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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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    Okay, thanks for your help again. Yes, I do think visualising these contour integrals geometrically would help alot.2012-05-24

1 Answers 1

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The reason why $\int_\gamma e^z=0$ was explained in comments by mrf:

  1. $\gamma$ is a closed curve because $\gamma(2\pi)=1=\gamma(0)$
  2. the function $e^z$ has an antiderivative, namely $e^z$ itself.

It is a consequence of the fundamental theorem of calculus (contour integration version) that every function with an antiderivative in some region $\Omega$ integrates to zero over every closed curve contained in $\Omega$.