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I'm trying to find the contour integral $$ \int_\gamma \frac{\mathrm{Im} (z)}{z - \alpha} dz $$ where $ \alpha $ is a complex number such that $ 0 < |\alpha| < 2 $ and $ \gamma $ is the circle oriented in the positive sense, centred at the circle with radius 3.

I can find that $$ \int_\gamma \frac{\mathrm{Im} (z)}{z - \alpha} dz = \int_0^{2\pi} \frac{e^{it}-e^{-it}}{2i} \frac{1}{e^{it}-\alpha} i e^{it} dz $$ but the denominator is making it difficult to find the value of the contour integral. How can I proceed in this?

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    $\begingroup$ Use that fact that $\text{Im} z = \frac{1}{2i}(z - \overline{z}) = \frac{1}{2i}(z - \frac{9}{z})$ on $\gamma$. $\endgroup$ – anomaly Mar 31 at 16:04
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Too long for a comment: wrong RHS. If $z = 3 e^{it} = 3(\cos t + i \sin t)$, $dz = 3i e^{it}dz = 3i(\cos t + i \sin t)dz$, $\mathrm{Im}(z) = 3\sin t$, $$\int_\gamma \frac{\mathrm{Im}(z)}{z - \alpha} dz = \int_0^{2\pi}\frac{3\sin t}{3 e^{it} - \alpha}3i e^{it}dt.$$ (three 3's missing).

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