### Author Topic: TT1 Problem 4 (morning)  (Read 1667 times)

#### Victor Ivrii ##### TT1 Problem 4 (morning)
« on: October 19, 2018, 04:01:31 AM »
\begin{problem}[4 pts]\label{pr-4}
Calculate integral
$\displaystyle{\int_L {(z+\bar{z})}}\,dz$ where $L$ is  shown on the figure below:

#### Hyun Woo Lee

• Newbie
• • Posts: 2
• Karma: 6 ##### Re: TT1 Problem 4 (morning)
« Reply #1 on: October 19, 2018, 11:35:25 AM »
This is a piece wise curve.
Let’s denote the first curve as $r_1$ And the second curve, line from $3+3i$ to $3$ as $r_2$
Parametrize both curve.
$$r_1(t) = 3e^{it} + 3i, 0 \geq t \geq -\frac{\pi}{2}$$
$$r_2(t) = -3ti + 3 + 3i, 0 \leq t \leq 1$$

Then, $$\int_{L}^{} (z+\bar{z}) dz = \int_{r_1}{} (z+\bar{z})dz + \int_{r_2}{} (z+\bar{z})dz$$

So, $$\int_{r_1} (z+\bar{z})dz =\int_{-\frac{\pi}{2}}^{0} [r(t) + \overline{r(t)}]r’(t) dt = \int_{-\frac{\pi}{2}}^{0} (3e^{it} + 3i + 3e^{-it} -3i)(3ie^{it}) dt$$
Then,
$$\int_{-\frac{\pi}{2}}^{0} \Bigl(9ie^{2it} + 9i \Bigr)dt = \frac{9}{2}e^{2it} + 9it \Big|_\frac{-\pi}{2}^{0}$$
This calculates to $$9 + \frac{9}{2}i\pi$$
Now we have to compute for $r_2$
$$\int_{r_2}{} (z+\bar{z})dz = \int_{0}^{1} [(3+i(3-3t))(3-i(3-3t))](-3i) dt$$
This is $$-3i\int{0}^{1} [1+i(1-t)][1-i(1-t)] dt = -3i\int_{0}^{1} 1 + (1-t)^2 dt = -3i\int_{0}^{1} t^2 - 2t + 2$$
Then,
$$-3i(\frac{t^3}{3} - t^2 + 2t) \Big|_{0}^{1} = -3i(\frac{1}{3} - 1 +2) = -4i$$
Adding the two integrals you get $$9 + \frac{9}{2}i\pi - 4i$$
« Last Edit: October 20, 2018, 03:05:21 PM by Victor Ivrii »

#### Victor Ivrii ##### Re: TT1 Problem 4 (morning)
« Reply #2 on: October 20, 2018, 03:14:36 PM »
Wrong limits for $t$ in the circular arc part, wrong calculations in the straight segment part.

#### Meng Wu ##### Re: TT1 Problem 4 (morning)
« Reply #3 on: October 21, 2018, 04:52:46 PM »
For $$\gamma_{1}(t)=3i+3e^{it}, \gamma_{1}(t) \text{ is negatively ortiented.}$$
Thus \begin{align}\int_{\gamma_{1}}(z+\bar z)dz&=\int_\frac{\pi}{2}^0(\gamma_{1}(t)+\overline {\gamma_{1}(t)})\gamma_{1}'(t)dt \\ &=\frac{9}{2}e^{2it}+9it \Bigg|_{\frac{\pi}{2}}^0 \\&=9-\frac{9}{2}\pi i\end{align}
For $$\gamma_2(t)=3+3i-3it=3+i(3-3t), 0\leq t\leq 1.$$
\begin{align}\int_{\gamma_{2}}(z+\bar z)dz&=\int_0^1 (\gamma_{2}(t)+\overline {\gamma_{2}(t)})\gamma_{2}'(t)dt \\&=\int_0^1 ((3+i(3-3it))+(3-i(3-3t)))(-3i)dt \\&=\int_0^1 -18idt \\&=-18it \Big|_0^1 \\&=-18i\end{align}
Therefore, $$\int_L(z+\bar z)dz=\int_{\gamma_{1}}(z+\bar z)dz+\int_{\gamma_{2}}(z+\bar z)dz=9-\frac{9}{2}\pi i -18i$$
« Last Edit: October 21, 2018, 05:08:58 PM by Meng Wu »