Toronto Math Forum
MAT3342018F => MAT334Tests => Term Test 1 => Topic started by: Victor Ivrii 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:

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}(ie^{it})^2(e^{it}) dt \\
&=3^3i\int_{\pi/2}^{\pi/2}(12ie^{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)(ii\pi)] \\
&= 3^3[\pi+\pi] \\
&= 54\pi
\end{align*}

54pi

Care to elaborate?
Where did I go wrong?

$3^{3}*2\pi$

54π

Min did everything right, Zihan and Fangqi are just flooding