Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $?

share|cite|improve this question
$e^{2\pi i} = e^0(\cos2\pi + i\sin2\pi) = \cos 0 + i\sin 0 = 1 + 0 = 1$, 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
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
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
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

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 $\Omega$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.