Let's solve the problem step by step: --- **Step 1: Identify the type of isolated singularity at \( z = 0 \) for \( g(z) = \sin(1/z) \).** **Explanation:** - The function \( g(z) = \sin(1/z) \) is not defined at \( z = 0 \), but is holomorphic elsewhere in any punctured neighborhood around \( z = 0 \). - Expand \( \sin(1/z) \) using its Taylor series: \[ \sin(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{1}{z}\right)^{2n+1} = \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \cdots \] - The Laurent series about \( z = 0 \) has infinitely many negative powers of \( z \) and **no** lowest power. - Such a singularity is called an **essential singularity**. **Step 1 Final Answer:** **\( z = 0 \) is an *essential singularity* for \( g(z) = \sin(1/z) \), because its Laurent expansion contains infinitely many negative powers of \( z \), and the function exhibits wild behavior near \( z = 0 \).** --- **Step 2: Calculate \( \text{Res}_{z=0} g(z) \), i.e., the residue of \( g(z) \) at \( z = 0 \).** **Explanation:** - The residue is the coefficient of \( 1/z \) in the Laurent expansion around \( z = 0 \). - From the expansion: \[ \sin(1/z) = \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \cdots \] - The coefficient of \( 1/z \) is **1**. **Step 2 Final Answer:** **The residue of \( g(z) \) at \( z = 0 \) is \( \boxed{1} \).** --- **Summary of answers:** 1. \( z = 0 \) is an essential singularity for \( g(z) = \sin(1/z) \). 2. The residue at \( z = 0 \) is \( 1 \).