Step 1: Simplify the integrand
\[ \frac{z e^{1/z}}{z} = e^{1/z} \] Step 2: Use Laurent series of $e^{1/z}$ around $z = 0$
\[ e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n! z^n} \] So, \[ e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2! z^2} + \dots \] Step 3: Use Cauchy's integral formula
\[ I = \frac{1}{2\pi i} \oint e^{1/z} dz = \text{coefficient of } \frac{1}{z} \text{ in } e^{1/z} = 1 \]
Let \( f: \mathbb{R} \to \mathbb{R} \) \(\text{ be any function defined as }\) \[ f(x) = \begin{cases} x^\alpha \sin \left( \frac{1}{x^\beta} \right) & \text{for } x \neq 0, \\ 0 & \text{for } x = 0, \end{cases} \] where \( \alpha, \beta \in \mathbb{R} \). Which of the following is true? \( \mathbb{R} \) denotes the set of all real numbers.