Calculation of the Residue at $z=0$ of $\sin(1/z)/(z-3)$?

Question

I am trying to calculate the residue of $\frac{\sin(1/z)}{(z-3)}$ at $z=0$. However I am not sure how to start.

My idea is that I write $\sin(\frac{1}{z})= (\frac{1}{z}) - \frac{(\frac{1}{z})^3}{3!} + \frac{(\frac{1}{z})^5}{5!} -+ o((\frac{1}{z})^7)$ and since the residue can be found as the coefficient of the Laurent series at $\frac{1}{z}$ therefore obtain the residue.

Thus $\frac{\sin(1/z)}{(z-3)} = (\frac{1}{z-3}) * (\frac{1}{z}) - \frac{(\frac{1}{z})^3}{3!} + \frac{(\frac{1}{z})^5}{5!} -+ o((\frac{1}{z})^7) = \frac{1}{z(z-3)} - \frac{1}{3!z^3(z-3)} + \frac{1}{5!z^5(z-3)} -+ o((\frac{1}{z})^8) = \frac{1}{z} * \frac{1}{(z-3)} - \frac{1}{3!z^2(z-3)} + \frac{1}{5!z^4(z-3)} -+ o((\frac{1}{z})^7) $ which would make my residue $\frac{1}{(z-3)} - \frac{1}{3!z^2(z-3)} + \frac{1}{5!z^4(z-3)} -+ o((\frac{1}{z})^7)$ However now I am stuck and not sure if I am even on the right path any help would be greatly appreciated.

Answer 1

Since, near $0$,$$\frac1{z-3}=-\frac13-\frac z{3^2}-\frac{z^2}{3^3}-\frac{z^3}{3^4}+\cdots$$and$$\sin\left(\frac1z\right)=\frac1z-\frac1{3!z^3}+\frac1{5!z^5}-\frac1{7!z^7}+\cdots,$$then, since the residue is the coefficient of $\frac1z$ of the Laurent expansion of $\frac{\sin\left(\frac1z\right)}{z-3}$ near $0$, it is equal to$$-\left(\frac13-\frac1{3!3^3}+\frac1{5!3^5}-\frac1{7!3^7}+\cdots\right)=-\sin\left(\frac13\right).$$Note: I had some help here; read the comments.

Answer 2

You want the coeff of the $\dfrac1z$ term.

Now, $\dfrac1{z-3}=-\dfrac 13\dfrac 1{(1-\frac z3)}=-\dfrac 13\sum_n(-1)^n(\frac z3)^n$.

So, we take the product: $\sin (\frac1z)(\frac1{z-3})=(\sum_n(-1)^n\frac{z^{-(2n+1)}}{(2n+1)!})\cdot(-\dfrac 13) (\sum_n(-1)^n(\frac z3)^n)$.

So the residue is $\sum_n\frac{(-1)^n}{(-3)^{2n+1}(2n+1)!}=\sin(-\frac13)$.