Kind of singularity with Laurent series of $f(z)=\frac{1}{1+2z}$

Question

I computed the power series of $f(z)=\frac{1}{1+2z}$ around point z = 0: $\sum_{n=0}^\infty (-2)^nz^n$. Hopefully it is right :D

Now i have to specify the kind of singularity at $z_0 = -1/2$ (With the consideration of the Laurent series $\sum_{n=-\infty}^\infty a_n(z-z_0)$ of $f$ around $z_0= -1/2$). I tried to compute first $a_{-1} = \pi i$.

My question is now should i go further and try to compute $a_{k}$ with $k≤-2$ and determine the pole order or is there a chance to get the Laurent series from my series above to determine the order from there?

Answer 1

At $-\frac12$, you have$$f(z)=\frac1{1+2z}=\frac12\cdot\frac1{z+1/2}.$$Therefore, the Laurent series of $f$ centered at $-\frac12$ is $\frac12\cdot\frac1{z+1/2}$ and so $f$ has a single pole at that point.