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I'm reading a passage on how to integrate $f(z)=z^\beta(1+z^2)^{-1}$ for $\beta\in(-1,1)$.

Take $\epsilon>0$, and let $\gamma_\epsilon$ be the arc $\{\epsilon e^{i\theta}:\theta\in(\delta,2\pi-\delta)\}$ for some small $\delta>0$. So $\gamma_\epsilon$ looks like a section of the circumference of a small circle of radius $\epsilon$. Over this arc, it's claimed that $|f(z)|\leq 2\epsilon^\beta$, but I don't see how they get this bound. All I see is that $ |f(z)|=\frac{|z^\beta|}{|z^2+1|}=\frac{\epsilon^\beta}{|z^2+1|} $ but I don't see how the factor of $2$ appears and where the denominator goes. For one thing, I know that $\epsilon$ is small enough so that the poles $\pm i$ of $f(z)$ are outside the circle of radius $\epsilon$. This is towards an eventual keyhole contour integration I believe. Could someone point out how that bound is reached?

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$\frac{\epsilon^{\space\beta}}{|z^2+1|} \leq \frac{\epsilon^{\space\beta}}{|1-|z^2||} = \frac{\epsilon^{\space\beta}}{1-\epsilon^{\space2}} \leq 2\epsilon^{\space\beta}$

The first inequality is the triangle inequality (the other side of it at least, and assuming $\epsilon < 1$ so that all is well-defined and to justify removing the surrrounding absolute value sign), and the second is simply from a slightly stricter assumption that $0<\epsilon < \frac 1 {\sqrt{2}}<1 \implies \epsilon^2 < \frac 1 2 \implies \frac 1 {1-\epsilon^{\space2}} < 2$. This assumption is reasonable as well, assuming at some point you will take $\epsilon \to 0$.

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    @MacDowell, a pleasure.2012-03-19