Proposition

Integrate the function $f(z) = \overline{z}$ over the following paths:

$\gamma(t) = t + it (0 \leq t \leq 1)$.
$\gamma(t) = t + it^2 (0 \leq t \leq 1)$.
$\gamma_1(t) = t$ with $0 \leq t \leq 1$, and $\gamma_2(t) = 1 + it$ with $0 \leq t \leq 1$.

Solution

1

\[\begin{align*} \int_{0}^{1} f(\gamma(t))\gamma'(t) dt &= \int_{0}^{1} (t - it)(1 + i) dt \\ &= \int_{0}^{1} (1 + 1)t dt \\ &= 1. \end{align*}\]

2

\[\begin{align*} \int_{0}^{1} f(\gamma(t))\gamma'(t) dt &= \int_{0}^{1} (t - it^2)(1 + 2it) dt \\ &= \int_{0}^{1} t + it^2 + 2t^3 dt \\ &= 1 + i/3. \end{align*}\]

3

\[\begin{align*} \int_{0}^{1} f(\gamma_1(t))\gamma_1'(t) dt + \int_{0}^{1} f(\gamma_2(t))\gamma_1'(t) dt &= \int_{0}^{1} t dt + \int_{0}^{1} (-i) \cdot i dt \\ &= 1/2 + 1 = 3/2. \end{align*}\]