Calculating the residue of $\frac{10z^4-10\sin(z)}{z^3}, z(0) = 0$

Question

$$\frac{10z^4-10\sin(z)}{z^3}, \quad z(0) = 0.$$

I've gotten that $$\operatorname{Res} = 0$$ but I'm not quite sure if that is correct, or if I have even used a correct pathway towards it. How should one work around sine (or cosine, for that matter) during residue calculations?

Answer 1

One way to work with sines and cosines in this problem is to reckon with their power series expansions found in many places, for instance here. You might also note that an even function has only even powers in its Laurent series whereas the residue comes from the $(-1)$ power, therefore even functions will have a residue of zero. Hmm, can you see why I would mention that last fact?

Answer 2

Your answer is correct. The residue of a function is the coefficient of $z^{-1}$ of the Laurent series expansion of the function.

We obtain \begin{align*} \color{blue}{[z^{-1}]}&\color{blue}{\frac{10z^4-10\sin z}{z^3}}\\ &=[z^{-1}]\left(10z-\frac{10}{z^3}\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots\right)\right)\\ &=-10[z^{-1}]\left(\frac{1}{z^2}-\frac{1}{3!}+\frac{z^2}{5!}-\cdots\right)\\ &\,\,\color{blue}{=0} \end{align*}