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Topics - Xuefen luo

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Test 4 / 2020F-Test4-MAIN-A-Q3
« on: December 09, 2020, 02:37:53 PM »
Problem 3. Consider $f(z)=\frac{3z}{(z+1)(z-2)}$.
(i)Decompose it into Laurent's series $f(z) = \sum_{n=?}^{n=?}a_nz^n$ (find r and R)
(a) In the disc {$z:|z|<r$};
(b) In the annulus {$z:r<|z|<R$};
(c) In the disc exterior {$z : |z| > R$}.
(ii) Calculate $Res(f(z),0)$ and $Res(f(z),\infty)$.

(i) Let $f(z)=\frac{A}{(z+1)} +\frac{B}{(z-2)}$, and let $r=1$, $R=2$.

Then, $A(z-2)+B(z+1)=3z$
$\left\{ \begin{align*}
\right.$ $\Rightarrow$ $\left\{ \begin{align*}

For $|z|<1: \frac{1}{z+1} = \sum_{n=0}^{\infty}(-z)^n$.

For $|z|>1: \frac{1}{z+1} = \frac{1}{z}\frac{1}{1+\frac{1}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(-\frac{1}{z})^n=\sum_{n=0}^{\infty}(-1)^nz^{-n-1}=\sum_{n=-\infty}^{-1}(-1)^{-n-1}z^n$

For $|z|<2: \frac{1}{z-2}=\frac{1}{2}\frac{1}{\frac{z}{2}-1}=-\frac{1}{2}\sum_{n=0}^{\infty}(\frac{z}{2})^n = - \sum_{n=0}^{\infty}2^{-n-1}z^n$

For $|z|>2: \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(\frac{2}{z})^n =  \sum_{n=0}^{\infty}2^nz^{-n-1}=\sum_{n=-\infty}^{-1}2^{-n-1}z^n$

f(z)&=\frac{1}{(z+1)} +\frac{2}{(z-2)}\\
\sum_{n=0}^{\infty}((-1)^n-2\cdot 2^{-n-1})z^n, & |z|< 1\\
\sum_{n=-\infty}^{-1}(-1)^{-n-1}z^n-2\cdot \sum_{n=0}^{\infty}2^{-n-1}z^n, &1<|z|<2\\
\sum_{n=-\infty}^{-1}((-1)^{-n-1}+2\cdot 2^{-n-1})z^n, &|z|>2\\

(ii)From part(i), we know the Laurent's series expression of $f(z)$.
For $z_0=0, f(z)=\sum_{n=0}^{\infty}((-1)^n-2\cdot 2^{-n-1})z^n$,
then $Res(f(z),0)=a_{-1}=(-1)^{-1}-2\cdot2^{1-1}=-1-2=-3$.
For $z_0=\infty, f(z)=\sum_{n=-\infty}^{-1}((-1)^{-n-1}+2\cdot 2^{-n-1})z^n$,
then $Res(f(z),\infty)=a_{-1}=(-1)^{1-1}+2\cdot2^{1-1}=1+2=3$

Test 4 / 2020F-Test4-MAIN-A-Q2
« on: December 09, 2020, 02:09:42 PM »
Problem 2. Calculate an improper integral $I=\int_{0}^{\infty} \frac{\sqrt{x}dx}{(x^2+2x+1)}$.

(a) Calculate  $ J_{R,\epsilon}=\int_{\Gamma_{R,\epsilon}} f(z)dz, \  f(z):= \frac{\sqrt{z}}{(z^2+2z+1)}$
 $\Gamma_{R,\epsilon}$ is the contour on the figure.

(b)Prove that $\int_{\gamma_R} f(z)dz \rightarrow 0$ and $\int_{\gamma_ \epsilon} f(z)dz \rightarrow 0$ as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$ where $\gamma_R$ and $\gamma_\epsilon$ are arcs.

(c) Express limit of $J_{R,\epsilon}$ as $R \rightarrow +\infty$, $\epsilon \rightarrow 0^+$ using $I$.

(a) Since $f(z) = \frac{\sqrt{z}}{(z^2+2z+1)} = \frac{\sqrt{z}}{(z+1)^2}$, $z=-1$ is the only singularity inside $\Gamma_{R,\epsilon}$ as $R>1$.

The residue is $Res(f(z),-1)=\frac{(\sqrt{z})'}{1!}|_{z=-1} = \frac{\frac{1}{2\sqrt{z}}}{1!}|_{z=-1} = -\frac{i}{2}$

Thus, by residue theorem $J_{R,\epsilon}= 2\pi i Res(f(z),-1)=2\pi i (-\frac{i}{2})=\pi$

(b) \begin{align*}
\int_{\gamma_R} f(z)dz &\leq \left| \int_{\gamma_R} f(z)dz \right|\\
&\leq 2\pi R\cdot max \left|\frac{\sqrt{z}}{(z^2+2z+1)} \right|\\
&\leq 2\pi R\cdot max \left| \frac{R^{\frac{1}{2}}e^{i\frac{1}{2}t}}{(Re^{it}+1)^2}\right| \ \ \  , \ \text{$z=Re^{it}, t\in [0,2\pi]$}\\
&\leq 2\pi R \cdot \frac{R^{\frac{1}{2}}}{(R+1)^2} \rightarrow 0 \ \ \ \ as \ R \rightarrow \infty\\
\int_{\gamma_ \epsilon} f(z)dz &\leq \left|\int_{\gamma_ \epsilon} f(z)dz\right|\\
&\leq 2\pi \epsilon \cdot max \left| \frac{\sqrt{z}}{(z^2+2z+1)} \right|\\
&\leq 2\pi \epsilon \cdot max \left| \frac{\epsilon^{\frac{1}{2}}e^{i\frac{1}{2}t}}{(\epsilon e^{it}+1)^2}\right| \ \ \ , \ \text{$z=\epsilon e^{it}, t\in [2\pi,0]$}\\
&\leq 2\pi \epsilon \cdot max (\frac{\epsilon^{\frac{1}{2}}}{(|1|-|\epsilon e^{it}|)^2})\\
&\leq 2\pi \epsilon \cdot \frac{\epsilon^{\frac{1}{2}}}{(1-\epsilon)^2} \rightarrow 0 \ \ \ \ as \ \epsilon \rightarrow 0\\

J_{R,\epsilon} &= \int_{\gamma_ R} f(z)dz +\int_{\gamma_ \epsilon} f(z)dz +\int_{\epsilon}^{\infty} f(z)dz+\int_{\infty}^{\epsilon}f(z)dz\\
\pi &= 0+0+\int_{\epsilon}^{\infty} f(z)dz+\int_{\infty}^{\epsilon}f(z)dz\\

As $R \rightarrow +\infty$, $\epsilon \rightarrow 0^+$,
\int_{\epsilon}^{\infty} f(z)dz &= \int_{0}^{\infty}\frac{\sqrt{x}dx}{(x^2+2x+1)}=I\\
\int_{\infty}^{\epsilon}f(z)dz &= \int_{\infty}^{0}\frac{\sqrt{z}}{(z^2+2z+1)}dz  \ \ \text{, $z=xe^{i2\pi},dz=e^{i2\pi} dx$}\\
&=\int_{\infty}^{0}\frac{\sqrt{x}e^{i\pi}}{(xe^{i2\pi}+1)^2} e^{i2\pi}dx\\
Then, the limit of $J_{R,\epsilon}$ as $R \rightarrow +\infty$, $\epsilon \rightarrow 0^+$ is $2I$. Thus, $2I=\pi \Rightarrow I=\frac{\pi}{2}$

Test 4 / 2020F-Test4-MAIN-A-Q1
« on: December 09, 2020, 01:16:20 PM »
Problem 1. (a) Find all zeroes of the function $f(z) = \frac{sin(2\pi \sqrt{z})}{sin(\pi z)}$ in domain $D=\mathbb{C}\ (-\infty,0]$.
(b)Also find all singular points of this function and determine their types(removable, pole (in which case what is it's order), essential singularity, not
isolated singularity).
(c) In particular, if $\infty$ is in the domain: check whether it is a zero.
(d) Draw these points on the complex plane.

Answer: Notice that $sin(2\pi \sqrt{z})$ has zeroes only at $z=\frac{n^2}{4}, n \in \mathbb{Z^+}$, and $sin(\pi z)$ has zeroes only at $z=m, m \in \mathbb{Z^+}$ in domain $D=\mathbb{C}\ (-\infty,0]$. Since all these zeroes are simple, we can conclude that:
-$z=\frac{n^2}{4}$ with $n \in \mathbb{Z^+}$ and $ n^2$ is a multiple of 4 such that $z=\frac{n^2}{4}=m$ for some $m\in \mathbb{Z^+}$, then $z$ are removable singularities;
-$z=\frac{n^2}{4}$ with $n \in \mathbb{Z^+}$  and  $n^2$ is not a multiple of 4 such that $z \notin \mathbb{Z^+}$, then $z$ are simple zeroes;
-$z=m$ for $m \in \mathbb{Z^+}$ and $m \neq \frac{n^2}{4}$ for any $n \in \mathbb{Z^+}$, then $z$ are simple poles.
-there are no essential singularities;
-$z=\infty$ is a non-isolated singularity.

Quiz 7 / LEC0101-Quiz7-ONE-E
« on: December 09, 2020, 12:21:13 PM »
Problem: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:

Answer: According to the picture, we divide the line into three parts, denoted $\gamma_1, \gamma_2, \gamma_3$.

For $\gamma_1$: $z=x, x\in [0,R]$ with $R \rightarrow \infty$.
$f(z) = x^4-3x^2+3$.
$f(z)$ is real number as $x$ traverses from $0$ to $R$.
Thus, change of argument of $f(z)$ is $0$.

For $\gamma_2$: $z=Re^{it}, t\in [o,\frac{\pi}{2}], R \rightarrow \infty$.
$f(z) = R^4e^{i4t}-3R^2e^{i2t}+3 =R^4(e^{i4t}-3\frac{R^2}{R^4}e^{i2t}+\frac{3}{R^4}) =R^4e^{i4t}$, as $R \rightarrow \infty$
Then, $4t\in [0,2\pi]$ and change of argument of $f(z)$ is $2\pi$.

For $\gamma_3$: $z=yi, y\in [R,0]$ with $R \rightarrow \infty$.
$f(z)= y^4+3y^2+3$
$f(z)$ is real number as $y$ traverses from $R$ to $0$.
Thus, change of argument of $f(z)$ is $0$.

Since $f(z)$ has no poles, we know the number of zeroes of $f(z)$ in the first quadrant is $\frac{1}{2\pi}(0+2\pi+0)=1$ by the argument principle.

Test 2 / 2020F Test2-MAIN-D Q3
« on: November 04, 2020, 06:49:45 AM »
Problem 3:Consoder the map $z \xrightarrow\ w = f(z) = \frac{1}{z}$ and compute the image $f(D)$ of $D = \{z:|z-1|>1, |z-2|<2\}$ is mapped to.
Check if $f:D\xrightarrow\ f(D)$ is a bijection.

Let $z=x+iy$ and $w=u+iv$.
$w=f(z)+\frac{1}{z}$, then $z=\frac{1}{w}$.

    |1-w|&> |w|\\
    |w-1|^2 &> |w|^2\\
    |w|^2 +|1|^2 - 2Re(w \cdot 1) &> |w|^2\\
    |w|^2 +1 - 2Re(w) &> |w|^2\\
    1 - 2Re(w) &>0\\
    2Re(w) &< 1\\
    Re(w) &< \frac{1}{2}
Hence, we have $u<\frac{1}{2}$ since $w=u+iv$.

|2w|^2 + |1|^2 - 2Re(2w \cdot 1) &< 4|w|^2\\
4|w|^2 + 1 - 4Re(w) &< 4|w|^2\\
1 - 4Re(w) &<0\\
4Re(w) &> 1\\
Re(w) &>\frac{1}{4}
Hence, we have $u>\frac{1}{4}$ since $w=u+iv$.
Now, we have $\frac{1}{4}<u<\frac{1}{2}$.
The figures of domain and codomian of f are attached below. For all $w$ in $f(D)$, there exists $z$ in $D$ such that $w=\frac{1}{z}$. Thus, $f$ is onto.

Then, we will show f is one to one.
Consider $z_1,z_2 \in D,$ and $f(z_1)=f(z_2)$,
    \frac{1}{z_1} &= \frac{1}{z_2} \ \ \ \  \text{$z_1,z_2 \neq 0$}\\
    z_1 &= z_2\\
Now, we have $f$ is one to one.
Therefore, $f:D \xrightarrow\ f(D)$ is bijection.

Test 2 / 2020F Test2-MAIN-D Q2
« on: November 04, 2020, 05:48:18 AM »
Problem 2: Calculate directly the integral $\int_{L} Re(z)\ dz$ where L is the path,consisting of two $\frac{1}{4}$-circles of radius 3 and 1 centered at 0 and two straight segment on the figure.

Consider $L_1$ be the line from 1 to 3:
$L_1: \gamma_1(x) = x$, $x \in [1,3]$, and $\gamma_1'(x) = 1$

Consider $L_2$ be the curve of circle centered at 0 with radius $r=3$:
$L_2: \gamma_2(t) = 3e^{it}$, $t \in [0,\frac{\pi}{2}]$, and $\gamma_2'(t) = 3ie^{it}$

Consider $L_3$ be the line from 3i to i:
$L_3: \gamma_3(y) = iy$, $y \in [3,1]$, and $\gamma_3'(y) = i$

Consider $L_4$ be the curve of circle centered at 0 with radius $r=4$:
$L_4: \gamma_4(t) = e^{it}$, $t \in [\frac{\pi}{2},0]$, and $\gamma_4'(t) = ie^{it}$

Let $f(z)=Re(z)$.

Now, we have
    \int_{L} Re(z)\ dz &= \int_{L_1} f(\gamma_1(x))\gamma_1'(x)\ dx+\int_{L_2} f(\gamma_2(t))\gamma_2'(t)\ dt+\int_{L_3} f(\gamma_3(y))\gamma_3'(y)\ dy+\int_{L_4} f(\gamma_4(t))\gamma_4'(t)\ dt\\
    &= \int_{1}^{3} x\ dx + \int_{0}^{\frac{\pi}{2}} 3cost \cdot3ie^{it}\ dt+\int_{3}^{1}0\cdot i\ dy+\int_{\frac{\pi}{2}}^{0} cost \cdot ie^{it}\ dt\\
    &= (\frac{1}{2}x^2)|_{1}^{3}+9i\int_{0}^{\frac{\pi}{2}} cos^2t+isintcost\ dt+i\int_{\frac{\pi}{2}}^{0} cos^2t+isintcost\ dt\\
    &= (\frac{9}{2}-\frac{1}{2}) +8i\int_{0}^{\frac{\pi}{2}} \frac{cos2t+1}{2}+i\frac{sin2t}{2}\ dt\\
    &= 4 +8i(\frac{sin2t}{4}+\frac{1}{2}-i\frac{cos2t}{4})|_{0}^{\frac{\pi}{2}}\\
    &=4 +8i(0+\frac{\pi}{4}+\frac{1}{2}i)\\
    &=4+2\pi i -4\\
    &=2\pi i
Therefore, $\int_{L} Re(z)\ dz =2\pi i$

Test 2 / 2020F Test2-MAIN-D Q1
« on: November 04, 2020, 04:08:54 AM »
Problem 1:
(a)Show that
\begin{align*} u(x,y)=sin(4x)e^{-4y}+3xy.\end{align*}
is a harmonic function.
(b) Find a harmonic conjugate function $v(x,y)$.
(c)Consider $u(x,y)+iv(x,y)$ and write it as a function $f(z)$ of $z=x+iy$.

 u_x &= 4cos(4x)e^{-4y}+3y\\
 u_{xx} &= -16sin(4x)e^{-4y}\\
 u_y &= -4sin(4x)e^{-4y}+3x\\
 u_{yy} &= 16sin(4x)e^{-4y}\\
 \Delta u &= u_{xx}+u_{yy}\\
Thus, $u(x,y)$ is a harmonic function.

(b) We know
v_x = -u_y = 4sin(4x)e^{-4y}-3x\\
v_y = u_x = 4cos(4x)e^{-4y}+3y

v(x,y) &= \int v_x\ dx\\
&=\int 4sin(4x)e^{-4y}-3x\ dx\\
&=-cos(4x)e^{-4y}- \frac{3}{2}x^2+\phi (y)\\
v_y &= 4cos(4x)e^{-4y} + \phi '(y)\\
&= 4cos(4x)e^{-4y}+3y\\
Now, we have
\phi '(y) &= 3y\\
\phi (y) &= \frac{3}{2}y^2 +C\\
v(x,y) &= -cos(4x)e^{-4y}- \frac{3}{2}x^2+\frac{3}{2}y^2 +C\\

(c) Consider $u(x,y)+iv(x,y)$, we have
u(x,y)+iv(x,y) &= sin(4x)e^{-4y}+3xy + i(-cos(4x)e^{-4y}- \frac{3}{2}x^2+\frac{3}{2}y^2 +C)\\
&= sin(4x)e^{-4y}+3xy -icos(4x)e^{-4y}- i\frac{3}{2}x^2+i\frac{3}{2}y^2 +iC\\
&= e^{-4y}(-i)(cos(4x)+isin(4x))-i\frac{3}{2}(x+iy)^2+iC\\
&= -ie^{4iz}-i\frac{3}{2}z^2+iC\ \ \ \text{for $z=x+iy$}\\
Therefore, $f(x) = -ie^{4iz}-i\frac{3}{2}z^2+iC$

Quiz 4 / Quiz4-problemSixE
« on: October 24, 2020, 02:50:58 AM »
Problem: Find a "Closed form" for the given power series:
\begin{align*} \sum_{n=1}^{\infty} z^{3n}.\end{align*}
Since we know that
\begin{align*} \sum_{n=0}^{\infty} z^{n} &= \frac{1}{1-z}\\
\sum_{n=0}^{\infty} (z^3)^n &= \frac{1}{1-z^3}\end{align*}
 \sum_{n=1}^{\infty} z^{3n} &=  \sum_{n=0}^{\infty} z^{3n} - z^{3*0}\\
 &=\sum_{n=0}^{\infty} z^{3n} - 1\\
 &= \frac{1}{1-z^3} - 1 \\
Therefore, the closed form for $ \sum_{n=1}^{\infty} z^{3n}$  is $\frac{1}{1-z^3} - 1 $.

Quiz 3 / Quiz3 problem 6E
« on: October 18, 2020, 03:52:58 PM »
Problem: Compute the following line integral:
\begin{align*}\int_{\gamma}^{} |z|^2 \,dz, \end{align*}
where $\gamma$ is the line segment from 2 to 3 + i
 \gamma(t)&= (1-t)z_0+tz_1\\
 &= 2(1-t)+(3+i)t\\
 &= 2-2t+3t+it\\
 &= 2+t+it \ \ \ \  (0 \leq t \leq 1)\\
 Let\ f(z)=|z|^2\\
 f(\gamma(t))&= |\gamma(t)|^2\\
 \int_{\gamma}^{} |z|^2 \,dz &= \int_{\gamma}^{} f(z) \,dz\\
 &=\int_{0}^{1} f(\gamma(t))\  \gamma'(t) \,dt\\
 &=\int_{0}^{1} (2t^2+4t+4) (1+i) \,dt\\
  &=(1+i)\int_{0}^{1} (2t^2+4t+4) \,dt \\
   &=(1+i)(\frac{2}{3}t^3 + 2t^2 +4t)|_{0}^{1}\\
    &=(1+i)(\frac{2}{3} + 2 +4)\\

Quiz 2 / Quiz 2 Section 6101
« on: October 06, 2020, 06:50:19 AM »
Find the limit of the function at the given point, or explain why it does not exist.
f(z)&=\frac{z^3-8i}{z+2i} \ (z \neq -2i) \ at \ z_0=-2i\\
\lim_{z \to -2i} f(z) &= \lim_{z \to -2i} \frac{z^3-8i}{z+2i}\\
 &=\lim_{z \to -2i} \frac{z^3+(2i)^3}{z+2i}\\
 &=\lim_{z \to -2i} \frac{(z+2i)(z^2-2iz-4)}{z+2i}   \ (By\ a^3+b^3=(a+b)(a^2-ab+b^2))\\
 &=\lim_{z \to -2i} (z^2-2iz-4)\\
 &= (-2i)^2-2i(-2i)-4\\
 &= -4-4-4\\
 &= -12\\

Quiz-4 / TUT0301
« on: October 18, 2019, 01:59:50 PM »

This is non-homogeneous differential equation, so to find the complimentary solution,
we need to consider $y''+2y'+y=0$.

We assume that $y=e^{rt}$ is a solution of this equation. Then the characteristic equation is:


Then, the complimentary solution is given by
$ y_c(t)=c_1e^{-t}+c_2te^{-t}$, where $c_1, c_2$ are constants.

To find the particular solution, we assume that $y_p(t)=Ae^{-t}$.
However, it fails because $e^{-t}$ is a solution of the homogeneous equation.
Also if we assume $y_p(t)=Ate^{-t}$, again it fails as $te^{-t}$ is also a solution of the homogeneous equation.

Then, we assume $y_p(t)=At^2e^{-t}$ is the particular solution,
then it satisfies the equation $y''+2y'+y=2e^{-t}$.

Since $y_p=At^2e^{-t}$,

Using these values in equation $y''+2y'+y=2e^{-t}$, we have:

i.e. $2Ae^{-t}=2e^{-t}$
i.e. $A=1$

Then the particular solution is

Hence the general solution of the equation is
i.e. $y=c_1e^{-t}+c_2te^{-t}+t^2e^{-t}$

Quiz-3 / TUT0301
« on: October 11, 2019, 02:14:55 PM »
Dividing both sides by $x^2$, then we have:
$y''+ \frac{1}{x}y'+\frac{(x^2-v^2)}{x^2}y=0$
Since $W=ce^{\int -p(x)dx}$, and $p(x)=\frac{1}{x}$ in this case, we have:
$W=ce^{-\int \frac{1}{x}dx}=ce^{-ln(x)+C}=ce^{ln(x^{-1})+C}=cx^{-1}e^C$
We know $e^C$ is just a constant, so we can just subsume it into $c$. Then the Wronskian is $W=\frac{c}{x}$

Quiz-2 / TUT0301
« on: October 04, 2019, 02:01:48 PM »
$1+(\frac{x}{y} - sin(y))y' = 0$
Let $M(x,y)=1$ and $N(x,y)=(\frac{x}{y} - sin(y))$
Then, $M_y = 0$ and $N_x = \frac{1}{y}$
We can see that this equation is not exact.
However, we know $\frac{N_x-M_y}{M}=\frac{1}{y}$, and $\mu =y$
Multiplying the original equation by $\mu(y)$, we have
Since $M_y=1$ and $N_x=1$, this equation is exact.
Thus, there exists a function $\psi(x,y)$ such that
$\psi_x(x,y)=y$ and $\psi_y(x,y)=x-ysin(y)$
Hence, $\psi(x,y) = \int y dx = xy + h(y)$
then $\psi_y(x,y)= x+h'(y) = x-ysin(y)$
We have $h'(y)=-ysin(y) \Rightarrow h(y)=- \int ysin(y)dy = ycos(y)-sin(y)$
Therefore, $\psi(x,y)=xy+ycos(y)-sin(y)$, and the solutions of the equation are given implicitly by $xy+ycos(y)-sin(y)=C$

Quiz-1 / TUT0301
« on: September 29, 2019, 04:48:10 PM »
$y'-y = 2te^{2t}, y(0) = 1$
y'-y &= 2te^{2t}\\
\mu &= e^{\int-1dt} = e^{-t}\\
e^{-t}y' - e^{-t}y &= 2te^t\\
e^{-t}y &= \int 2te^t dt\\
e^{-t}y &= 2(te^t - \int e^tdt)\\
e^{-t}y &= 2te^t - 2e^t + c\\
y &= 2te^2t - 2e^2t + ce^t\\
Since $y(0) = 1, y(0) = 0-2+c = 1$, then $c = 3$
Therefore, $y = 2te^2t - 2e^2t + 3e^t$.

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