Holomorphic function that has a triple pole at $0$, a simple pole at $1$, an essential singularity at $i$ and at $-i$

Question

Find a holomorphic function that has a triple pole at $0$, a simple pole at $1$, an essential singularity at $i$ and at $-i$. I know that $f(z)=\frac{1}{z^3}+\frac1{z-1} + e^{\frac{1}{z^2+1}}$ is such a function but is $g(z)=\frac{1}{z^3(z-1)} e^{\frac{1}{z^2+1}}$ also one ? I would tend to yes because $\lvert f(z)\rvert \to +\infty $ when $z\to 1$ or $z \to 0$ but I am not sure for the orders of poles and if taking a product would change things for Laurent series development since it is not the same as taking a sum.

Answer 1

Yes, that will work. Near $0$, $g$ is the product of $\frac1{z-1}$ with a holomorphic function of which $1$ is not a zero; therefore, $g$ has a simple pole at $1$. And the same argument shows that $0$ is a triple pole of $g$.

What could go wrong at $\pm i$? It could happen that $g$ had a pole or a removable singularity there. In other words, it could be a meromorphic function near $\pm i$. But that cannot happen, since$$e^{1/(z^2+1)}=z^3(z-1)g(z)$$and the product of two meromorphic functions is again a meromorphic functions. And meromorphic functions do not have essential singularities.

Answer 2

The function $e^{1\over 1+z^2}$ is holomorphic in $|z|<1$ and in $|z-1|<1.$ Therefore the order of the poles in the product is the same as in ${1\over z^3(z-1)}.$ For example the order of a pole $0$ is the smallest natural number $k$ such that $z^kg(z)$ extends to a holomorphic function in a neighborhood of $0.$