To solve the integral of the function $f(z) = \exp(-z^2)$ counterclockwise around the unit circle, we need to consider both the integral and whether Cauchy's Integral Theorem applies.

1. Definition and Context

Let's consider the contour integral around the unit circle $C$ in the complex plane: $\oint_C f(z) \, dz = \oint_C \exp(-z^2) \, dz$

Here, $C$ is the unit circle centered at the origin, described by $|z| = 1$.

2. Cauchy's Integral Theorem

Cauchy's Integral Theorem states that if a function $f(z)$ is analytic (holomorphic) throughout a simply connected domain $D$ that contains the contour $C$ and its interior, then the integral of $f(z)$ around $C$ is zero: $\oint_C f(z) \, dz = 0$ This theorem only applies if $f(z)$ is analytic in the domain containing $C$ and the interior of $C$.

3. Analyticity of $f(z) = \exp(-z^2)$

The function $f(z) = \exp(-z^2)$ is an entire function, meaning it is analytic everywhere in the complex plane, including within and on the unit circle $|z| = 1$.

4. Applying Cauchy's Integral Theorem

Since $f(z)$ is analytic everywhere, particularly within the domain containing the unit circle $C$ and its interior, Cauchy's Integral Theorem directly applies. Therefore, the contour integral is zero: $\oint_C \exp(-z^2) \, dz = 0$

5. Conclusion

The integral of $\exp(-z^2)$ around the unit circle is zero by Cauchy's Integral Theorem, because the function is analytic everywhere in the complex plane.

Questions for Further Exploration

1. What is the general form of Cauchy’s Integral Theorem?
2. How would the integral change if the function had a singularity inside the unit circle?
3. Can the result be extended to other contours besides the unit circle?
4. What happens if we integrate around a contour that encloses a singularity?
5. How does the result relate to the residue theorem?

Tip: Always check whether a function is analytic in the domain before applying Cauchy's Integral Theorem.