### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Jeffery Mcbride

Pages: [1] 2
1
##### Quiz-4 / Re: Q4 TUT 0202
« on: December 02, 2018, 01:45:54 PM »
Never got any feedback on this answer, is it correct?

2
##### Quiz-7 / Re: Q7 TUT 5101
« on: December 01, 2018, 08:49:16 PM »
Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$

3
##### Quiz-7 / Re: Q7 TUT 5101
« on: December 01, 2018, 04:54:19 PM »
You're right, $\displaystyle f( iy)$ is totally irrelevant here. My mistake. Do you mean the fact that f([-R, R]) lies completely in the upper half plane. Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$

When $\displaystyle z\ \rightarrow \ \pm \infty$, Re(f(z)) $\displaystyle \rightarrow \ \infty$ and Im(f(z)) $\displaystyle \rightarrow \ \infty \$and the endpoints of $\displaystyle f([ -R,\ R]) \$lie in the first quadrant for R really big. So the change in argument is going to be 0.

This can be coupled with what I said about the curve.

$\displaystyle f\left( Re^{it}\right)$ =  $\displaystyle e^{4it} \$from [0, $\displaystyle \pi$] because the 4 term dominates. So the change in argument is 4$\displaystyle \pi$.

So in total we get 2 zeros as the total change in argument is 4$\displaystyle \pi$.

4
##### Quiz-7 / Re: Q7 TUT 5101
« on: November 30, 2018, 04:24:14 PM »
$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ z^{4} \ +\ 3iz^{2} \ +\ z\ -\ 2\ +\ i\\ \\ f( iy) \ is\ always\ positive\\ \\ f( x) \ =\ x^{4} \ +\ 3ix^{2} \ +\ x\ -\ 2\ +\ i\\ \\ Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2 \end{array}$

This is positive at x = R, then negative when it hits x = 1 to x = -1 then it becomes positive again as x approaches -R
.
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$ This is always positive.

So, we are always in the upper half plane. Change in Arg = 0.

$\displaystyle f\left( Re^{it}\right) \ =\ e^{4it} \$from [0, $\displaystyle \pi$] because the 4 term domindates. So the change in argument is 4$\displaystyle \pi$.

$\displaystyle \frac{0+4\pi }{2\pi } \ =\ 2$ zeros total

5
##### Quiz-7 / Re: Q7 TUT 0101
« on: November 30, 2018, 04:01:32 PM »
Using Rouche's Theorem, which is a corollary of the Argument Principle.

$\displaystyle f( z) =\ z^{4} \ -\ 2z\ -\ 2$ in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$

$\displaystyle Let\ g( z) \ =\ -2$

For the smaller curve we have $\displaystyle z=\frac{1}{2} e^{i\theta }$,
$\displaystyle |f( z) -( -2) |\ \leq \ z^{4} \ +\ 2z\ \ =\ \frac{17}{16} \ < \ |g( z) |\$

So, f(z) and g(z) have the same number of zeros in the smaller circle, which is 0.

$\displaystyle Now,\ let\ g( z) \ =\ z^{4}$
For the larger curve we have $\displaystyle z=\frac{3}{2} e^{i\theta }$
$\displaystyle |f( z) \ -\ z^{4} |\ \leq |2z\ +\ 2\ |\ \ =\ 5\ < \ |g( z) |\ =\ \frac{81}{16} \ \$

So, f(z) and g(z) have the same number of zeros in the larger circle, which is 4.
Thus, all four zeros of f(z) are in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$

6
##### Term Test 2 / Re: TT2A Problem 5
« on: November 29, 2018, 05:00:07 PM »
$\displaystyle f( z) \ =\ \frac{3z}{( z-2)( z+1)}$

By partial fractions:

$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ \frac{A}{z-2} \ +\ \frac{B}{z+1}\\ \\ f( z) \ =\ \frac{2}{z-2} \ +\ \frac{1}{z+1} \end{array}$

a) as $\displaystyle |z|\ < \ 1$

$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{1\ -\ ( -z)}\\ \\ =\ -\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \sum ^{\infty }_{n\ =\ 0}( -z)^{n}\\ \\ =\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{2^{n}} \ +( -1)^{n}\right) z^{n} \end{array}$

b) as $\displaystyle 1\ < \ |z|\ < \ 2$

$\displaystyle f( z) \ =\$$\displaystyle -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}}$

$\displaystyle \begin{array}{{>{\displaystyle}l}} =-\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\ \\ =-\sum ^{\infty }_{n\ =\ 0}\frac{z^{n}}{2^{n}} \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\ \\ =\sum ^{\infty }_{n\ =\ 0} -\frac{1}{2^{n}} \ z^{n} \ +\ \sum ^{1}_{n\ =\ -\infty }\frac{1}{( -1)^{n-1}} \ z^{n} \end{array}$

c) as $\displaystyle |z|\ >\ 2$

$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ \frac{2}{z} \ \frac{1}{1\ -\ \frac{2}{z}} \ +\frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}} \ \\ =\ \frac{2}{z} \ \sum ^{\infty }_{n\ =\ 0}\left(\frac{2}{z}\right)^{n} +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\ \\ =\sum ^{\infty }_{n\ =\ 0}\left(\frac{2^{n+1}}{z^{n+1}}\right) \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\ \\ =\sum ^{-1}_{n\ =\ -\infty }\left(\frac{1}{2^{n-2}} \ +\ \frac{1}{( -1)^{n-1}}\right) z^{n}\\ \\ \end{array}$

7
##### Quiz-6 / Re: Q6 TUT 5301
« on: November 17, 2018, 05:39:12 PM »
No problem!

8
##### Quiz-6 / Re: Q6 TUT 5301
« on: November 17, 2018, 05:24:26 PM »
Why 2k? It should just be k.

$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ \pi \ cot\ ( \pi z)\\ \\ =\frac{\pi cos( \pi z)}{sin( \pi z)} \end{array}$

So clearly, f(z) has a pole when sin($\displaystyle \pi z) \ =\ 0$ which happens whenever z is any integer
This pole is order 1.

9
##### Quiz-6 / Re: Q6 TUT 0203
« on: November 17, 2018, 04:32:23 PM »
$\displaystyle f( z) \ =\ \frac{g( z)}{( z-z_{0})^{l}}$

where $\displaystyle g( z)$ is analytic in $\displaystyle |z\ -\ z_{0} |\ < \ R\ and\ g( z_{0}) \ \neq 0$
Then,
$\displaystyle \frac{f'( z)}{f( z)} \ =\ \frac{g'( z)}{g( z)} \ -\ \frac{l}{z\ -\ z_{0}}$

Since $\displaystyle g( z_{0}) \ \neq \ 0,\ then\ \frac{g'( z)}{g( z)}$ is analytic at $\displaystyle z_{0}$ , so the residue of that first term is 0. So for the residue of the whole thing, we read the coefficent of the second term which is -l. So we get

$\displaystyle Res\left(\frac{f'( z)}{f( z)} ;l\right) \ =-l$

10
##### Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P4M
« on: November 10, 2018, 04:53:13 PM »

\begin{equation*}
I( b) \ =\int ^{\infty }_{0}\frac{e^{-bx} sin( x)}{x} \ dx\ for\ b\ \geq 0.\\
\\
I( 0) \ =\ \int ^{\infty }_{0}\frac{sin( x)}{x} \ ( This\ is\ the\ integral\ of\ interest) .\\
\\
I'( b) \ =\ -\ \int ^{\infty }_{0} \ e^{-bx} sinx\ dx\\
\\
=\ -\ \frac{1}{1\ +\ b^{2}}\\
\end{equation*}

Integrate this with respect to b and we get:

\begin{equation*}
I( b) \ =\ \pi /2\ -\ arctan( b) ,\ setting\ b\ =\ 0,\ gives\ us\ \pi /2\\
so,\\
\\
\int ^{\infty }_{0}\frac{sin( x)}{x} \ \ =\ \pi /2\ \\
\end{equation*}

11
##### Quiz-4 / Re: Q4 TUT 0202
« on: November 07, 2018, 06:24:36 PM »
By Cauchy's theorem, we get:
\begin{equation*}
0\ =\ \int _{\gamma } e^{iz^{2}} dz\ =\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} ie^{i\left( Re^{i\theta }\right)^{2}} Re^{i\theta } d\theta \ +\ \int ^{0}_{R} e^{ir^{2} \ e^{i\pi /2}} e^{\pi /4} \ dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{ir^{2}( cos\pi /2\ _{\ } +\ isin\pi /2)} dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
\\
\\
\end{equation*}

Now, just handling the middle integral, we get:

\begin{equation*}
\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \\
=R\int ^{\pi /4}_{0} e^{iR^{2} cos2\theta \ -\ R^{2} sin( 2\theta ) \ } ie^{i\theta } d\theta \\
This\ is\ estimated\ above\ by:\\
\\
R\int ^{\pi /4}_{0} e^{-R^{2} sin( 2\theta ) \ } d\theta \ \leq R\int ^{\pi /4}_{0} e^{-2R^{2}\frac{\theta }{\pi } \ } d\theta \ < \ \frac{\pi }{2R}\\
\end{equation*}

\begin{equation*}
So\ the\ limit\ as\ R\ \rightarrow \ \infty ,\ this\ will\ be\ 0,\ and\ we\ will\ get:\\
0=\int ^{R}_{0} e^{ix^{2}} dx\ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
0=\int ^{\infty }_{0}\left( cos\left( x^{2}\right) +\ isin\left( x^{2}\right) dx\right) \ -\ \frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\\
\frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{1+i}{\sqrt{2}} \ \frac{\sqrt{\pi }}{2} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{\sqrt{2\pi }}{4} \ +\ i\ \frac{\sqrt{2\pi }}{4} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\\
\end{equation*}

So, we separate imaginary and real parts and we get

\begin{equation*}
\int ^{\infty }_{0} cos\left( x^{2}\right) dx\ =\ \int ^{\infty }_{0} sin\left( x^{2}\right) dx\ =\frac{\sqrt{2\pi }}{4} \
\end{equation*}

12
##### Quiz-4 / Re: Q4 TUT 0203
« on: October 27, 2018, 04:20:42 PM »
Like so?

\begin{equation*}
f( z) \ =\ z\ +\ \frac{1}{z} \ is\ the\ derivative\ of\ F( z) \ =\ \frac{z^{2}}{2} \ +\ Log( z)\\
\\
This\ is\ valid\ only\ where\ the\ function\ Log( z) \ is\ analytic.\ This\ domain\ includes\ the\ domain\ \\
Im\ z\  >\ 0.\ Hence,\\
\\
\int _{\gamma }\left( z\ +\ \frac{1}{z} \ \right) dz\ =\ \int _{\gamma } f( z) dz\ =\ \ \int _{\gamma } F'( z) dz\\
\\
=\ F( endpoint) \ -\ F( initial\ point)\\
\\
=F( 6+2i) \ -\ F( -4\ +\ i)\\
\\
=\left(\frac{( 6+2i)^{2}}{2} \ +\ Log( 6\ +\ 2i)\right) \ -\ \left(\frac{( -4+i)^{2}}{2} \ +\ Log( -4\ +\ i)\right)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ Log( 6\ +\ 2i) \ -\ Log( -4\ +\ i)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ \left( log\left(\sqrt{40}\right) \ +\ iArg( 6\ +\ 2i)\right) \ -\ \left( log\left(\sqrt{17}\right) \ +\ iArg( -4\ +\ i)\right)\\
\\
=\frac{17}{2} \ +\ 16i\ +\ log\left(\sqrt{40}\right) \ \ -\ log\left(\sqrt{17}\right) \ +\ i\ tan^{-1}\left(\frac{1}{3}\right) \ -\ i\ \ \left( \pi \ -\ tan^{-1}\frac{1}{4}\right)
\end{equation*}

13
##### Quiz-4 / Re: Q4 TUT 5301
« on: October 26, 2018, 06:05:27 PM »

\begin{equation*}
F( z) \ =\ \frac{-1}{z}\\
\\
The\ function\ is\ analytic\ in\ all\ of\ Re\ z\  >\ 0\\
\\
So\ we\ just\ want\ F( end\ point) \ -\ F( first\ point)\\
\\
F( 1-i) \ -\ F( 1\ +\ i)\\
\\
=\ \frac{-1}{1-i} \ +\ \frac{1}{1+i}\\
\\
=\ \frac{-1\ -\ i\ +\ 1\ -\ i\ }{( 1-i)( 1+i)} \ \\
\\
=\ \frac{-2i}{-2}\\
\\
=\ i
\end{equation*}

14
##### Quiz-4 / Re: Q4 TUT 5102
« on: October 26, 2018, 06:00:06 PM »

\begin{equation*}
\int _{|z|\ =\ 2} \ \frac{e^{z}}{ \begin{array}{l}
z( z-3)\\
\end{array}}\\
\\
=\ \int _{|z|\ =\ 2} \ \frac{e^{z}}{ \begin{array}{l}
( z-0)( z-3)\\
\end{array}}\\
\\
=\ \int _{|z|\ =\ 2} \ \frac{\zeta ( z)}{ \begin{array}{l}
( z-0)\\
\end{array}} \ ,\ \zeta ( z) \ =\ \frac{e^{z}}{( z-3)}\\
\\
=( 2\pi i) \zeta ( 0) \ =\ 2\pi i\ \frac{e^{0}}{( 0\ -\ 3)} \ =\ \frac{-2\pi i}{3}
\end{equation*}

15
##### Quiz-4 / Re: Q4 TUT 5101
« on: October 26, 2018, 05:55:29 PM »

\begin{equation*}
\int _{|z|\ =\ 1}\frac{sin( z)}{z}\\
\\
=\int _{|z|\ =\ 1} \ \frac{sin( z)}{z\ -\ 0}\\
\\
Set\ \zeta ( z) \ =\ sin( z)\\
\\
So,\ by\ Cauchy's\ formula,\\
\\
f( z) \ =\ \frac{1}{2\pi i}\int _{\gamma } \ \frac{\zeta ( z)}{\zeta \ -\ z}\\
\\
\int _{|z|\ =\ 1} \ \frac{sin( z)}{z\ -\ 0} \ =\ ( 2\pi i) \zeta ( 0) \ \\
\\
=\ ( 2\pi i)( sin\ 0) \ =\ 0\ \ \\
\end{equation*}

Pages: [1] 2