This is a problem from my past QUal: "Prove that all singularity of $$\frac{1}{e^z+3z}$$ is of order 1. You don't need to find the singularities."
Usually this kind of problem is easy to me. My procedure is to find the singularities, and use Taylor series on it. That's why this problem throws me off. I cannot find the poles here: $e^z+3z=0$. And the comment at the end clearly indicates that I took the wrong path. But then how can I prove this?
If $f$ is analytic in a neighbourhood of $z=a$ and $1/f(z)$ has a singularity at $z=a$ that is not a pole of order $1$, then both $f(a) = 0$ and $f'(a)=0$. But here $f(a) = e^a + 3a$ and $f'(a) = e^a + 3$, so...