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1

Quiz-1 / TUT5103 quiz1

« on: September 26, 2019, 11:50:11 PM »

\begin{aligned}
t y^{\prime}+2 y&=\sin (t), t>0\\
y^{\prime}+\frac{2}{t} y&=\frac{\sin (t)}{t}\\
y^{\prime}+p(t) y&=g(t)
\end{aligned}
$$
\therefore P(t)=\frac{2}{t} \quad g(t)=\frac{\sin (t)}{t}
$$
$$
\mu(t)=e^{\int \frac{2}{t} d t}=e^{2 \ln (t)}=t^{2}
$$
Muttiply $\mu(t)$ to standard equation
$$
\begin{array}{c}{t^{2} y^{\prime}+2 t y=t \sin t} \\ {\mathrm{and}\left(t^{2} y\right)^{\prime}=t \sin t}\end{array}
$$
Integrating  both  sides:
$$
\int\left(t^{2} y\right)^{\prime}=\int t \sin t
$$
Thus, $$t^{2} y=\int t \sin t$$
Integrating by parts:
$$
\begin{aligned}
u&=t, \quad d V=\sin t\\
d u&=d t \quad V=-\cos t
\end{aligned}
$$
$$
\begin{aligned}
\therefore \int t \sin t &=u v-\int v d u \\
 &=-t \cos t-\int-\cos t d t \\
  &=-t \cos t+\int \cos t d t \\
   &=-t\cos t+\sin t+C \\
   \end{aligned}

$$
\begin{aligned}
\therefore t^{2} y&=-t \cos t+\sin t+C\\
y&=\displaystyle\frac{-t \cos t+\sin t+C}{t^{2}}
\end{aligned}

since given $t>0$, $y \rightarrow 0$ as $t \rightarrow \infty$

2

Final Exam / Re: FE-P3

« on: December 18, 2018, 11:13:44 AM »

f(z) = $\frac{\mathrm{sin}\mathrm{}(z)}{\mathrm{cos}\mathrm{}(z)}+z\frac{\mathrm{cos}\mathrm{}\wedge 2(z)}{\mathrm{sin}\mathrm{}\wedge 2(z)}$
      =$ \frac{{{\mathrm{sin}}^{\mathrm{3}}\left(z\right) +\ }{\mathrm{zcos}}^{\mathrm{3}}\left(z\right)\ \ \ \ =g}{{\mathrm{cos} \left(z\right)\ }{sin}^2\left(z\right)\ \ \ \ =h}$
     

Cos(z)sin$\mathrm{\wedge}$2(z) = 0

cos(z) = 0 or sin$\mathrm{\wedge}$2(z) = 0

so z =k$\pi $ or $z=\frac{\pi }{2}+k\pi $
 
1, when z = k$\pi $

g = ${sin}^3(z)+z{cos}^3\left(z\right)\neq 0$                            h= ${\mathrm{cos} \left(z\right)\ }{sin}^2\left(z\right)=0$
                                                                     $h^`=-{sin}^3\left(z\right)+2{cos}^2\left(z\right){\mathrm{sin} \left(z\right)\ }=0$                    ${\ h}^{''}\neq 0$


So pole of order = 2

 2, when z =$\frac{\pi }{2}+k\pi $               

g = ${sin}^3(z)+z{cos}^3\left(z\right)\neq 0$                            h= ${\mathrm{cos} \left(z\right)\ }{sin}^2\left(z\right)=0$
                                                                    $h^`=-{sin}^3\left(z\right)+2{cos}^2\left(z\right){\mathrm{sin} \left(z\right)\ }\ \neq 0$
                               
So pole of order = 1

3

Final Exam / Re: FE-P5

« on: December 18, 2018, 10:48:54 AM »

P(z)=z^3 -3 + 2z -i = (z^3 – 3)+(2z- i)

a, |z-1|<1
     0 < z < 2
Using Rouche Theorem, let z = 2, z^3-3= 5 > 2z - i, z^3 -3 is the  dominant,
z^3-3 =0 and  z has 3 roots.
So it means when z<2, P(z) has 3 roots.
                                   let z = 0, z^3-3 = -3 < 2z - i, 2z - i is the  dominant,
2z - i =0 and  z has 1 root.
So it means when z<0, f(z) has 1 root.

So when 0 < z < 2, f(z) has 3-1 =2 roots.

b, |z-1|>1, |z|<2
     z < 0, z > 2, -2 < z < 2 so -2 < z < 0
when -2 < z < 0, it has 1 root.

c,  z > 2
 when z < 2, it has 3 roots already, so when z > 2, it has 0 root.

4

Final Exam / Re: FE-P4

« on: December 18, 2018, 10:48:14 AM »

$f\left(z\right)=\lambda \cdot \frac{a-z}{1-\overline{a}z}$
$\left|\lambda \right|=1,\ so\ \lambda =e^{it}$
f(0) = 1/2, so $\lambda \cdot \frac{a}{1}=\frac{1}{2}\ $, $\lambda a=\frac{1}{2}$ , $\left|\lambda \right|\cdot \left|a\right|=\frac{1}{2}$ , so $\left|a\right|=\frac{1}{2}\ \ $
We let a = $\frac{1}{2}\cdot e^{i\theta }$, wo can have $\lambda a=\ \frac{1}{2}\cdot e^{i(\theta +t)}$
Since         
        f(1) = $\lambda \cdot \frac{a-1}{1-\overline{a}}=-1$
$\lambda \left(a-1\right)=\overline{a}-1 $
$\lambda a-\lambda =\ \frac{1}{2}e^{-i\theta }-1$
$\frac{1}{2}-e^{it}=\frac{1}{2}e^{-i\theta }-1$
$\frac{1}{2}-\left[cost+isint\right]=\frac{1}{2}\left[cos\theta -isin\theta \right]-1$
From the above, we can have 1, cost =$\ -\frac{1}{2}cos\theta +\frac{3}{2}$   2, sint = $\frac{1}{2}sin\theta \ $, 3, $\theta +t=2k\pi $

Solving these equation, can have $t=\ \theta =0,\ so\ a=\frac{1}{2\ }\ ,\ \ \lambda =1\ $
$f\left(z\right)=\frac{\frac{1}{2}-z}{1-\frac{1}{2}z}=\frac{1-2z}{2-z}$

b, 

let $f\left(z\right)=$ $\frac{1-2z}{2-z}=z$,  we can have ${(z-2)}^2=3,\ z1=\sqrt{3}+2,\ \ z2=-\sqrt{3}+2\ $
$(f')$=$\frac{-2\left(2-z\right)-(1-2z)(-1)}{{(z-2)}^2}$= $\frac{-3}{{(z-2)}^2}$
$|f'|$=$\left|\frac{-3}{{(z-2)}^2}\right|$=$\frac{3}{{(z-2)}^2}$
                                         $arg(f')$ = arg($\frac{-3}{{(z-2)}^2}$) = arg(-1) - arg($\frac{{(z-2)}^2}{3}$)
                                                                              =$\ \ \pi -\ \mathrm{arg(}\frac{{(z-2)}^2}{3}\mathrm{)\ }$

5

Final Exam / Re: FE-P1

« on: December 18, 2018, 10:47:02 AM »

f(z)=$\frac{1}{z^2+2z+2}$
Let $z^2+2z+2=0,$ we can have $\ {(z+1)}^2=i^2,{\ z}_1=i-1,\ z_2=-i-1$
let $f\left(z\right)=\ \frac{1}{z^2+2z+2}=\ \frac{A}{z-\left(i+1\right)}+\frac{B}{z-\left(-i-1\right)}$
  = $\frac{A\left(z-\left(-i+1\right)\right)+B(z-\left(i+1\right))}{(z-\left(i+1\right))(z-\left(-i+1\right))}$
  =$\frac{\left(A+B\right)z-A-B+\left(A-B\right)i}{(z-\left(i+1\right))(z-\left(-i+1\right))}$
      After calculation, we can have $A=-i/2,\ B=i/2$
     So f(Z) = $\frac{-i}{2}\bullet \frac{1}{z-\left(i+1\right)}+\frac{i}{2}\bullet \frac{1}{Z-(-i+1)}$
     a,      f(z) = $-\frac{i}{2}\bullet \frac{1}{i+1}\bullet \frac{1}{\frac{z}{i+1}-1}+\frac{i}{2}\bullet \frac{1}{-i+1}\bullet \frac{1}{\frac{z}{-i+1}-1}$
                    = $\frac{i}{2(i+1)}\bullet \sum^{\infty }_{n=0}{{(\frac{z}{i+1})}^n-\frac{i}{2(1-i)}\sum^{\infty }_{n=0}{{(\frac{z}{-i+1})}^n}}$
   
                  For convergence, we need $\left|\frac{z}{i+1}\right|<1,\ \left|\frac{z}{-i+1}\right|<1,\ \ so\ we\ can\ have\ z<\sqrt{2}.$

     b,       f(z) = $-\frac{i}{2}\cdot \frac{1}{z}\cdot \frac{1}{1-\frac{i+1}{z}}$ + $\frac{i}{2}\cdot \frac{1}{Z}\cdot \frac{1}{1-\frac{-i+1}{z}}$
                      = $\frac{i}{2z}\cdot \sum^{\infty }_{n=0}{{(\frac{i+1}{z})}^n-\frac{i}{2z}\sum^{\infty }_{n=0}{{(\frac{1-i}{z})}^n}}$
                  For convergence, we need  $\left|\frac{i+1}{z}\right|<1,\ \left|\frac{1-i}{z}\right|<1,\ so\ we\ can\ have\ z>\sqrt{2}$
                  When z =$\sqrt{2}$ , for $\sum^{\infty }_{n=0}{a_n},\ {\mathop{\mathrm{lim}}_{n\to \infty } a_n\neq 0\ }$, so we can conclude it is geometric divergent.

6

Quiz-7 / Re: Q7 TUT 5201

« on: November 30, 2018, 04:13:22 PM »

$f\left(z\right)=z^4-3z^2+3$
When z lines in x axis, z = x+ yi = x

So                                                             $f\left(x\right)=x^4-3x^2+3$

Because the domain of f(x) is $\left[0,\left.R\right]\right.$, so arg(f(z)) = 0

Let  $z=Re^{it}$, when $0\le t\le \frac{\pi }{2}$,

So                                                            f($Re^{it}$) = $R^4e^{4it}-3R^2e^{2it}+3$, $\mathrm{0}\mathrm{\le }\mathrm{4}\mathrm{t}\le \2pi $,

Hence                                                   arg(f(z)) = 2$\pi$

When z = yi  $0\le y\le R$                    arg(f(z)) = 0

So the net change of the angle is 0 + 2$\pi$ + 0 = 2$\pi$, and $\frac{1}{2\pi }\bullet 2pi\ =1$

There are 1 zero of the function.

7

Quiz-7 / Re: Q7 TUT 0102

« on: November 30, 2018, 04:00:01 PM »

$f\left(z\right)=2z^4-2iz^3+z^2+2iz-1$
When z lies in x axis, z = x + yi = x

So$f\left(x\right)=2x^4-3ix^3+x^2+2ix-1$

Because the domain of f(x) is $\left[-\infty ,\infty \right]$,  $f\left(-\infty \right)\to \infty ,\ f\left(\infty \right)\to \infty ,\ $so arg(f(z)) = 0

Let $z=Re^{it}$,  , when 0 $\mathrm{<}$= t $\mathrm{<}$= $\pi$
$f\left(t\right)=2R^4e^{i4t}-2iR^3e^{i3t}+R^2e^{i2t}+2iRe^{it}-1$
And $0\le 4t\le 4\pi $

So arg(f(z)) = 4$\pi$


So the net change of the angle is $0+4\pi =4\pi $ , and  $\frac{1}{2\pi }\bullet 4\pi =2$

There are 2 zeroes of the function.

8

Quiz-7 / Re: Q7 TUT 0201

« on: November 30, 2018, 03:58:30 PM »

Since $ze^z-\frac{1}{4}=0\ {{\mathop{\Leftrightarrow}\limits_{}}}\ 4z-e^{-z}=0$
$\ f\left(z\right)=4z,\ s\left(z\right)=e^{-z},\ for\ 0<\left|z\right|<2.$
When $\left|z\right|=2,\ \left|s(z)\right|=\left|e^{-z}\right|=e^{-Re(z)}\le e^2\cong 7.387\dots <8$

And                   $\left|4z\right|=4\left|z\right|=4\bullet 2=8$

So $\left|s(z)\right|<\left|f(z)\right|,\ \ for\ \left|z\right|=2.$

Hence $g\left(z\right)=f\left(z\right)-s\left(z\right)=4z-e^{-z}$ has the same number of zeros

As $f\left(z\right)=4z\ \ in\ \left|z\right|<2$, this is 1 zero.

And when z = 0,$g\left(0\right)=4\bullet 0-e^0=-1\neq 0$

Hence z = 0 is not a zero of g(z).

We can conclude that $ze^z-\frac{1}{4}=0\ has\ 1\ zero\ in\ 0<\left|z\right|<2$.