Finding contour integral $\int_\gamma \frac{\mathrm{Im} (z)}{z - \alpha} dz$

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?

• Use that fact that $\text{Im} z = \frac{1}{2i}(z - \overline{z}) = \frac{1}{2i}(z - \frac{9}{z})$ on $\gamma$. – anomaly Mar 31 at 16:04

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).