Laurent series coefficient $a_{-1} $ for $f(z)=\frac{1}{2-e^z} $ I am still a beginner in complex analysis. I am trying to compute the coefficient $a_{-1}$ of the Laurent series $\sum_{m=-1}^{\infty}a_n(z-z_0) $ of the function $$f(z)=\frac{1}{2-e^z} $$ near the singularity $z_0=log(2)$.

Here is what I did so far, since the singularity $log(2)$ is a pole of order 1 and the fact that there are no other singularities in the unit circle centered at z=0 the other nearest singularities are at $log(2) \pm 2i\pi$ and their modulus is $\gt 1 $.

I tried to compute $a_{-1} $ using contour integration around the unit circle centered at z=0 $$a_{-1}=\frac{1}{2i\pi} \int_{0}^{2\pi}\frac{ie^{it}}{2-e^{e^{it}}}dt $$

Then I get $$ \frac{1}{2i\pi}[\frac{1}{2}(e^{it}-log(2-e^{it})) ]_{0}^{2\pi} $$

Which is equal to 0. From what I know, I should get 0 if there was no singularity inside my contour but $z_0=log(2) $ has a modulus 0.301.

I already know that the expansion of $f(z)$ near $z_0=log(2) $ is of the form 
$$ R(z)=-\frac{1}{2}\frac{1}{z-log(2)}+...$$ 

Then I should get $a_{-1}=-\frac{1}{2} $. I did not find my mistake..
$a_{-1}=\operatorname {Res} (f,\log 2)=\lim _{z\to \log 2}\, \dfrac{z-\log 2}{2-e^z}$

Using L'Hopital rule

$a_{-1}=\lim _{z\to \log 2}\, \dfrac{1}{-e^z}=-\dfrac{1}{2}$
You're right, we're looking for the coefficient, but you have to delimit an outline or you have at least one pole inside like |z-\ln{2}|\leq\epsilon after you make your parametrisation with z = \epsilon exp{(i\theta)}