Author Topic: TT1 Problem 4 (night)  (Read 944 times)

Victor Ivrii

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TT1 Problem 4 (night)
« on: October 19, 2018, 04:16:57 AM »
Calculate integral $\displaystyle{\int_L {\bar{z}^2}}\,dz $ where $L$ is a circular arc shown on the figure below:

Min Gyu Woo

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Re: TT1 Problem 4 (night)
« Reply #1 on: October 19, 2018, 08:55:00 AM »
The equation for the curve is defined to be

$$L(t) = 3i + 3e^{it}$$ for $\frac{\pi}{2} \geq t\geq \frac{-\pi}{2}$

Also, $L'(t)=3ie^{it}$

We're given that $f(z)=\overline{z}^2$ so all we have to do is parametrize the equation:

\begin{align*}

\int_{L}f(z)dz&=\int_{\pi/2}^{-\pi/2}(\overline{3i+3e^{it}})^2(3ie^{it})dt \\
&=3i\int_{\pi/2}^{-\pi/2}(-3i+e^{-it})^2(e^{it})dt \qquad\color{red}{\text{Misprint}}\\
&=3^3i\int_{\pi/2}^{-\pi/2}(i-e^{-it})^2(e^{it}) dt \\
&=3^3i\int_{\pi/2}^{-\pi/2}(-1-2ie^{-it}+e^{-2it})(e^{it})dt \\
&=3^3i\int_{\pi/2}^{-\pi/2}(-e^{it}-2i+e^{-it})dt \\
&=-3^3i\int_{\pi/2}^{-\pi/2}(e^{it}-e^{-it}+2i)dt \\
&=-3^3i\left[\frac{1}{i}e^{it}+\frac{1}{i}e^{-it}+2it\right]^{-\pi/2}_{\pi/2} \\
&=\frac{-3^3i}{i}\left[e^{it}+e^{-it}+2i^2t\right]^{-\pi/2}_{\pi/2} \\
&= -3^3\left[e^{it}+e^{-it}-2t\right]^{-\pi/2}_{\pi/2} \\
&= -3^3[(e^{-i\pi/2}+e^{i\pi/2}-2(-\pi/2))-(e^{i\pi/2}+e^{-i\pi/2}-2(\pi/2))] \\
&= -3^3[(-i+i+\pi)-(i-i-\pi)] \\
&= -3^3[\pi+\pi] \\
&= -54\pi
\end{align*}
« Last Edit: October 20, 2018, 03:27:09 PM by Victor Ivrii »

Zihan Wan

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Re: TT1 Problem 4 (night)
« Reply #2 on: October 19, 2018, 10:55:58 AM »
-54pi

Min Gyu Woo

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Re: TT1 Problem 4 (night)
« Reply #3 on: October 19, 2018, 11:05:30 AM »
Care to elaborate?
Where did I go wrong?

Zihan Wan

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Re: TT1 Problem 4 (night)
« Reply #4 on: October 19, 2018, 11:10:10 AM »
$-3^{3}*2\pi$

Fangqi Lu

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Re: TT1 Problem 4 (night)
« Reply #5 on: October 19, 2018, 12:47:03 PM »
-54π

Victor Ivrii

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Re: TT1 Problem 4 (night)
« Reply #6 on: October 20, 2018, 03:32:14 PM »
Min did everything right, Zihan and Fangqi are just flooding