Messages

1

MAT334--Lectures & Home Assignments / Re: 3.2 Q2

« on: December 10, 2018, 04:13:17 PM »

$x^2+y^2\leq4, |z|^2=x^2+y^2\Rightarrow|z|^2\leq4\Rightarrow|z|\leq2$
so,$|z||e^z|=\sqrt{x^2+y^2}e^x$since $x\geq0,y\geq0$

If go through x-axis then y=0 and |f(z)|=$2e^x(0\leq x\leq2)$
then, x=$r\cos t\Rightarrow|f(z)|=2e^{2\cos t},0\leq t\leq \frac{\pi}{2}$around of circle, and then $x=0, |f(z)|=y,0\leq y\leq2$

On the x-axis, the max of |f|is $2e^2$, on the circle is t=0$\Rightarrow|f(z)=2e^2|$, on the y-axis is 2

so, Max${(2e^2,2e^2,2)}=2e^2 $

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MAT334--Lectures & Home Assignments / Re: FE Sample Question 4 (a)

« on: December 07, 2018, 10:18:10 PM »

Hanyu，I think you made a mistake when you calculated $e^{it}$,

since $e^{it}=e^{-i\theta}=-1$

then $\lambda=-1$

And you can also check the value of $\lambda$ and a since $\lambda$a=5 in your third line of calcualtion

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MAT334--Lectures & Home Assignments / Re: Mobius tranformation

« on: December 04, 2018, 03:53:39 PM »

For example, z=2+i, f(z)=$z^2$
f'(z)=2(2+i)
arg(f'(z))=arctan$\frac{1}{2}$
because4+2i represents the point (4,2) and the arg for (4,2) is arctan$\frac{1}{2}$
Hope it works for you.

4

Term Test 2 / Re: TT2A Problem 5

« on: November 24, 2018, 08:35:43 AM »

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

(a) |z|<1

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

         $= \sum_{n=0}^{\infty}-(\frac{z}{2})^n+(-z)^n$

         $= \sum_{n=0}^{\infty}\frac{-z^n}{2^n}+(-1)^nz^n$

         $= \sum_{n=0}^{\infty} (\frac{-1}{2^n}+(-1)^n)z^n$

(b) 1<|z|<2

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

      $= \sum_{n=0}^{\infty} -(\frac{z}{2})^n+\sum_{n=0}^{\infty}\frac{1}{z}(\frac{-1}{z})^n$

      $= \sum_{n=0}^{\infty}\frac{-z^n}{2^n}+\sum_{n=0}^{\infty}\frac{(-1)^n}{z^{n+1}}$

      $= \sum_{n=0}^{\infty}\frac{-z^n}{2^n}+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{z^{n}}$

      $= \sum_{n=0}^{\infty}\frac{-1}{2^n}z^n+\sum_{n=-\infty}^{1}(-1)^{-n-1}z^{n}$

(c)|z|>2

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

      $=\sum_{n=0}^{\infty}\frac{2}{z}(\frac{2}{z})^n+\frac{1}{z}(\frac{-1}{z})^n$

     $=\sum_{n=0}^{\infty}\frac{2^{n+1}}{z^{n+1}}+\frac{(-1)^n}{z^{n+1}}$

     $=\sum_{n=-\infty}^{0}(2^{n+1}+(-1)^n)z^{n-1}$

     $=\sum_{n=-\infty}^{1}(2^{n+2}+(-1)^{n+1})z^{n}$

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MAT334--Lectures & Home Assignments / Re: 2.3 Q7

« on: November 21, 2018, 10:10:15 PM »

Fixed it. Thx. And my answer is different from the textbook answer but I cannot see mistakes in my steps.

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MAT334--Lectures & Home Assignments / Re: 2.3 Q7

« on: November 21, 2018, 06:23:05 PM »

 $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$
Let z=$e^{i\theta}$$\Rightarrow$$\cos\theta=\frac{z^2+1}{2z}$ $\Rightarrow$ $a+b\cos\theta=a+\frac{b}{2}(z+\frac{1}{z})$,$d\theta=\frac{dz}{iz}$

$\int_{0}^{2\pi}\frac{d\theta}{a+b\cos\theta}$

=$\int_{|z|=1}\frac{dz}{iz(a+\frac{b}{2}(z+\frac{1}{z}))}$

=$\int_{|z|=1}\frac{2dz}{i(2az+bz^2+b)}$

$2az+bz^2+b=0$$\Rightarrow$ $z=\frac{-a\pm\sqrt{a^{2}-b^{2}}}{b}$

a>b>0$\Rightarrow$$\frac{-a}{b}<-1$ $\Rightarrow \frac{-a+ \sqrt{a^{2}-b^{2}}}{b} >-1 $which is inside |z|=1 and $\frac{-a- \sqrt{a^{2}-b^{2}}}{b}$<-1 which is outside |z|=1

$\therefore$ $\int_{|z|=1}\frac{2dz}{i(2az+bz^2+b)}$
     
      =$\int_{|z|=1}\frac{2/i(z+\frac{a+\sqrt{a^{2}-b^{2}}}{b})}{(z-\frac{-a+\sqrt{{a^2}-{b^2}}}{b})}dz$

      =$2\pi i f(z_0)$
     
      =$2\pi i \frac{2}{\frac{2i\sqrt{{a^2}-{b^2}}}{b}}$
     
      =$\frac{2\pi b}{\sqrt{{a^2}-{b^2}}} $

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MAT334--Lectures & Home Assignments / Re: 2.2 Q11

« on: November 19, 2018, 09:59:15 PM »

    $\frac{1}{4-z}=\frac{1}{4}\frac{1}{1-\frac{z}{4}}$
    $\frac{1}{4-z}=\frac{1}{4}\sum_{n=0}^{\infty}(\frac{z}{4})^n$
   
   Take derivatives on both sides, then
    $\frac{1}{(4-z)^2}=\sum_{n=1}^{\infty}\frac{1}{4^{n+1}}nz^{n-1}$  (Here, use the hint)
   
    $\frac {z^2}{(4-z)^2}=\sum_{n=1}^{\infty}\frac{1}{4^{n+1}}nz^{n+1}$
             
              $=\sum_{n=1}^{\infty}n(\frac{z}{4})^{n+1}$


OK Fiexd it.

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Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P5

« on: November 09, 2018, 10:56:09 AM »

Calculate the coefficient at $z^{-2}$. The answer is rather obvious because $f(z)$ decays as $z^{-4}$ as $z\to \infty$

So I should start from n=-1 because the coefficient is 0 when n=0?

While starting from the term with $0$ coefficient is not technically an error, it is definitely a bad bad practice.

And now is there any other mistake I have made?

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Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P5

« on: November 09, 2018, 10:30:36 AM »

Calculate the coefficient at $z^{-2}$. The answer is rather obvious because $f(z)$ decays as $z^{-4}$ as $z\to \infty$

So I should start from n=-1 because the coefficient is 0 when n=0?

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Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P5

« on: November 09, 2018, 09:46:03 AM »

I have modified but for |z|>5, the largest power I got is -2.

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Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P5

« on: November 08, 2018, 12:19:47 PM »

Let $w=z^2$, then $f(w)=\frac{16}{(w-16)(w+25)}$
                                 
                                  $=\frac{16}{41}\frac{1}{w-16}-\frac{16}{41}\frac{1}{w+25}$

               plug $w=z^2$, $f(z)=\frac{16}{41}\frac{1}{z^2-16}-\frac{16}{41}\frac{1}{z^2+25}$


(a) $|z|<4 ,so |w|<16 $
 
    $ f(w)=\frac{1}{41} \frac{1}{-1+\frac{w}{16}}-\frac{16}{41\cdot25} \frac{1}{1-\frac{-w}{25}}$
           
           $=\frac{-1}{41}\sum_{n=0}^{\infty}(\frac{w}{41})^n-\frac{16}{41\cdot25}\sum_{n=0}^{\infty}(\frac{-w}{25})^n$
   
          $=\sum_{n=0}^{\infty}(\frac{-1}{41\cdot16^n}-\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n$
   
   $f(z)=\sum_{n=0}^{\infty}(\frac{-1}{41\cdot16^n}-\frac{(-1)^n16}{41\cdot25^{n+1}}\,)z^{2n}$


(b) $4<|z|<5 ,so 16<|w|<25$
   
     $ f(w)=\frac{16}{41w} \frac{1}{1-\frac{16}{w}}-\frac{16}{41\cdot25} \frac{1}{1-\frac{-w}{25}}$

            $=\frac{16}{41w}\sum_{n=0}^{\infty}(\frac{16}{w})^n-\frac{16}{41\cdot25} \sum_{n=0}^{\infty}(\frac{-w}{25})^n$
   
            $  =\sum_{n=0}^{\infty}\frac{16^{n+1}}{41w^{n+1}}-  (\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n  $

            $= \sum_{n=1}^{\infty} \frac{16^n}{41w^n}-  (\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n  $
   
            $=\sum_{n=-\infty}^{-1} \frac{16^{-n}}{41}w^n-\sum_{n=0}^{\infty}(\frac{(-1)^n16}{41\cdot25^{n+1}}\,){w}^n  $

 $ f(z)=\sum_{n=-\infty}^{-1} \frac{16^{-n}}{41}z^{2n}-\sum_{n=0}^{\infty}(\frac{(-1)^n16}{41\cdot25^{n+1}}\,)z^{2n}  $


(c) $5<|z| ,so 25<|w|$

   $ f(w)=\frac{16}{41w} \frac{1}{1-\frac{16}{w}}-\frac{16}{41w} \frac{1}{1-\frac{-25}{w}}$

           $ =\frac{16}{41w}\sum_{n=0}^{\infty}(\frac{16}{w})^n- \sum_{n=0}^{\infty} \frac{16}{41w}(\frac{-25}{w})^n  $

           $= \sum_{n=0}^{\infty}\frac{16^{n+1}}{41w^{n+1}}-\sum_{n=0}^{\infty} \frac{16(-25)^n}{41w^{n+1}}$

           $= \sum_{n=0}^{\infty} (\frac{16^{n+1}}{41}-\frac{16(-25)^n}{41})w^{-n-1}$
   
           $= \sum_{n=-\infty}^{0} (\frac{16^{-n+1}}{41}-\frac{16(-25)^{-n}}{41})w^{n-1}$
 
    $f(z)= \sum_{n=-\infty}^{-1} (\frac{16^{-n+1}}{41}-\frac{16(-25)^{-n}}{41})z^{2n-2}$

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MAT334--Lectures & Home Assignments / 2.5Q14

« on: November 05, 2018, 09:16:40 PM »

I have no idea about the proof. Can anyone help me with this question？
Q: If f is analytic in $|z-z_0|<R$ and has a zero of order m at $z_0$, show that $Res(\frac{f’}{f};z_0)=m$

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MAT334--Lectures & Home Assignments / Re: 2.4 question 15

« on: November 04, 2018, 04:53:47 PM »

Q15: valid everywhere
Q16: $|z|=\frac{\pi}{2}$
 
             

                 

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Term Test 1 / Re: TT1 Problem 3 (noon)

« on: October 19, 2018, 09:41:04 AM »

(a) WTS $v_{xx} + v_{yy}=0$
     $v_x = e^xsiny+xe^xsiny+ye^xcosy$, $v_{xx}=e^xsiny+e^xsiny+xe^xsiny+ye^xcosy$
     $v_y =xe^xcosy+e^xcosy-ye^xsiny$, $v_{yy}=-xe^xsiny-e^xsiny-e^xsiny-ye^xcosy$
     so, $v_{xx} + v_{yy}=0$

(b) Since v is harmonic, then it is analytic.
     $v_x=u_y, -v_y=u_x$
     $u=\int v_x dy=\int e^xsiny+xe^xsiny+ye^xcosy dy$
       $= -e^xcosy-xe^xcosy+e^x(ysiny+cosy)+h(x)$
     $u_x=-e^xcosy-e^xcosy-xe^xcosy+e^x(ysiny+cosy)+h^{'}(x)$
           $=-e^xcosy-xe^xcosy+e^xysiny+h^{'}x$
     so,$-xe^xcosy-e^xcosy+ye^xsiny=-e^xcosy-xe^xcosy+e^xysiny+h^{'}(x)$
     so, $h^{'}(x)=0$
     h(x)=c
    $u=-xe^xcosy+e^xysiny+c$
   
(c) $f(z)=u+iv= -xe^xcosy+e^xysiny+c+ixe^xsiny+iye^xcosy$
                       $=-e^xcosy(x-iy)+e^xsiny(y+ix)+c$
                       $=-e^xcos(y)\bar{z}+e^xsin(y)i\bar{z}+c$
                       $=\bar{z}e^x(cosy+isiny)+c$
                       $=\bar{z}e^{Rez}e^{Imz}+c$
                       $=\bar{z}e^z+c$


     

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Quiz-2 / Re: Q2 TUT 0102

« on: October 05, 2018, 06:54:08 PM »

Since\begin{equation*} |z|^2=x^2+y^2\end{equation*},
then  \begin{equation*}g(z)= \frac{1}{[1-(x^2+y^2)]^3}\end{equation*}   
Hence, g(z) is not continuous at all points of circle \begin{equation*}x^2+y^2=1\end{equation*}