Let's solve the problems step by step: --- ### (a) What type of singularity is \( z = 0 \)? Provide reasons for your answer. #### **Step 1: Analyze the function near \( z = 0 \)** - The function is \( f(z) = \frac{1}{\sin(1/z)} \). - At \( z = 0 \), \( 1/z \) becomes unbounded (goes to infinity). - The sine function, \(\sin(1/z)\), oscillates infinitely as \( z \to 0 \), and is not defined at \( z = 0 \). **Explanation:** \( f(z) \) does not have a limit as \( z \to 0 \) because \(\sin(1/z)\) oscillates infinitely often between -1 and 1, so the denominator never settles to a value or zero in any regular way. #### **Step 2: Classify the singularity** - At \( z = 0 \), the function does not have a pole (since there is no finite order zero or pole at \( z = 0 \)). - The function also cannot be made analytic by removing the point. - The singularity is **essential**, as the behavior is highly oscillatory and not removable or a pole. **Explanation:** A singularity is essential if the limit does not exist and the function exhibits extreme oscillatory behavior, as happens here due to \( \sin(1/z) \). **Final answer for (a):** > **The singularity at \( z = 0 \) is an essential singularity, because \( \sin(1/z) \) oscillates infinitely as \( z \to 0 \), making the function's behavior highly non-regular.** --- ### (b) What type of singularity is \( z = \frac{1}{2\pi} \)? Provide reasons for your answer. #### **Step 1: Analyze the function at \( z = \frac{1}{2\pi} \)** - Substitute \( z = \frac{1}{2\pi} \): \( 1/z = 2\pi \). - \( \sin(2\pi) = 0 \), so the denominator becomes zero. **Explanation:** At this value, the denominator is zero, which may indicate a pole if the numerator is nonzero. #### **Step 2: Classify the singularity** - Near \( z = \frac{1}{2\pi} \), set \( z = \frac{1}{2\pi} + \epsilon \). - Then, \( 1/z \approx 2\pi - 2\pi^2 \epsilon \) (using Taylor expansion). - \( \sin(1/z) \approx \sin(2\pi - 2\pi^2\epsilon) \approx -2\pi^2 \epsilon \) (since \( \sin(x) \approx x \) near 0). - So, \( f(z) \approx \frac{1}{-2\pi^2 \epsilon} \), which diverges as \( \epsilon \to 0 \). **Explanation:** This is the behavior of a simple pole. **Final answer for (b):** > **The singularity at \( z = \frac{1}{2\pi} \) is a simple pole, because the denominator vanishes linearly as \( z \to \frac{1}{2\pi} \), causing the function to diverge as \( 1/(z - \frac{1}{2\pi}) \).** --- ## **Summary of Answers** - (a) \( z = 0 \) is an **essential singularity**. - (b) \( z = \frac{1}{2\pi} \) is a **simple pole**.