Author Topic: Q3 TUT 0203  (Read 639 times)

Victor Ivrii

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Q3 TUT 0203
« on: October 12, 2018, 06:17:09 PM »
Directly compute the following line integral:
$$
\int_\gamma(z^2 +3z +4)\,dz
$$
where $\gamma$ is the circle $\{z\colon |z|=2\}$, oriented counterclockwise. Draw the picture.

Min Gyu Woo

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Re: Q3 TUT 0203
« Reply #1 on: October 12, 2018, 06:17:20 PM »
Our $\gamma (t)$ in this case is

 

\begin{equation}

 

\gamma(t) = 2e^{it}\\ 

 

\gamma'(t) = 2ie^{it}

 

\end{equation}

 

With $ 0\leq t \leq 2\pi$

 

Writing the integral in terms of $\gamma$, $\int_{a}^{b} f(\gamma(t))\gamma'(t)dt$

 

\begin{equation}

 

\begin{aligned}

 

\int_{\gamma}(z^2+3z+4)dz &= \int_{0}^{2\pi}(2e^{2it}+6e^{it}+4)(2ie^{it})dt \\

 

&= 4i\int_{0}^{2\pi}(e^{3it}+3e^{2it}+4e^{it})dt \\

 

&= 4i\left[\frac{1}{3i}e^{3it}+\frac{1}{2i}e^{2it}+4\frac{1}{i}e^{it}\right]_0^{2\pi} \\

 

&= 4\frac{i}{i}\left[\left(\frac{1}{3}e^{6i\pi}+\frac{1}{2}e^{4i\pi}+4e^{2i\pi}\right)-(\frac{1}{3}+\frac{1}{2}+4)\right]\\

 

&= 0\\

\end{aligned}

 

\end{equation}

 

Since $e^{i2\pi k} = 1, k\in\mathbb{Z}$ 
« Last Edit: October 12, 2018, 07:15:21 PM by Min Gyu Woo »

oighea

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Re: Q3 TUT 0203
« Reply #2 on: October 12, 2018, 06:23:43 PM »
Please note that the curve is a circle of radius 2 centered on z=0. The parametric equation of the curve is $\gamma(t) = 2e^{it}$, and the derivative is $\gamma'(t) = 2ie^{it}$. Otherwise, it is mostly correct as it is a line integral of a simple-closed curve around a holomorphic function $f(z) = z^2 + 3z + 4$. Otherwise, your answer and steps are mostly correct.

Also note that the integrand $\int_a^bf(\gamma(t))\gamma'(t)\,dt$ is of the form when you differentiate a function composition. Hence, as a shorthand, you can calculate or verify your answer as $F(\gamma(b)) - F(\gamma(a))$

To verify, $F(z) = \frac{1}{3}z^3 + \frac{3}{2}z^2 + 4z$.
Then $F(\gamma(z)) = \frac{1}{3}(2e^{it})^3 + \frac{3}{2}(2e^{it})^2 + 4(2e^{it})$
$F(\gamma(z)) = \frac{8}{3}e^{3it} + \frac{12}{2}e^{2it} + 8e^{it}$, and $f(\gamma(z))\gamma'(z) = (2^2e^{2it} + 3\cdot2e^{it} + 4)(2ie^{it}) = 8ie^{3it} + 12ie^{2it} + 8ie^{it}$
$F(\gamma(z)) = \frac{8}{3}e^{3it} + 6e^{2it} + 8e^{it}$.
The integral is then expressed as $F(2\pi) - F(0)$.
We found out that $e^{2it} = e^{4it} = e^{6it} = e^{kit} = e^0 = 1$ for all $k \in \mathbb{Z}$.
Hence \frac{8}{3}e^{3it} + 6e^{2it} + 8e^{it} = \frac{8}{3} + 6 +8$ when $t = 0$ and $t = 2\pi$.
As $F(2\pi) = $F(0)$, it follows the integral is zero.
« Last Edit: October 12, 2018, 07:07:30 PM by oighea »