Holomorphic Function $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$

Question

Let $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$

Prove that this function is holomorphic in $\mathbb{C} \setminus \{0\}$ and calculate $$\int_{|z|=2} g(z)\,\mathrm{d}z$$

Edit:

What I have done: \begin{align} \begin{split} g(z) & = \sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}\\ & = (2-3i)\cdot \sum_{n=1}^\infty \dfrac{1}{n! z^n}\\ & = (2-3i)\cdot \sum_{n=1}^{\infty} \dfrac{z^{-n}}{n!}\\ & = (2-3i)\cdot \sum_{n=0}^{\infty} \dfrac{z^{-n}}{n!} - \dfrac{z^{-0}}{0!}\\ & = (2-3i)\cdot (e^{z^{-1}} -1)\\ & = (2-3i) e^{z^{-1}} -(2-3i) \\ & = 2e^{z^{-1}} - 3ie^{z^{-1}} +3i - 2 \end{split} \end{align}

let $z^{-1}=z^*$ for a $z^* = x+ iy$

Then let's see $2e^{z^{*}} - 3ie^{z^{*}} +3i - 2$ is holomorphic

Proof:

$ 2e^{z^{*}} - 3ie^{z^{*}} +3i - 2$

$ = 2e^{x + iy} - 3ie^{x + iy} +3i - 2 $

$ = 2e^{x} e^{iy} - 3ie^{x}e^{iy} +3i - 2 $

$ = 2e^{x} [ \cos(y) + i \sin(y) ] - 3 ie^{x} [ \cos(y) + i \sin(y) ] +3i - 2 $

$ = 2e^{x}\cos(y) + 2 i e^{x} \sin(y) - 3 ie^{x} \cos(y) - 3 i^2 e^{x} \sin(y)+3i-2 $

$ = 2e^{x}\cos(y) + 2 i e^{x} \sin(y) - 3 ie^{x} \cos(y) + 3 e^{x} \sin(y) +3i - 2 $

$ = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2 + 2 i e^{x} \sin(y) - 3 i e^{x} \cos(y) +3i $

$ = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2 + i[2 e^{x} \sin(y) - 3 e^{x} \cos(y) +3] $

Now let $u(x,y) = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2$

And let $v(x,y) = 2 e^{x} \sin(y) - 3 e^{x} \cos(y) +3$

Answer 1

It was discussed in the comments that $g(z)=(2-3i)(e^{1/z}-1)$ is holomorphic in the complex plane excluding the origin. The function $e^{1/z}$ is a composition of holomorphic functions, thus holomorphic.

The residue is $2-3i$ so the integral evaluates to $2 \pi i \cdot(2-3i)$.