Singularity - Removable or Pole?

Question

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.

Answer 1

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.

Answer 2

The easiest thing in this cases (for me) is just to calculate the principal part of the Laurent expansion at zero.

$\sin (3z) = 3z-9z^3/2+...$ so $f(z)= 3/z-9z/2-3/z +h.o.t. = -9z/2 +h.o.t.$

In particular, the principal part of the Laurent expansion is zero and hence there is a removable singularity at zero (residue $= 0$).

Answer 3

Or simply Taylor ... $\sin(3z)=3z+o(z^2)$, so $$\lim_{z\to0}\frac{\sin(3z)-3z}{z^2}=\lim_{z\to0}\frac{o(z^2)}{z^2}=0\;.$$ Hence, the function extends holomorphically to $z=0$.