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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 $?

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$e^{2\pi i} = e^0(\cos2\pi + i\sin2\pi) = \cos 0 + i\sin 0 = 1 + 0 = 1$, right? – Dylan Moreland May 24 '12 at 22:11
Ahh okay, thanks. I forgot about euler's formula! – Derrick May 24 '12 at 22:15
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Note that your curve is the unit circle, so in particular it is closed, i.e $\gamma(2\pi) = \gamma(0) = 1$. This curve will return over and over and you need to learn to recognize it. – mrf May 24 '12 at 22:20
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That depends on the interval. Note that $e^{it} = \cos t + i\sin t$, so the point $e^{it}$ is the point on the unit circle for which the angle (counting from the positive $x$-axis) is $t$. To trace out the complete circle, let $t$ vary from $0$ to $2\pi$. To get the semi-circle in the upper half plane, let $t$ vary from $0$ to $\pi$. Try to figure out a few other variations on your own. – mrf May 24 '12 at 22:33
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Yes, assuming that $R > 0$. If $R < 0$, the radius is $-R$. (But you wouldn't call this a unit circle.) – mrf May 24 '12 at 22:44
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