# Toronto Math Forum

## MAT334--2020F => MAT334--Tests and Quizzes => Test 2 => Topic started by: RunboZhang on November 04, 2020, 07:03:43 PM

Title: 2020F Test2-ALT-F Q2
Post by: RunboZhang on November 04, 2020, 07:03:43 PM
\textbf{Problem 2:} \\\\ \text{Calculate directly (that means, without Cauchy's or Green's theorem the integral)} \\\\ \begin{gather} \begin{aligned} \int_{L}{Re(z) \,dz} \end{aligned} \end{gather}
$\text{where L is a path, consisting of two acrs of the radii 2, from 1 to -1 and to 1 on the figure (fig is attached below).}$

\textbf{Solution: } \\\\ \text{Denote the path on upper-half plane } L_1 \text{ and the lower-half plane } L_2 \ \text{, we have:} \\\\ \begin{gather} \begin{aligned} L_1 = -i\sqrt{3}+2e^{it_1}, \ t_1 \in [\pi/3, 2\pi/3] \\\\ L_2 = i\sqrt{3}+2e^{it_2}, \ t_2 \in [4\pi/3, 5\pi/3] \end{aligned} \end{gather}

\text{Now we calculate } L_1 \ \text{ and } \ L_2 \text{respectively: }\\\\ \begin{gather} \begin{aligned} \int_{L_1}{Re(-i\sqrt{3}+2e^{it}) \cdot d(-i\sqrt{3}+2e^{it})} \,dt &= \int_{\pi/3}^{2\pi/3}{2cos(t)\cdot i2e^{it}}\,dt\\\\ &= i\cdot 4 \int_{\pi/3}^{2\pi/3}{cos(t)e^{it}} \,dt\\\\ &= i\cdot 4 \int_{\pi/3}^{2\pi/3}{\frac{1}{2}\cdot (e^{it}+e^{-it})\cdot e^{it}} \,dt\\\\ &= i\cdot 2 \int_{\pi/3}^{2\pi/3}{(e^{2it}+1)} \,dt\\\\ &= i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2}) \end{aligned} \end{gather}

\begin{gather} \begin{aligned} \int_{L_2}{Re(i\sqrt{3}+2e^{it}) \cdot d(i\sqrt{3}+2e^{it})} \,dt &= \int_{4\pi/3}^{5\pi/3}{2cos(t)\cdot i2e^{it}}\,dt\\\\ &= i\cdot 4 \int_{4\pi/3}^{5\pi/3}{cos(t)e^{it}} \,dt \\\\ &= i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2}) \end{aligned} \end{gather}

\text{Lastly,} \\\\ \\\\ \begin{gather} \begin{aligned} \int_{L}{Re(z)} \,dz &= \int_{L_1}{Re(z)} \,dz + \int_{L_2}{Re(z)} \,dz\\\\ &= i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2}) + i\cdot 2(\frac{\pi}{3} - \frac{\sqrt{3}}{2})\\\\ &= i\frac{4}{3}\pi - i2\sqrt{3} \end{aligned} \end{gather}