Author Topic: 2020F Test2-ALT-F Q2  (Read 247 times)

RunboZhang

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2020F Test2-ALT-F Q2
« on: November 04, 2020, 07:03:43 PM »
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\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).}
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\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}
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$
\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}
$

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\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}
$