**Step 1: Write the power series for \( e^z \) and form \( f(z) \)** Recall the Taylor (Maclaurin) series for \( e^z \) about \( z=0 \): \[ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{z^n}{n!} \] So, \( e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots = \sum_{n=1}^{\infty} \frac{z^n}{n!} \). Now, \( f(z) = \frac{e^z - 1}{z} \): \[ f(z) = \frac{1}{z} \sum_{n=1}^{\infty} \frac{z^n}{n!} = \sum_{n=1}^{\infty} \frac{z^{n-1}}{n!} \] **Explanation:** We expressed \( e^z \) as a power series and subtracted 1 to get \( e^z - 1 \), then divided each term by \( z \) to find the series for \( f(z) \). --- **Step 2: Write the explicit form of the Laurent expansion and analyze the singularity** Let’s write a few terms of the expansion: \[ f(z) = \frac{z^0}{1!} + \frac{z^1}{2!} + \frac{z^2}{3!} + \cdots = 1 + \frac{z}{2} + \frac{z^2}{6} + \cdots \] There are **no negative powers of \( z \)**. Thus, the Laurent expansion is actually a Taylor series (regular power series) about \( z=0 \). **Explanation:** Since the series contains no negative powers of \( z \), \( f(z) \) is analytic at \( z=0 \); there is no pole or essential singularity. --- ### Final Answer The Laurent expansion of \( f(z) = \frac{e^z - 1}{z} \) about the origin is: \[ f(z) = 1 + \frac{z}{2} + \frac{z^2}{6} + \cdots = \sum_{n=1}^{\infty} \frac{z^{n-1}}{n!} \] **\( f(z) \) does *not* have a singularity at \( z=0 \);** it is analytic there. In fact, \( z=0 \) is a removable singularity, and the function is entire (analytic everywhere in \( \mathbb{C} \)).