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So the example he gave was using the ML-Inequality. Let $v = \{e^{i \theta} \mid 0 \leq \theta \leq 2\pi \}$.

$\left|\int(1/z)\,dz\,\right|$ with respect to $v$ is less than or equal to $\int \left|1/z\right| \left|dz\right|$ with respect to $v^*$. Then he says that is equal to the integral with respect to $v$ of $\left|dz\right|$ which is equal to the $l(v) = 2 \pi$. ( he says that $\left|1/z\right|$ is the unit circle. ) I don't understand why $\left|1/z\right|$ is the unit circle and any of the steps after he makes this claim.

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    your $V=\{e^{i\theta} : 0<\theta<\pi\}$ is the upper half of the circle. the inverse of these elements $\{1/z : z\in V\}=\{e^{-i\theta} : 0<\theta<\pi\}$ is the bottom half of the circle. im not sure what youre asking, i think there are some mistakes in there, maybe some $\pi$'s instead of $2\pi$'s or something...hard to read2011-04-01
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    Sorry. That should have been 2 $\pi$. Yeah that makes sense though. Basically you are mapping the top half of the circle into the bottom half and vice versa.2011-04-01
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    This is a silly remark, but nobody would call a curve $v$ in complex analysis. I suspect that it's a $\gamma$ (Greek letter "gamma").2011-04-01

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$|1/z|$ is a number, so it can't be the unit circle. :)

But if $z$ lies on the unit circle, then $|z|=1$ and therefore $|1/z| = 1/|z|=1/1=1$, so $\int_\gamma |1/z| |dz| = \int_\gamma 1 |dz|$.