^N}2\pi r_0\\
&=\frac{Mr_0}{r_0-r}\left(\frac{r}{r_0}\right)^N.
\end{align*}
Since $r<r_0$, $\displaystyle\lim_{N\to\infty}|\rho_N(z)|=0$ i.e. $\displaystyle\lim_{N\to\infty}\rho_N=0$. Hence, we have proved that
$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n.$$
Now, suppose that $f(z)$ is analytic when $|z-z_0|<R_0$. Then $g(z):=f(z+z_0)$ is analytic when $|z|=|(z+z_0)-z_0|<R_0$. So, $g(z)$ has the Maclaurin series representation
$$g(z)=\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}z^n\ (|z|<R_0)$$
or
$$f(z+z_0)=\sum_{n=0}^\infty\frac{f^{(n)}(z_0)}{n!}z^n\ (|z|<R_0).$$
Replacing $z$ by $z-z_0$, we finally obtain the Taylor series representation
$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(z_0)}{n!}(z-z_)^n\ (|z-z_0|<R_0).$$

Example. The function $f(z)=e^z$ is entire, so it has a Maclaurin series representation for all $z\in\mathbb{C}$.
$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}\ (|z|<\infty).$$

Example. $\sin z$ is defined as
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}.$$
So,
\begin{align*}
\sin z&=\frac{1}{2i}\left[\sum_{n=0}^\infty\frac{(iz)^n}{n!}-\sum_{n=0}^\infty\frac{(-iz)^n}{n!}\right]\\
&=\frac{1}{2i}\sum_{n=0}^\infty[1-(-1)^n]\frac{i^nz^n}{n!}\\
&=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}\ (|z|<\infty).
\end{align*}
Using the definition $\cos z=\frac{e^{iz}+e^{-iz}}{2}$, one can similarly obtain the Maclaurin series representation for $\cos z$
$$\cos z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}.$$