For the complex-valued function
$$f(z) = \left(\frac{\sin 3z}{z^2}-\frac{3}{z}\right)$$
classify the singularity at $z=0$ and calculate its residue.
Attempt at Solution
Rewriting $f(z) = \left(\frac{\sin (3z) – 3z}{z^2}\right)$, I'm not sure whether the singularity at 0 is removable or a pole because although both numerator and denominator vanish at $z=0$, the sine function is involved and the degree in the denominator is $2$.
Assuming it's a double pole at $z=0$, I calculated the residue to be $0$.
Comments & clarifications welcome. Thank you.
Best Answer
Another way:
$$\lim_{z\to 0}\left(\frac{\sin 3z}{z^2}-\frac{3}{z}\right)=\lim_{z\to 0}\frac{\sin 3z-3z}{z^2}\stackrel{\text{L'Hospital}}=\lim_{z\to 0}\frac{3\cos 3z-3}{2z}\stackrel{\text{L'H}}=\lim_{z\to 0}\frac{-9\sin 3z}{2}=0$$
So the singularity is a removable one.