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How do I prove for $r>0$ the estimation $\left |\int_{K_r(0)}\frac{1}{z}dz \right |\leq \frac{1}{r}2\pi r=2\pi$?

I know that this is estimation lemma, I tried to google, but I don't understand some parts.

$L(\gamma)=\int_{\alpha}^{\beta}=\left | \gamma'(t) \right |dt$ - that is length.

3.1 Estimation Lemma Let $f : U → C$ be continuous (where U is some subset of C), let $γ$ be a path in $U$, and suppose $|f(z)| < M$ for all $z ∈ γ$.

$$\left | \int_{\gamma}f(z)dz \right |\leq ML$$

But I still don't understand how they get $2\pi$?

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  • $\begingroup$ Is $\;K_r(0)\;$ the circle of radius $\;r\;$ and center the origin? $\endgroup$ – DonAntonio Apr 16 '17 at 22:22
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$$\left|\oint_{K_r(0)}\frac1zdz\right|\le\oint_{K_r(0)}\frac{dz}{|z|}=\frac1{|z|=r}\oint_{K_r(0)}dz=\frac1r2\pi r=2\pi$$

assuming $\;K_r(0)=\;$ the circle of radius $\;r\;$ and center at the origin of the complex plane.

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