Messages

1

MAT334--Lectures & Home Assignments / Re: Section 2.6

« on: November 21, 2018, 12:18:03 PM »

Here is my solution to 5):
$I =\int_{-\infty}^{\infty}\frac{xsinx}{x^4+1}dx$
$Let f(z)=\frac{ze^iz}{z^4+1}$
Solve$z^4+1=0$:$z=e^{i\frac{\pi+2k\pi}{4}},k=0,1,2,3$
f has pole of order 1 at:
$z_1=\frac{1+i}{\sqrt2}$
$z_2=\frac{-1+i}{\sqrt2}$
Then,calculate the residues:
$Res(f;z_1)=\frac{ze^iz}{4z^3}$ at $z=\frac{1+i}{\sqrt2}$= $-\frac{1}{4}z_1^2e^{iz_1}$
Similarly:
$Res(f;z_2)=\frac{ze^iz}{4z^3}$ at $z=\frac{-1+i}{\sqrt2}$= $-\frac{1}{4}z_2^2e^{iz_2}$

Then,calculate the sum of residues
=$-\frac{1}{4}z_1^2e^{iz_1}-\frac{1}{4}z_2^2e^{iz_2}$
=$\frac{1}{2}e^{-\frac{1}{\sqrt2}}sin(\frac{1}{\sqrt2})$

Then,$\int_0^\pi f(Re^{i\theta})e^{i\theta}d\theta|$ = $|\int_0^\pi \frac{Rexp(i\theta)exp(iRe^{i\theta})}{R^4e^{4i\theta}+1}Re^{i\theta}d\theta|$
$\leq \frac{R^2}{R^4+1}$, as $R \rightarrow \infty$

Finally:
$2\pi i(\frac{1}{2}e^{-\frac{1}{\sqrt2}}sin(\frac{1}{\sqrt2}))$ = $\int_{-\infty}^{\infty}\frac{xe^{ix}}{x^4+1}dx + 0$
=$\int_{-\infty}^{\infty} \frac{xcosx+ixsinx}{x^4+1}dx$
Thus, $I=\pi (e^{-\frac{1}{\sqrt2}}sin(\frac{1}{\sqrt2}))$

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MAT334--Lectures & Home Assignments / Re: 2.5 - Q23

« on: November 20, 2018, 10:40:57 PM »

I was also wondering if the answer key is wrong.
I don't understand how to get the part $a_n=\frac{-4}{3^{n+2}}$.

3

MAT334--Lectures & Home Assignments / Re: 2.5 - Q23

« on: November 20, 2018, 11:22:03 AM »

Here is what I did for part (a):
For $1<|z|<2$, if we follow equation (12) and (13), we have
$Res(\frac{z+2}{\frac{z-2}{z+1}};-1) = \frac{z+2}{z-2}|_{z=-1}= -\frac{1}{3}$

$Res(\frac{z+2}{\frac{z+1}{z-2}};2) = \frac{z+2}{z+1}|_{z=2}= \frac{4}{3}$

By (12) and (13):
$\sum_{-\infty}^{\infty}a_nz^n$,where
$a_n = \sum\frac{1}{3}(-1)^n,n=-1,-2,...$
$and$
$a_n = -\sum\frac{4}{3}(2)^{-n-1}=-\sum\frac{4}{3}(\frac{1}{2^{n+1}}),n=0,1,2,...$

Similar for when $2<|z|<\infty$
You can read example 12 from 2.5

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MAT334--Lectures & Home Assignments / Re: 2.6 Q11

« on: November 19, 2018, 09:39:32 PM »

$\newcommand{\Res}{\operatorname{Res}}$
Here is my solution：
$I=\int_0^{2\pi} \frac{cos2\theta dz}{1-2acos\theta+a^2}~dx$
By substitution:   let $z=e^{i\theta}$, then $dz=ie^{i\theta}, d\theta= \frac{dz}{ie^{i\theta}}=\frac{dz}{iz}$
$cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{1}{2} (z+z^{-1})$
Similarly, $\cos2\theta=\frac{1}{2} (z^2+z^{-2})$
\begin{align*}
I=&\int_{|z|=1}\frac{\frac{1}{2} (z^2+z^{-2})}{1-a(z+z^{-1})+a^2}\frac{dz}{iz}\\
=&\frac{1}{2i}\int_{|z|=1}\frac{z^2+\frac{1}{z^2}}{z-az^2-a+a^2z}dz\\
=&\frac{1}{2i}\int_{|z|=1}\frac{z^4+1}{z^2(z-az-a+a^2z)}dz\\
=&\frac{1}{2i}\int_{|z|=1}\frac{z^4+1}{z^2(z-a)(1-az)}dz
\end{align*}
Let $f(z)=\frac{z^4+1}{z^2(z-a)(1-az)}$
$f(z)$ has simple pole at $z=a, z=\frac{1}{a}$, and pole of order 2 at $z=0$, but since $-1<a<1$, we only consider $z=a$ and $z=1$.z=0
\begin{align*}
&\Res(f(z);a)=\Res(\frac{\frac{z^4+1}{z^2(1-az)}}{z-a};a)=\frac{a^4+1}{a^2(1-a^2)},\\
&\Res(f(z);0)=\Res(\frac{\frac{z^4+1}{(z-a)(1-az)}}{(z-0)^2};0)=\frac{d}{dz}\frac{z^4+1}{(z-a)(1-az)}
\end{align*}
 at $z=a$, $=-\frac{1+a^2}{a^2}$
By Residue Theorem, $I=2\pi i\frac{1}{2i}(\frac{a^4+1}{a^2(1-a^2)}-\frac{1+a^2}{a^2}=\frac{2\pi a^2}{1-a^2}$

5

Quiz-6 / Re: Q6 TUT 0201

« on: November 17, 2018, 06:32:49 PM »

$\frac{z^2}{z^2-1}$$=\frac{z^2-1+1}{z^2-1}$
$=1+\frac{1}{z^2-1}$
$=1+\frac{1}{(z-1)(z+1)}$
$=1+\frac{1}{2}\frac{1}{z-1}-\frac{1}{z+1}$, by partial fractions
$=1+\frac{1}{2}(\frac{1}{z-1}-\frac{1}{2+(z-1)})$
$=1+\frac{1}{2}(\frac{1}{z-1}-\frac{1}{2}\frac{1}{1+\frac{z-1}{2}})$
$=1+\frac{1}{2}((z-1)^{-1}-\frac{1}{2}\frac{1}{1-\frac{1-z}{2}})$
$=1+\frac{1}{2}((z-1)^{-1}-\frac{1}{2}\sum_{0}^{\infty} (\frac{1-z}{2})^n)$
Residue=$\frac{1}{2}$

6

MAT334--Lectures & Home Assignments / Re: 2.3 Q6

« on: November 08, 2018, 02:46:03 AM »

$\int_0^{2\pi} \frac{dθ}{3+sinθ+cosθ}~dx$$=\int_{|z|=1} \frac{dz}{iz(3+\frac{1}{2i}(z-\frac{1}{z}))+\frac{1}{2}(z+\frac{1}{z}))}$
where $iz(3+\frac{1}{2i}(z-\frac{1}{z})+\frac{1}{2}(z+\frac{1}{z})) = i(3z+(\frac{z^2}{2i}-\frac{1}{2i})+(\frac{z^2}{2}+\frac{1}{2}))$
$=i(\frac{6zi+z^2-1+z^2i+i}{2i})$
$=\frac{(i+1)z^2+6iz+(i-1)}{2}$
The integral $=\int_{|z|=1} \frac{dz}{\frac{(i+1)z^2+6iz+(i-1)}{2}}$
$=\int_{|z|=1} \frac{2dz}{(i+1)z^2+6iz+(i-1)}$
$=\int_{|z|=1} \frac{2dz/(z-(\frac{-3i+i\sqrt{7}}{1+i}))}{z-(\frac{-3i-i\sqrt{7}}{1+i})}$
$So,f(\frac{-3i-i\sqrt{7}}{1+i})=\frac{2}{\frac{-2i\sqrt{7}}{1+i}}$
The integral$=\frac{1+i}{-\sqrt{7}}2\pi$

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MAT334--Lectures & Home Assignments / Re: 2.5Q14

« on: November 05, 2018, 09:51:14 PM »

Assume f is analytic at $z_0$ and has a zero of order m at $z_0$.
There exists $p>0$ and an analytic function g which does not vanish at $z_0$ on $D(z_0 ,p)$ such that $f(z)=(z-z_0)^m g(z)$ for $z$ in $D(z_0 ,p)$.
Then we can take derivatives:
$f'(z)=m(z-z_0)^{m-1} g(z) + (z-z_0)^m g'(z)$
So $\frac{f'(z)}{f(z)}$ = $\frac{m(z-z_0)^{m-1} g(z) + (z-z_0)^m g'(z)}{(z-z_0)^m g(z)}$
$=\frac{m(z-z_0)^{m} g(z)}{(z-z_0)^m (z-z_0)g(z)}$ + $\frac{(z-z_0)^{m} g'(z)}{(z-z_0)^m g(z)}$
$=\frac{m}{z-z_0}$ + $\frac{g'(z)}{g(z)}$
Now, $\frac{g'}{g}$ is analytic on D, thus $\frac{f'}{f}$ has simple pole at $z_0$
This implies that $Res(\frac{f'}{f};z_0) = m$