Solution: The function \( f(z) = \frac{e^z}{z^2 + 1} \) has singularities where \( z^2 + 1 = 0 \), i.e., at \( z = \pm \iota \). Among these, only \( z = \iota \) lies inside the unit circle \( |z| = 1 \). Thus, the integral can be evaluated using the residue theorem which say that the integral of a function \( f(z) \) around a simple closed curve \( C \) is given by: \[ \oint_C f(z) \, \mathrm{d} z = 2\pi \iota \sum_{k=1}^{n} \text{Res}(f, z_k), \] where \( z_k \) are the poles of \( f(z) \) inside \( C \).

In this problem, since there is only one pole $z = \iota $ inside the curve $\vert z \vert = 1$, the integral will be \[ \oint_C \frac{e^z}{z^2 + 1} \, dz = 2\pi \iota \cdot \text{Res}(f, \iota ). \]

Let us find the residue at \( z = \iota \). Note that $z = \iota $ is a simple pole of the function. We recall that the residue of a function \( f(z) \) at a simple pole \( z = z_0 \) is given by: \[ \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z). \] Therefore, \[ \text{Res}(f, \iota ) = \lim_{z \to \iota } (z - \iota ) \frac{e^z}{(z - \iota )(z + \iota )}. \] Simplifying, \[ \text{Res}(f, \iota ) = \lim_{z \to \iota } \frac{e^z}{z + \iota } = \frac{e^{\iota } }{2\iota }. \] By the residue theorem, \[ \oint_C \frac{e^z}{z^2 + 1} \, dz = 2\pi \iota \cdot \frac{e^{\iota } }{2\iota } = \pi e^{\iota } . \]