Proposition

Integrate the following functions over the circle $C[0, 2]$:

$f(z) = z + \overline{z}$.
$f(z) = \frac{1}{z^4}$.
$f(z) = z^2 - 2z + 3$.
$f(z) = xy$.

Solution

The circle can be parametrized as $\gamma(t) = 2\exp(it) = 2\cos(t) + 2i\sin(t)$. Then $\gamma’(t) = 2i\exp(it) = -2\sin(t) + 2i\cos(t)$.

1

\[\begin{align*} \int_{0}^{2\pi} f(\gamma(t))\gamma'(t)dt &= \int_{0}^{2\pi} -8\sin t \cos t + 8i \cos^2(t) dt \\ &= \int_{0}^{2\pi} -4\sin 2t + 4i (\cos(2t) + 1) dt \\ &= 2\cos(2t) + 2i\sin(2t) + 4ti \Big\vert^{2\pi}_0 \\ &= 8\pi i. \end{align*}\]

2

\[\begin{align*} \int_{0}^{2\pi} f(\gamma(t))\gamma'(t)dt &= \int_{0}^{2\pi} \frac{2i}{16e^{3it}} dt \\ &= \frac{ie^{-3it}}{-24i}\Big\vert^{2\pi}_0 \\ &= 0. \end{align*}\]

3

\[\begin{align*} \int_{0}^{2\pi} f(\gamma(t))\gamma'(t)dt &= \int_{0}^{2\pi} (4e^{2it} - 4e^{it} + 3)2ie^{it} dt \\ &= \int_{0}^{2\pi} 8ie^{3it} - 8ie^{2it} + 6ie^{it} dt \\ &= 2e^{4it} - 4e^{2it} + 6e^{it} \Big\vert^{2\pi}_0 = 0. \end{align*}\]

4

\[\begin{align*} \int_{0}^{2\pi} f(\gamma(t))\gamma'(t)dt &= 2\int_{0}^{2\pi} - \cos t \sin^2t + i\cos^2 t \sin t dt \\ &= 2(-\frac{\sin^3t}{3}\Big\vert^{2\pi}_0 + i\frac{\cos^3t}{-3}\Big\vert^{2\pi}_0) \\ &= 2 \cdot 0 = 0. \end{align*}\]