Calculate $\int_\gamma \frac{1}{(z-z_0)^2}dz$ This is the definition of the fundamental theorem of contour integration that I have:

If $f:D\subseteq\mathbb{C}\rightarrow \mathbb{C}$ is a continuous function on a domain $D \subseteq \mathbb{C}$ and $F:D\subseteq \mathbb{C} \rightarrow \mathbb{C}$ satisfies $F'=f$ on $D$, then for each contour $\gamma$ we have that:
$\int_\gamma f(z) dz =F(z_1)-F(z_0)$
where $\gamma[a,b]\rightarrow D$ with $\gamma(a)=Z_0$ and $\gamma(b)=Z_1$. $F$ is the antiderivative of $f$.

Let $\gamma(t)=Re^{it}, \ 0\le t \le 2\pi, \ R>0$. In my example it said $\int_\gamma \frac{1}{(z-z_0)^2}dz=0$. Im trying to calculate it out myself, but I got stuck.
I get that $f(z)=\frac{1}{(z-z_0)^2}$ has an antiderivative $F(z)=-\frac{1}{(z-z_0)}$.
Thus by the fundamental theorem of contour integration:
$\int_\gamma \frac{1}{(z-z_0)^2}dz =F(z_1)-F(z_0)\\F(\gamma(2\pi))-F(\gamma(0))\\F(Re^{2\pi i})-F(R)\\-\frac{1}{Re^{2\pi i}-z_0} +\frac{1}{R-z_0}\\-\frac{1}{Re^{i}-z_0} +\frac{1}{R-z_0}$
How does $\int_\gamma \frac{1}{(z-z_0)^2}dz=0$?

$\gamma(2\pi)=Re^{2\pi i}=R=Re^0=\gamma(0)$
