Proposition

Compute the following integrals, where $\gamma$ is the boundary of the square with vertices at $\pm 4 \pm 4i$, positively oriented:

$\int_{\gamma}\frac{\exp(z^2)}{z^3} dz$
$\int_{\gamma}\frac{\exp(3z)}{(z - \pi i)^2} dz$
$\int_{\gamma}\frac{\sin(2z)}{(z - \pi)^2} dz$
$\int_{\gamma}\frac{\exp(z)\cos(z)}{(z - \pi)^3} dz$.

Solution

1

Let $f(z) = \exp(z^2)$. By Theorem 5.1 [A first course in complex analysis], $f’‘(0) = \frac{1}{\pi i} \int_{\gamma} \frac{f(z)}{(z - 0)^3} dz$. Therefore, $\int_{\gamma}\frac{\exp(z^2)}{z^3} dz = \pi i f’‘(0) = 2$.

2

Let $f(z) = \exp(3z)$. Using the same idea as above, the integral equals $2\pi i f’(\pi i) = 6\pi i\exp(3\pi i)$.

3

Let $f(z) = \sin(2z)$. Using the same idea, the integral equals $2\pi i f’(\pi) = 2$.

4

Let $f(z) = \exp(z)\cos(z)$. $f’(z) = \exp(z)(\cos(z) - \sin(z))$ and $f’‘(z) = -2\sin(z)\exp(z)$. Therefore, the integral equals $2\pi i f’’(\pi) = 0$.