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If $a \in \mathbb{C}$ with $|a| < 1$, then the values of $\displaystyle \frac{(1-|a|^2)}{\pi} \int_{\gamma} \frac{|dz|}{|z+a|^2}$, where $\gamma$ is the simple closed curve $|z| = 1$ taken with the positive orientation. What will be the integration?

I know this theorem: $$\int_{\gamma} f |dz| = \int_{a}^{b} f(\gamma(t))d|\gamma|(t)$$ where $f$ is continuous on ${\gamma}$.

How do I use this theorem?

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    Related: https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula – 2017-01-18
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    @GDumphart How do I deal with $|dz|$? – 2017-01-18

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On the positive oriented contour over the unit circle, we have $\displaystyle\;|dz| = \frac{dz}{iz}$ and $$\frac{1}{|z+a|^2} = \frac{1}{(z+a)(\bar{z}+\bar{a})} = \frac{1}{(z+a)(\frac{1}{z} + \bar{a})} = \frac{z}{\bar{a}}\frac{1}{(z+a)(z+\frac{1}{\bar{a}})} $$ This means $$\mathcal{I} \stackrel{def}{=} \frac{1-|a|^2}{\pi}\oint_{|z|=1} \frac{|dz|}{|z+a|^2} = \frac{1-|a|^2}{\pi\bar{a}i}\oint_{|z|=1} \frac{dz}{(z+a)(z+\frac{1}{\bar{a}})} $$ The last integral is an ordinary contour integral. Since the integrand has only one root $-a$ inside the unit circle, we can evaluate it by taking the residue at $-a$. The end result is $$\mathcal{I} = \frac{1-|a|^2}{\pi\bar{a}i}\times \frac{2\pi i}{-a + \frac{1}{\bar{a}}} = 2$$

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    I know only Lebesgue and Riemann-Stieltjes integrals. What is the meaning od $|dz$ here? – 2017-01-18
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    @Santiago, it means the arc length of the contour, On the unit circle with the parametrization $z = e^{i\theta}$, $|dz|$ is simply $d\theta$. – 2017-01-18