I want to know as to whether a singularity can be removable as well as a pole. If we take the complex function $f(z)= (e^z)/(z+ \sin(z))$ and consider the singularity at $z=0$ then on the face of it, it looks like a simple pole as the denominator has a simple zero at $z=0$.
Now to evaluate the residue we will have to consider the limit $(ze^z/(z+ \sin(z))$ as $z$ tends to $0$. To compute this limit we will have to define $\sin(z)/z$ as $z$ tends to $0$ as $1$. The limit comes out to be $1/2$. Am I being correct in calling this singularity a pole. My confusion arises
from the aforementioned definition of $\sin(z)/z$ as $z$ tends to $0$ which is done in case of removable singularities.
Type of singularity at $z=0$ in $(e^z)/(z+ \sin(z))$
complex-analysissingularity
Best Answer
It sounds to me as if you’re slightly confused about limits. In particular, the sentence “To compute this limit we will have to define $\sin(z)/z$ as $z$ tends to $0$ as $1$” makes no sense. No definition is required here (beyond the general definition of the term “limit”); the limit of $\frac{\sin z}z$ for $z\to0$ is $1$. The expression $\frac{\sin z}z$ is undefined at $z=0$. You can define it to be $1$ there if you like, but that’s not necessary in order to find that its limit for $z\to0$ is $1$.
Everything else that you wrote is correct. The denominator has a simple zero at $z=0$, and thus the function has a simple pole there, and the residue at this pole is $\frac12$, as you correctly determined by finding the limit of $zf(z)$.