1
$\begingroup$

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.

  • 0
    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

1 Answers 1

1

$|1/z|$ 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 $\int_\gamma |1/z| |dz| = \int_\gamma 1 |dz|$.