Singularities of $f(z) = \frac{e^z - e}{(z^2 - 2z + 1)} + z^3\sin(\frac1z).$

Question

In the solution section of my textbook it is said that the function

$$f(z) = \frac{e^z - e}{(z^2 - 2z + 1)} + z^3\sin(\frac1z).$$

has a pole of order $1$ at $z_0 = 1$. I don't understand why this is so. The function $f(z)$ can be written as

$$f(z) = \frac{e^z - e}{(z-1)^2} + z^3\sin(\frac1z)$$

and therefore, in my opinion, $f(z)$ has a pole of order $2$ at $z_0 = 1$. I did find the correct answer that $0$ is an essential singularity by expanding $\sin(\frac1z)$ in a Laurent series but I don't understand why $1$ is a pole of order $1$. Is it possible that it's a mistake?

Answer 1

The solution in your textbook is correct.

Recall that $z_0$ is a pole of order $k$ if and only if $$\lim_{z \to z_0}(z - z_0)^k f(z) \neq 0 \quad \text{and} \quad \lim_{z \to z_0}(z - z_0)^{k+1} f(z) = 0.$$

One has

$$\lim_{z \to 1}(z - 1)f(z) = \lim_{z \to 1}\frac{e^z - e}{(z-1)} = e \neq 0$$

and

$$\lim_{z \to 1}(z - 1)^2f(z) = 0.$$

So $z_0 = 1$ is a pole of order $1$.

Answer 2

Note that the Lauren expansion of $\frac{e^z-e}{(z-1)^2}$ is given by

$$\frac{e^z-e}{(z-1)^2} =\frac{e}{z-1}+e\sum_{n=0}^\infty \frac{(z-1)^n}{(n+2)!}$$