We are tasked with finding and classifying the singularities of the function $f(z) = \sin\left(\frac{1}{z}\right) \cos\left(\frac{1}{z}\right)$. Let’s go through the steps one by one.

Step 1: Analyze the function

We start by noting that the given function can be rewritten using a trigonometric identity. Recall the identity:

$\sin(A) \cos(A) = \frac{1}{2} \sin(2A)$

Applying this to $A = \frac{1}{z}$, we get:

$f(z) = \sin\left(\frac{1}{z}\right) \cos\left(\frac{1}{z}\right) = \frac{1}{2} \sin\left(\frac{2}{z}\right)$

Thus, the function becomes:

$f(z) = \frac{1}{2} \sin\left(\frac{2}{z}\right)$

Step 2: Identify the singularities

To identify the singularities of $f(z)$, we look for points where the function is not analytic. In this case, the function $\sin\left(\frac{2}{z}\right)$ involves $\frac{2}{z}$, which becomes problematic when $z = 0$, as division by zero occurs.

Thus, the point $z = 0$ is a singularity of the function. No other singularities exist because the sine function $\sin\left(\frac{2}{z}\right)$ is analytic everywhere else in the complex plane, except at $z = 0$.

Step 3: Classify the singularity at $z = 0$

Now we need to classify the singularity at $z = 0$. Singularities can generally be classified as one of the following:

1. Removable singularity
2. Pole
3. Essential singularity

To classify the singularity at $z = 0$, we investigate the behavior of $f(z)$ as $z \to 0$.

Step 3.1: Behavior of $f(z)$ as $z \to 0$

The function $f(z) = \frac{1}{2} \sin\left(\frac{2}{z}\right)$ oscillates wildly as $z \to 0$. The sine function $\sin\left(\frac{2}{z}\right)$ has no limit as $z$ approaches 0; instead, it exhibits an essential singularity at $z = 0$ because the function does not tend to infinity or a finite limit, and instead oscillates between all possible values in an unpredictable way near $z = 0$.

Step 3.2: Laurent series near $z = 0$

We can further confirm that $z = 0$ is an essential singularity by examining the Laurent series for $\sin\left(\frac{2}{z}\right)$. The Laurent series expansion of $\sin\left(\frac{1}{z}\right)$ around $z = 0$ involves an infinite number of negative powers of $z$, which is a characteristic of an essential singularity.

Thus, $z = 0$ is an essential singularity.

Conclusion

The function $f(z) = \sin\left(\frac{1}{z}\right) \cos\left(\frac{1}{z}\right)$ has a singularity at $z = 0$, and this singularity is classified as an essential singularity.

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Would you like a detailed explanation of any of the steps or a clarification on singularity classifications? Here are five related questions you might find helpful:

1. How do you compute the Laurent series of a function around a singularity?
2. What is the difference between an essential singularity and a pole?
3. Can you classify singularities using the residue of a function?
4. How does the behavior of $\sin\left(\frac{1}{z}\right)$ as $z \to 0$ confirm an essential singularity?
5. What are examples of other functions with essential singularities?

Tip: A function with an essential singularity can take on almost every complex value infinitely often near that point (Picard's Theorem).