Messages

16

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*}

17

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*}

18

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*}

19

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*}

20

Quiz-4 / Re: Q4 TUT 0203

« on: October 26, 2018, 05:54:28 PM »

\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\ in\ cludes\ 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)
\end{equation*}

21

Quiz-4 / Re: Q4 TUT 0201

« on: October 26, 2018, 05:52:26 PM »

\begin{equation*}
This\ formula\ cannot\ be\ re-written\ with\ Cauchy's\ formula,\ so\ we\ use\ Cauchy's\ theorem.\\
\\
\int _{|z|\ =\ 1} f( z) dz\ =\ 0\\
\\
\int _{|z|\ =\ 1} \ \frac{z}{( z-2)^{2}} \ dz\ =\ 0\\
\end{equation*}

22

Quiz-4 / Re: Q4 TUT 0101

« on: October 26, 2018, 05:48:20 PM »

\begin{equation*}
cos\ \theta \ =\ \frac{1}{2} \ \left( z\ +\ \frac{1}{z} \ \right)\\
\\
d\theta \ =\ \frac{dz}{iz}\\
\\
\frac{d\theta }{a\ +\ bcos\theta } \ =\ \frac{dz}{iz\ \left( a\ +\ \frac{bz}{2} \ +\ \frac{b}{2z} \ \right)}\\
\\
=\ \frac{dz}{\frac{i}{2}\left( bz^{2} \ +\ 2az\ +\ b\right)}\\
\\
=\ \frac{2\ dz}{\ i\left( bz^{2} \ +\ 2az \ +\ b\right)}\\
\\
Complete\ the\ square:\\
=\ \frac{-2a \ \pm \sqrt{4a^{2} \ -\ 4b^{2}}}{2b} \ \\
\\
=\ \frac{-a \ \pm \sqrt{a^{2} \ -\ b^{2}}}{b}\\
\\
\left( z\ +\frac{a}{b} \ +\sqrt{\frac{a^{2}}{b^{2}} \ -\ 1} \ \right)\left( z\ +\frac{a}{b} \ -\sqrt{\frac{a^{2}}{b^{2}} \ -\ 1} \ \right)\\
\\
\ ( z\ -\ i\left( -\ \frac{a}{b} \ -\sqrt{\frac{a^{2}}{b^{2}} \ -\ 1} \ \right) \ is\ outside\ of\ |z|\ =\ 1\ =\ z\ -\ q\\
\\
( z\ +\ i\left( \ \frac{a}{b} \ \ -\sqrt{\frac{a^{2}}{b^{2}} \ -\ 1} \ \right) \ is\ inside\ of\ |z|\ =\ 1\ =\ z\ -\ p\\
\\
So\ Cauchy's\ formula\ gives\ us\\
\frac{1}{2\pi i} \ \int _{|z|\ =\ 1} \ \frac{dz}{( z\ -\ q)( z\ -\ p)} \ \ =\ \frac{1}{( p\ -\ q)} \ =\ \frac{1}{2ib\ \sqrt{\frac{a^{2}}{b^{2}} \ -\ 1}}\\
=\frac{1}{2i\ \sqrt{a^{2} \ -\ b^{2}}}\\
\\
\\
But\ the\ integral\ is\ just\\
\\
\frac{1}{2\pi i}\int _{|z|\ =\ 1} \ \frac{dz}{( z\ -\ q)( z\ -\ p)}\\
\\
=\ \frac{1}{2\pi i}\int ^{2\pi }_{0} \ \frac{ie^{i\theta } d\theta }{2ie^{i\theta }( a\ +\ bcos\theta )}\\
\\
=\ \frac{1}{4\pi i}\int ^{2\pi }_{0} \ \frac{d\theta }{( a\ +\ bcos\theta )}\\
\\
so,\\
\\
\int ^{2\pi }_{0} \ \frac{d\theta }{( a\ +\ bcos\theta )} \ =\ \frac{4\pi i}{2i\sqrt{a^{2} -b^{2}}}\\
\\
=\frac{2\pi }{\sqrt{a^{2} \ -\ b^{2}}}
\end{equation*}

23

Quiz-2 / Re: Q2 TUT 5301

« on: October 09, 2018, 08:49:26 PM »

PDF Attachment contains solution

24

Quiz-2 / Re: Q2 TUT 5201

« on: October 08, 2018, 04:33:05 PM »

Every power series has a convergence radius R, where Sum[a_nxⁿ] converges if |x| < R.

The first summation is the power series equal to e^z

and we have a_n = 1/n!

lim |a_n+1| / |aⁿ| = 1 / (n + 1)

= 0. So our convergence radius R is infinity and the power series converges for all z.

The second summation is the power series equal to cos(z) and we know:

cos(z) = (1/2)*(e^iz + e^-iz)

From the first series, we know the e^z is convergent on all z, so cos(z) is also convergent on all z.