I have to evaluate:
$$I=\int_{\gamma} \frac{\Re(z)}{2z-i} dz$$
where $\gamma$ is the unit circle centered at $0$. Cauchy's integral formula tells us that:
$$f^{(n)}(a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} dz$$
But this only works for $f$ holomorphic, and I'm finding it tough to work out how to apply it to this problem, since $\Re$ isn't a holomorphic function anywhere, let alone in the disc enclosed by $\gamma$.
Is there some sort of trick to pull out some nice factors so I can get the integral into a form I can work with?
Note that $\text{Re}(z)=\frac12(z+\bar z)$. Furthermore, note that for $z\in \gamma$, $\bar z=\frac1z$. Then, we can write
$$\begin{align} \oint_\gamma \frac{\text{Re}(z)}{2z-i}\,dz&=\frac12 \oint_\gamma \frac{z}{2z-i}\,dz+\frac12 \oint_\gamma \frac{\bar z}{2z-i}\,dz\\\\ &=\frac12\oint_\gamma \frac{z}{2z-i}\,dz+\frac12 \oint_\gamma \frac{1}{(2z-i)z}\,dz\\\\ &=\frac14\oint_\gamma \frac{z}{z-i/2}\,dz+\frac i2 \oint_\gamma \frac{1}{z}\,dz-\frac i2\oint_\gamma \frac{1}{z-i/2}\,dz \tag 1 \end{align}$$
Now apply Cauchy's Integral Formula to each of the terms on the right-hand side of $(1)$.