I am supposed to find the Laurent expansion of
$$g(z) = f(z)/z^3 = (e^{3iz} - 3 e^{iz} +2)/z^3 $$
Now using Taylor I get that
$$ f(z) = \sum_{n=0}^\infty \frac{(3iz)^n}{n!} - 3\frac{(iz)^n}{n!} + 2 = \sum_{n=0}^\infty \frac{(3iz)^n}{n!} - 3\frac{(iz)^n}{n!} + \left( \frac{1}{2}\right)^n $$
Now when $|z|<1$ we can write the Laurent series as
$$ g(z) = \overbrace{\sum_{n=0}^{\infty} (1-z^3)^n}^{\Large z^3} \left( e^{3iz} - 3 e^{iz} +2\right)$$
But I do not know how to continue, the book gives the answer as
$$ g(z) = \sum_{n=2}^\infty \frac{3(3^{n-1}-1)}{n!}i^n z^n = \sum_{n=-1}^\infty \frac{3(3^{n+2}-1)}{(n+3)!}i^{n+3}z^n \quad 0 < |z| < \infty$$
but I do not quite see how to obtain this.