24-10-2023
Problem: Let $\gamma $ be the counterclockwise oriented triangle with vertices $0,2$ and $1 + \iota $.
- Write the parametrization for each of the three smooth parts of $\gamma$.
- Compute $\displaystyle \int\limits_{\gamma }\mathop{\mathrm{Re}}(z)^2 \mathrm{d}z$.
Solution: Look at the figure below. This represents the given curve with counterclockwise orientation.
- The parametrization for each of the three smooth parts of $\gamma $ is given by \begin{align*} & \gamma _1: [0,1] \to \mathbb{C}, \ t \mapsto 2t \\ & \gamma _2: [0,1] \to \mathbb{C}, \ t \mapsto (2-t) + t \iota \\ & \gamma _3: [0,1] \to \mathbb{C}, \ t \mapsto (1 + \iota)(1 - t) . \end{align*}
- The integral will be \begin{equation}\label{eq:24Oct2023-1} \int\limits_\gamma \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z = \int\limits_{\gamma_1} \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z + \int\limits_{\gamma_2} \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z + \int\limits_{\gamma_3} \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z \end{equation} Let us compute the integral over each path. \begin{align*} \int\limits_{\gamma_1} \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z & = \int\limits_{0}^{1} (2t)^2(2 \cdot \mathrm{d}t) = 8 \int\limits_{0}^{1}t^2 \mathrm{d}t = 8 \left[ \frac{t^3}{3} \right]_0^1 = \frac{8}{3}. \end{align*} \begin{align*} \int\limits_{\gamma_2} \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z & = \int\limits_{0}^{1} (2-t)^2((-1 + \iota ) \cdot \mathrm{d}t) \\ & = (-1 + \iota )\int\limits_{0}^{1}(t-2)^2 \mathrm{d}t \\ & = (-1 + \iota )\left[ \frac{(t-2)^3}{3} \right]_0^1 \\ & = (-1 + \iota ) \left( -\frac{1}{3} + \frac{8}{3} \right) = \frac{7}{3}(-1 + \iota). \end{align*} Finally, \begin{align*} \int\limits_{\gamma_3} \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z & = \int\limits_{0}^{1} (1-t)^2((-1 - \iota ) \cdot \mathrm{d}t) \\ & = (-1 - \iota )\int\limits_{0}^{1}(t-1)^2 \mathrm{d}t \\ & = (-1 - \iota )\left[ \frac{(t-1)^3}{3} \right]_0^1 \\ & = (1 + \iota ) \left( 0 + \frac{1}{3} \right) = \frac{1}{3}(-1 - \iota). \end{align*} Therefore, using \eqref{eq:24Oct2023-1}, \[ \int\limits_\gamma \mathop{\mathrm{Re}}(z)^2 \mathrm{d}z = \frac{8}{3} + \frac{7}{3}(-1 + \iota ) + \frac{1}{3}(-1 - \iota ) = 2\iota . \]