I want to find the singularities of $$\frac{z-3}{(z^2-9)z}e^{\frac{1}{1+z}}$$ Which at first sight are $-3, 3, 0, -1$.
The expression simplifies to $$\frac{1}{(z+3)z}e^{\frac{1}{1+z}}$$ leading me to my first question: Was $3$ a removable singularity? Or was it never a singularity to begin with?
Now, to classify $-3, 0, -1$, I wanted to use the Laurent series. I'm stuck. Expanding the exponential into a power series gives us the correct form for discussing the singularity at $-1$, i.e., $\sum_{k = 0}^\infty\frac{1}{k!(z+1)^k}$. But how can we incorporate the fraction $\frac{1}{(z+3)z}$? I'm having similar problems when expanding the fraction into a power series. Basically, because the terms are multiplied, I don't know how to handle them.
Is there a clever way to work around even having to use the Laurent series? Or am I approaching calculating it wrong?
Since the function is undefined at $3$, yes, it is a singularity. And, in fact, it is a removable singularity.
Since the limit $0$ is not a removable singularity of $\frac{z-3}{(z^2-9)z}e^{\frac1{1+z}}$, but the limit $\lim_{z\to0}z\times\frac{z-3}{(z^2-9)z}e^{\frac1{1+z}}$ exists (in $\mathbb C$), $0$ is a pole of $\frac{z-3}{(z^2-9)z}e^{\frac1{1+z}}$.
Finally, $-1$ is an essential singularity of $e^{\frac1{1+z}}$. Since $\frac{z-3}{(z^2-9)z}$ is analytic near $-1$, $-1$ is still an essential singularity of $\frac{z-3}{(z^2-9)z}e^{\frac1{1+z}}$.