Write the Laurent series around zero for the entire function $f(z)=z^2e^{3z}$
I'm a little confused on how to represent the complex functions by series, as I did in the calculation of real functions, but do not know if it's right
$$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}\rightarrow e^{3z}=\sum_{n=0}^\infty \frac{3^nz^n}{n!}\rightarrow z^2e^{3z}=\sum_{n=0}^\infty \frac{3^nz^{n+2}}{n!}$$
ii) Find the Laurent series representation for $f(z)=z^2\sin(\frac{1}{z^2})$ where $0<|z|<\infty$
For the second part, you may write $$ \sin\left(\frac{1}{z^2}\right)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!\:z^{2(2n+1)}},\quad z \neq0, $$ giving $$ f(z)=z^2\sin\left(\frac{1}{z^2}\right)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!\:z^{4n}},\quad z \neq0. $$ since $\displaystyle u \longmapsto \sin(u) $ has a power series with an infinite radius of convergence.