I've come across this problem on my lecture book of Complex Variable Function,
Expand $f(z) =$ $\frac{e^z}{z^3}$ into a power series on the $0 < |z| < \infty$ domain
so far I've been learning about the Taylor and Laurent series, so I'm thinking to turn the function to this form
$\frac{e^z}{z^3}$ $= e^z .$ $\frac{1}{z^3}$ then by expanding the $e^z$ I get
$\frac{e^z}{z^3}$ $=$ $( z +$ $\frac{z^2}{2!}$ $+$ $\frac{z^3}{3!}$ $+ ... +$ $\frac{z^n}{n!}$ $). $ $\frac{1}{z^3}$
Is this correct? since I'm not really sure about the next step.
any idea of starting point on how to solve this problem would really help, thank you
You are on the right track, but $$e^z=\color{red}1+z +\frac{z^2}{2!}+\frac{z^3}{3!}+ \cdots +\frac{z^n}{n!}+\color{red}\cdots ,$$ so it's
$$\frac{e^z}{z^3}=\left(1+ z +\frac{z^2}{2!}+\frac{z^3}{3!}+ \cdots +\frac{z^n}{n!}+\cdots\right)\cdot \frac{1}{z^3}.$$
Then just distribute the multiplication:
$$\frac{e^z}{z^3}=\dfrac1{z^3}+ \dfrac1{z^2} +\frac{1}{2!z}+\frac{1}{3!}+ \cdots +\frac{z^{n-3}}{n!}+\cdots .$$