Evaluating the Contour Integral ∮C e^1/z dz

This problem requires evaluating a contour integral using the Residue Theorem from complex analysis. The contour C is the unit circle traversed counter-clockwise.

Identifying the Singularity and Contour

• The function is $f(z) = e^{1/z}$.
• The function has a singularity at $z=0$.
• The contour C is the unit circle, $|z|=1$.
• The singularity $z=0$ lies inside the contour C.

Applying the Residue Theorem

The Residue Theorem states that $\oint_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k)$, where $z_k$ are the singularities inside C.

In this case, the only singularity inside C is at $z=0$. We need to find the residue of $e^{1/z}$ at $z=0$.

Finding the Residue at z=0

1. Find the Laurent series expansion of $e^{1/z}$ around $z=0$.
2. Recall the Maclaurin series for $e^w$: $e^w = \sum_{n=0}^\infty \frac{w^n}{n!} = 1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \dots$
3. Substitute $w = 1/z$: $e^{1/z} = \sum_{n=0}^\infty \frac{(1/z)^n}{n!} = \sum_{n=0}^\infty \frac{1}{n! z^n} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots$
4. The residue is the coefficient of the $z^{-1}$ term in the Laurent series.
5. The coefficient of $z^{-1}$ (when $n=1$) is $\frac{1}{1!} = 1$.
6. Therefore, $\text{Res}(e^{1/z}, 0) = 1$.

Calculating the Contour Integral

Using the Residue Theorem:

Since $i = \sqrt{-1}$, the result can be written as $2\pi\sqrt{-1}$.

Conclusion

The value of the contour integral ∮C e^1/z dz is $2\pi i$, which matches option C, $2\pi\sqrt{-1}$.