$$\int_{\gamma}\frac{\log(z)\ dz}{(z-1)^2(z-1+3i)}$$
where $\gamma$ encloses the branch cut of $\log(z)$ in counter-clockwise direction with infinitesimal distance, starting at $-\infty+i\epsilon$ and ending at $-\infty+i\epsilon$.
So i'm not quite sure about what to do about the branch cut.
Proceed as normal we would first notice the two singularities at $z=1$ and $z=1-3i$. Normally the log function would have a branch cut along the negative real axis. But i assume i need one of these singularities to be enclosed by this contour? It seems i'd have to assume that $z=1$ is enclosed within this contour? Then i'd simply proceed with residue theorem to calculate the value of the integral.
In this case:
$$\mathrm{Res}_{z=1}=\lim_{z\to1}\frac{d}{dz}{(z-1)^2f(z)}=\lim_{z\to1}\frac{d}{dz}\left(\frac{\log(z)}{z-1+3i}\right)=-\frac{i}{3}$$
So the integral would be $$\int_Cf(z)dz=2\pi i\left(-\frac{i}{3}\right)=\frac{2\pi}{3}$$
I am guessing and assuming lots of things here, would appreciate if someone could clear things up!
Many thanks.
You will firstly need the contour to be closed to apply the residue theorem.
I assume that you meant starting at $-\infty-i\epsilon$ and ending at $-\infty+i\epsilon$, so that the countour is indeed counter-clockwise?
To enclose the countour, let the integrated function go around a sufficiently big circle, i.e., $Re^{i\theta}$, $\theta \in [-\pi,\pi]$, and $R$ sufficiently large.
Note that $\theta \in [-\pi,\pi]$: the branch cut of the $log$ function is at the negative axis!! That is, the imaginary part of the $log$ function must lie within the range of $\theta$ too.
Now that observe the integral is $0$ on the big circle, we can finally apply the residue theorem. Note that we have a clockwise rotation on the big circle, actually $-2\pi i \: Res$ is the result.
Now that you can calculate the Residues and obtain the integral :)