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$$\int_{\gamma} ze^{-z} dz$$ where ${\gamma}$ is the unit circle centered at the origin.

By Cauchy's Theorem it is the composition of functions analytic in C and so is analytic on and inside ${\gamma}$, therefore it is equal to 0.

But I'm looking for how you would answer this question using the FTC?

Edit: fixed the question

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Just look for an indefinite integral of $ze^{-z}$. – Harald Hanche-Olsen Mar 18 '12 at 22:09
This looked like a typo to me. Maybe $\displaystyle\int_\gamma \Big( f(z)=ze^{-z} \Big)\;dz$ would have been clear, even if perhaps unprecedented. – Michael Hardy Mar 18 '12 at 22:12
Fixed the question. – Jim_CS Mar 18 '12 at 22:20
up vote 1 down vote accepted

Take any smooth parametrization $\gamma(t)$, $t\in[0,1]$. Then $$ \int_{\gamma} ze^{-z} dz=\int_0^{1}\gamma(t)e^{-\gamma(t)}\gamma'(t)\,dt=\left.e^{-\gamma(t)}-\gamma(t)e^{-\gamma(t)}\right|_0^{1}=0 $$ since $\gamma(0)=\gamma(1)$.

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But how can you say it has an anti-derivative...dont you have to prove it has an anti-derivative before you can make your statement? – Jim_CS Mar 19 '12 at 0:22
It depends on what the symbol $\int_\gamma f(z)dz$ means to you. The way I understand it, the first equal sign is the definition. After that, I have a usual Riemman integral: $$\int \gamma(t)e^{-\gamma(t)}\gamma'(t)dt=\int ue^{-u}du=e^{-u}-ue^{-u},$$ with $u=\gamma(t)$. – Martin Argerami Mar 19 '12 at 0:27
But the second equal sign can't be automatically valid? Some functions mightn't have an anti-derivative. That's why it seems to me that I should have to prove a function has an anti-derivative before I can apply your statement? – Jim_CS Mar 19 '12 at 0:52
You lost me there. If you differentiate $e^{-u}-ue^{-u}$ you get $ue^{-u}$, so the former is an antiderivative of the latter. And, in any case, any continuous function has an antiderivative, whether you can write it as an explicit formula or not. – Martin Argerami Mar 19 '12 at 0:56
ok ty, i have to get my head around it. – Jim_CS Mar 19 '12 at 10:24

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