# A question about complex integration using the residues theorem

I have a question about complex analysis. I am trying to get some intuition about residues and integration.

Let $c(t) = e^{2\pi it}$ be the unit circle of the complex plane.

I consider the following functions :
$f(z) = z$
$g(z) = 1/z$

Integrating $f$ around the $c(t)$ circle yields $0$ because f has no pole inside the unit disk.
Integrating $g$ around the $c(t)$ circle yields $2\pi i$ (residues theorem).

But $g(c(t)) = f(c(-t))$ : intuitively, I should be integrating the same function around the same circle, but in opposite directions. I would expect the integral values to be opposite, but I don't understand why one is null and the other not!

There is surely an obvious reasoning or calculus error, and I can't see where.

Thanks Yuufo

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Remember that for complex contour integration on a smooth curve $\gamma$ with endpoints $a$ and $b$, you have:$$\int_\gamma f(z)\,dz=\int_a^b f(\gamma (t))\gamma^\prime (t)\,dt.$$ The $\gamma^\prime (t)\,dt$ is the "path element" that joriki alludes to. So if you parametrize the unit circle by $\gamma (t)=e^{it}$ from 0 to $2\pi$, you get $\gamma^\prime (t)=ie^{it}$, and

\begin{align*} &\int_\gamma z\,dz=\int_0^{2\pi} ie^{2it}\,dt=0\text{, and}\\ &\int_\gamma \frac{1}{z}\,dz=\int_0^{2\pi} \frac{ie^{it}}{e^{it}}\,dt=2\pi i. \end{align*}

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Thanks to you both, Jorikis answer gives me the intuition and Josephs the calculus details :) –  user8329 Mar 16 '11 at 9:51
@Yuufo: You're welcome :-). By the way, if you ping people like I'm pinging you at the start of this comment, they get notified about a comment. I only came across this comment by chance because it wasn't addressed to me. –  joriki Mar 16 '11 at 19:31

I think what you're missing here is the essential difference between real and complex integration. It seems you're thinking of this as just "adding up" all the function values along the way, so it shouldn't make a difference in which order you add them up.

However, complex integration doesn't work that way, because you integrate along a path and the "path element" $\mathrm{d}z$ is itself complex and its "direction" enters into the result. So it matters not only which function values you encounter along the way, but also which "path element" they get multiplied by. In the two different integrations you describe, the same function values get multiplied by different path elements. If you draw a diagram with little arrows showing the "direction" of the integrand and the path element at some points, you'll probably realize why they add up to $0$ in one case and not in the other.

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