Evaluating $\oint_{|z|=2} \frac{\log{(z-\alpha)}}{z}-\frac{\log z}z \, \mathrm dz$ 
Let $\alpha \in (-1,0)$ and define $$f(z)= \frac{\log{(z-\alpha)}}{z}-\frac{\log z}z$$ where the logarithms take their principal values, i.e. the arguments of $z$ and $z-\alpha$ are between $-\pi$ and $\pi$. Evaluate $$\oint_{|z|=2}f(z) \, \mathrm dz$$

I get $0$ but I am not sure if I have calculated the Laurent series correctly. I said  $$f(z) = \frac {1}{z}\log\left(1-\frac {\alpha} z\right)=-\frac{\alpha}{z^2} - \frac{\alpha^2}{z^3} + \cdots$$ which would imply the integral is $0$. Laurent series is valid as $f$ is analytic on an annulus $1 \le |z| \le 3$ taking the branch cut from $\alpha$ to $0$, and the series is convergent as $|\alpha/z| < 1$. I am just a bit concerned about the use of the identity $\log z - \log \omega = \log(z/\omega)$ as this is only true modulo $2 \pi i$, but I am pretty sure (from a rough sketch) that there is no added constant there. Other solutions would also be appreciated.
 A: Your solution is correct. To justify it, it is sufficient to justify the formula
$$\log(1-\alpha)-\log z=\log(1-\alpha/z)$$
for $|z|=1$, and where all branches are principal. The right hand side is analytic
in the region $|z|>|\alpha|$ which contains the unit circle. The summands in the 
left hand side are analytic on the unit circle except the point $-1$, where they
are discontinuous. Also LHS coincides with the RHS when $\arg z$ is sufficiently small. Thus the equality follows from uniqueness property of analytic functions.
The discontinuities at $-1$ of the summands at in the LHS cancel because they
have opposite jumps.
A: The integral
$$I(\alpha) = \oint\limits_{|z|=2}\dfrac1z \log\left(1-\dfrac\alpha z\right)dz$$
has the singularity in the point $z=0.$ 
If $\underline{|\alpha|<2},$ then, in accordance with OP, $I(\alpha)= 0.$
On the other hand, for $|\alpha|<2$
$$I'(\alpha) = -\oint\limits_{|z|=2}\dfrac1{z^2\left(1-\dfrac\alpha z\right)} = -\oint\limits_{|z|=2}\dfrac1{z(z-\alpha)} = -2\pi i\, \left(\mathop{\mathrm{Res}}\limits_{z=0}\dfrac1{z(z-\alpha)}+\mathop{\mathrm{Res}}\limits_{z=\alpha}\dfrac1{z(z-\alpha)}\right) = -2\pi i\left(\lim\limits_{z\to 0}\dfrac1{2z-\alpha}+\lim\limits_{z\to \alpha}\dfrac1{2z-\alpha}\right) = 0.$$
If $\underline{|\alpha| \ge 2},$ then the proposed Laurent series doesn't converge on the integration curve.
At the same time, for $\alpha>2$
$$I'(\alpha) = -\oint\limits_{|z|=2}\dfrac1{z^2\left(1-\dfrac\alpha z\right)} = -\oint\limits_{|z|=2}\dfrac1{z(z-\alpha)} = -2\pi i\, \mathop{\mathrm{Res}}\limits_{z=0}\dfrac1{z(z-\alpha)} = -2\pi i\lim\limits_{z\to 0}\dfrac1{2z-\alpha},$$
$$I'(\alpha) = \dfrac{2\pi}\alpha,$$
$$I(\alpha) = 2\pi i\int\limits_2^\alpha\dfrac{d\alpha}{\alpha} = 2\pi i \log\alpha = \pi i \log\left(\dfrac{\alpha^2}4\right),\tag1$$
wherein 
$$I(2) = I(-2) = 0,$$
so formula $(1)$ is valid also for $\alpha<-2$.
Therefore,
$$\boxed{I(\alpha) = 
\begin{cases}
0,\text{ if } |\alpha|\le 2\\[4pt]
\pi i \log\left(\dfrac{\alpha^2}4\right),\text{ otherwise. }
\end{cases}}\tag2$$

Note that substitution
$$z = 2e^{it}$$
allows to get
$$I(\alpha) = \int\limits_{-\pi}^\pi\dfrac12e^{-it}\cdot\log\left(1-\dfrac\alpha2 e^{-it}\right)\cdot2ie^{it}dt = i\int\limits_{-\pi}^\pi\log\left(1-\dfrac\alpha2e^{-it}\right)dt,$$
and this allows to check obtained result, using Wolfram Alpha, for the arbitrary values of $\alpha.$
For example, $I\left(\dfrac52\right) = \pi i\log\dfrac{25}4.$ 
