Integral of $e^\left(1/z^2\right)$ around $|z|=1$ in the complex plane

Question

$e^\left(1/z^2\right)$ has an essential singularity at $0.$ Don't know how to do this integral.

Answer 1

Draw the circle $z=\exp{i\theta}$. For any given value of $\theta$ between $0$ and $\pi$ compare $\exp(1/z^2) \Delta z$ between $\theta$ and $\theta+\Delta \theta$ with $\exp(1/z^2) \Delta z$ between $\theta+\pi$ and $\theta+\pi+\Delta \theta$. You should see they are negatives of each other, since the integrand is an even function, forcing a zero Riemann sum. Thus, the contour integral must be zero.

We can adapt this symmetry based argument to prove that when you integrate a Laurent series around an isolated singularity at $z=z_0$ all terms will give zero, except for the term with $(z-z_0)^{-1}$. That's how contour integrals get reduced to residue values based on the coefficient of $(z-z_0)^{-1}$. In this problem that coefficient is zero so the contour integral is zero, too.