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 $f$.
I was reading an example that said:
Let $\gamma(t)=e^{it}$ where $0\le t \le 2\pi$. We have that $\int_\gamma e^z dz=0$ by the fundamental theorem of contour integration.
The part I'm not sure is, how did they get that $\int_\gamma e^z dz=0$? I tried working it out myself and I got: $F'=e^z=f$. Also, $F(z)=e^z$
$F(z_1)=F(\gamma(b))=F(\gamma(2\pi))=F(e^{2\pi i})=e^{e^{2\pi i}}\\ F(z_0)=F(\gamma(a))=F(\gamma(0))=F(1)=e^1$.
But how does $\int_\gamma e^z dz =F(z_1)-F(z_0) = e^{e^{2\pi i}} - e^1 =0 $?