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I don't understand how does $ \epsilon $ come at the end of this formula in article from Wikipeida

\begin{align} \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-a} \,dz - f(a) \right | &= \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)-f(a)}{z-a} \,dz \right |\\[.5em] &\leq \frac{1}{2 \pi} \int_0^{2\pi} \frac{ |f(z(t)) - f(a)| } {\varepsilon} \,\varepsilon\,dt\\[.5em] &\leq \max_{|z-a|=\varepsilon}|f(z) - f(a)| \xrightarrow[\varepsilon\to 0]{} 0. \end{align}

$$ \frac{1}{2 \pi} \int_0^{2\pi} \frac{ |f(z(t)) - f(a)| } {\varepsilon} ,\varepsilon \text{ (<--this epsilon) }\,dt $$ And also $dz$ is changed in to $dt$ shouldn't there be something to balance it.?

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up vote 1 down vote accepted

The proof defines $z = a + \epsilon e^{it}$ since $z$ is a point on a circular contour of radium $\epsilon$ centered at $a$. Hence, $dz = \epsilon i e^{it} dt$. So the integral's denominator becomes $\epsilon e^{it}$, and this cancels with what we have from the change of variables in the numerator. Also, the $i$ is absorbed by the absolute value sign.

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