Integral of $\sqrt{1+\frac{1}{z}}$ on $\partial B(0,2)$ We are supposed to take $g$ as the branch of $\sqrt{1+\frac{1}{z}}$ on $\Bbb{C}\setminus [-1,0]$. Then we are supposed to find the integral of this g on $\partial B(0,2)$. The hint we were given is to use Laurent series.

Edit: I forgot to mention that for our branch $g$ we want $g(1) = \sqrt{2}$. So I defined my branch as $g(re^{i\theta}) = \sqrt{r}e^{i(\operatorname{Arg}(z+1)-\operatorname{Arg}(z))}$, where $\operatorname{Arg}$ is the principal branch.
So this is my current understanding: I know that with $g$ defined as above, $g$ can be represented as a Laurent series $g = \sum_{-\infty}^{\infty} a_kz^k$ that is convergent in $1 < |z| < \infty$. I also know that the coefficients of the Laurent series are given by
$$ a_k = \frac{1}{2\pi i} \int_{\partial B(0,R)} \frac{g(z)}{z^{k+1}} dz,$$
where $R$ is in the annulus of convergence. So if I can determine $a_{-1}$ then I will have
$$ 2\pi ia_{-1} =\int_{\partial B(0,R)} g(z) dz,$$
which is the integral I am looking for since $2$ is in the annulus of convergence. But I am stuck on how to determine the value of $a_{-1}$.
 A: This function $g$ is analytic in $\mathbb C\setminus [-1,0]$, and hence expressible as a Laurent series of the form
$$
g(z)=\sum_{k\in\mathbb Z}a_kz^k,
$$
and this series converges whenever $|z|>1$.
First observation is that $a_k=0$, for all $k> 0$, since
$$
a_k=\frac{1}{2\pi i}\int_{|z|=R}\frac{g(z)}{z^{k+1}}\,dz,
$$
for all $R>1$. To see this, as $g$ is bounded, for $|z|>1$, then the integral above tends to zero, as $R\to\infty$.
Hence,
$$
g(z)=a_0+\frac{a_{-1}}{z}+\frac{a_{-2}}{z^2}+\cdots. \tag{1}
$$
and
$$
1+\frac{1}{z}=a_0^2+\frac{2a_0a_{-1}}{z}+\frac{2a_0a_{-2}+a_{-1}^2}{z^2}+\cdots
\tag{2}
$$
Thus $a_0=1$ or $a_0=-1$. At this point we shall use the condition that $g(1)=\sqrt{2}$, which implies that:
a. $g(x)$ is real and for $x>0$, since $\big(g(x)\big)^2=1+\dfrac{1}{x}>0$, and thus $g(x)=\pm1\sqrt{1+\dfrac{1}{x}}\in\mathbb R$.
b. $g(x)$ is positive and for $x>0$. This is since $g(1)>0$ and if it was negative or zero, for some $\xi>0$, then it would have to vanish for some $\eta\in (1,\xi]$, due to IVT. Impossible, since $1+\dfrac{1}{x}\ne 0$, for $x>0$.
Since $g(x)>0$, for $x>0$, letting $x\to\infty$ is $(1)$, we obtain that $a_0>0$, and consequently $a_0=1$.
Finally, from $(2)$ we obtain that $2a_0a_1=1$ and thus
$$
a_{-1}=\frac{1}{2},
$$
and hence
$$
\int_{|z|=2}g(z)\,dz=2\pi i a_{-1}=\pi i.
$$
