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: $a_{-1}=\operatorname {Res} (f,\log 2)=\lim _{z\to \log 2}\, \dfrac{z-\log 2}{2-e^z}$
Using L'Hopital rule
$a_{-1}=\lim _{z\to \log 2}\, \dfrac{1}{-e^z}=-\dfrac{1}{2}$
A: Your contour integration should have been, conveniently parameterised:
$$a_{-1} = \lim_{\epsilon \to 0}\frac{1}{2\pi i}\oint_{|z-\ln{2}|\leq\epsilon}{f(z)\,dz} =\lim_{\epsilon \to 0}\frac{1}{2\pi i}\int_{0}^{2\pi}{f(\epsilon\exp{(i\theta)}+\ln{2})\,\epsilon i\exp{(i\theta)}\,d\theta} \tag{*}$$
If now, one puts your $f$ into $(*)$ it turns to:
 $$a_{-1} = \lim_{\epsilon \to 0}\frac{1}{2\pi i}\int_{0}^{4\pi}{\frac{\epsilon i\exp{(i\theta)}}{2-2\exp{(\epsilon\exp{(i\theta}))}}\,d\theta}\tag{**}$$
We know that:
$$1-\exp{(\epsilon\exp{(i\theta)})}=-\epsilon\exp{(i\theta)}+\mathcal{O}(\epsilon^2)$$
Pluging this expansion into $(**)$ one has:
$$ a_{-1} = -\lim_{\epsilon \to 0}\frac{1}{4\pi}\int_{0}^{2\pi}{\frac{\epsilon \exp{(i\theta)}}{\epsilon \exp{(i\theta)}+\mathcal{O}(\epsilon^2)}\,d\theta}=-\frac{1}{2}$$
A: 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)}
A: Since $f$ is holomorphic at $z=0$ it is no surprise that $\int_{0}^{2\pi}\frac{ie^{it}}{2-e^{e^{it}}}dt=0$.
You have to compute $a_{-1}$ of the expansion $\sum_{m=-1}^{\infty}a_n(z-z_0)^n$ with $z_0=\log 2$ !!
$a_{-1}=-\frac{1}{2}$ is correct.
