The given problem asks for the nature of the point $z = 0$ for the function:

$f(z) = \frac{z - \sin(\sin(z))}{z^3}.$

Step 1: Identifying potential issues at $z = 0$

The function is expressed as a fraction, and we are evaluating the behavior around $z = 0$. At $z = 0$, the denominator is 0, which suggests a singularity or an issue in defining the function directly. To classify the nature of the singularity, we need to analyze the behavior of both the numerator and denominator near $z = 0$.

Step 2: Expanding the numerator using Taylor series

For small $z$, we can approximate $\sin(z)$ and $\sin(\sin(z))$ using their Taylor series expansions:

1. Taylor expansion of $\sin(z)$: $\sin(z) = z - \frac{z^3}{6} + \mathcal{O}(z^5).$
2. Taylor expansion of $\sin(\sin(z))$: Substituting the first few terms of $\sin(z)$ into itself: $\sin(\sin(z)) = \sin\left(z - \frac{z^3}{6} + \mathcal{O}(z^5)\right) = z - \frac{z^3}{6} + \mathcal{O}(z^5).$

Thus: $z - \sin(\sin(z)) = z - \left(z - \frac{z^3}{6} + \mathcal{O}(z^5)\right) = \frac{z^3}{6} + \mathcal{O}(z^5).$

Step 3: Simplifying the function $f(z)$

We now substitute the above expansion into the function:

$f(z) = \frac{\frac{z^3}{6} + \mathcal{O}(z^5)}{z^3} = \frac{1}{6} + \mathcal{O}(z^2).$

As $z \to 0$, the function $f(z)$ approaches $\frac{1}{6}$, which means it has a finite limit at $z = 0$.

Step 4: Classifying the singularity

Since the function $f(z)$ has a finite value at $z = 0$, the singularity at this point is removable. That is, we can redefine the function at $z = 0$ to be $f(0) = \frac{1}{6}$, making the function continuous.

Final Answer

The correct classification for $z = 0$ is:

Removable singularity.

Would you like further details or additional examples of different types of singularities?

Related Questions:

1. What is the difference between a pole and a removable singularity?
2. How do you compute Laurent series for complex functions?
3. What are essential singularities and how do they differ from poles?
4. Can a singularity be both isolated and removable?
5. How is the behavior of a function near a singularity related to its residue?

Tip: Removable singularities occur when the limit of a function exists at a point where it is not originally defined.