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Topics - yuxuan li

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1
Test 2 / Test 2 - E - Q1
« on: November 06, 2020, 05:09:31 PM »

Question:
(a) Show that $u(x,y)=-8x^{3}+24xy^{2}+4xy$ 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 $z=x+iy$.

Answer:
(a)
$
\begin{align*}
&u_x=-24x^{2}+24y^{2}+4y\\
&u_{xx}=-48x\\
&u_y=48xy+4x\\
&u_{yy}=48x\\
\Rightarrow &u_xx+u_yy=-48x+48x=0\\
\Rightarrow &\text{It's harmonic.}\square
\end{align*}
$

(b)
$
\begin{align*}
v_x&=-u_y=-48x-4x\\
v_y&=u_x=-24x^{2}+24y^{2}+4y\\
\Rightarrow & v(x,y)=\int{(-48x-4x)}dx+\phi(y)=-(24x^{2}y+2x^{2})+\phi(y)\\
v_y&=-24x^{2}+\phi'(y)\\
\Rightarrow &-24x^{2}+24y^{2}+4y=-24x^{2}+\phi'(y)\\
\Rightarrow &\phi(y)=\int{(24y^{2}+4y)}dy=8y^{3}+2y^{2}+C\\
\Rightarrow &v(x,y)=-24x^{2}y-2x^{2}+8y^{3}+2y^{2}+C\\
\end{align*}
$

(c)
$
\begin{align*}
u(x,y)& + iv(x,y)\\
&=-8x^{3}+24xy^{2}+4xy-24ix^{2}y-2ix^{2}+8iy^{3}+2iy{2}+iC\\
&=-8(x^{3}-iy^{3}-3xy^{2}+3ix^{2}y)-2i(x^{2}-y^{2}+2ixy)+iC\\
&=-8(x+iy)^{3}-2i(x+iy)^{2}+iC\\
&=-8z^{3}-2iz^{2}+iC\\
\Rightarrow & f(z)=-8z^{3}-2iz^{2}+iC
\end{align*}
$

2
Quiz 5 / Quiz 5 - LEC0101 - D
« on: November 06, 2020, 11:38:17 AM »

Question: Give the order of each of the zeros of the given function: $(z^{2}+z-2)^{3}$
Answer:
$
\begin{align*}
&f(z) = (z^{2}+z-2)^{3} = (z-1)^{3}(z+2)^{3}\\
&\text{Let } f(z) = (z^{2}+z-2)^{3}=0 \text{, we have two zeros: }\\
&z_1=1\text{, }z_2=-2\\
&f'(z)=3(2z+1)(z^{2}+z-2)^{2}\\
&f''(z)=6(z^{2}+z-2)(5z^{2}+5z-1)\\
&f'''(z)=120z^{3}+180z^{2}-72z-66\\
\\
&\text{Case 1: }z_1=1\text{, }\\
&f(z_1)=0\\
&f'(z_1)=0\\
&f''(z_1)=0\\
&f'''(z_1)=162\neq0\\
&\text{order of zero at 1 is 3.}\\
\\
&\text{Case 2: }z_2=-2\text{, }\\
&f(z_2)=0\\
&f'(z_2)=0\\
&f''(z_2)=0\\
&f'''(z_2)=-162\neq0\\
&\text{order of zero at -2 is 3.}\\
\end{align*}
$

3
Quiz 4 / LEC0101-1-B
« on: October 23, 2020, 02:29:32 PM »

Question: Find the power series about the origin for the given function: $\frac{z^{3}}{1-z^{3}}\text{ , }|z|<1$
Answer:
$\begin{align*}
\frac{z^{3}}{1-z^{3}}&=\frac{z^{3}+1-1}{1-z^{3}}\\
&=\frac{z^{3}-1}{1-z^{3}}+\frac{1}{1-z^{3}}\\
&=-1+\frac{1}{1-z^{3}}\\
&=-1+\sum_{n=0}^{\infty} (z^{3})^{n}\\
&=-1+1+\sum_{n=1}^{\infty} z^{3n}\\
&=\sum_{n=1}^{\infty} z^{3n}\text{ , where }|z|<1\\
\end{align*}$

4
Quiz 3 / Quiz 3 LEC0101-1D
« on: October 09, 2020, 01:15:54 PM »

Problem: Show that the function $w=g(z)=e^{z^{2}}$ maps the lines {$x = y$} and {$x = -y$} onto the circle {$|w|=1$}.
Show further that g maps each of the two pieces of the region {$x+iy: x^{2}>y^{2}$} onto the set {$w: |w|>1$}.

$
\begin{align}
\text{when }x^{2}=y^{2}\text{:}
\text{    when x = y: } \notag\\
\text{    }g(z)&=g(x+ix)\notag\\
&=e^{x^{2}+2ix^{2}-x^{2}}\notag\\
&=e^{2ix^{2}}\notag\\
&=\cos{(2x^{2})}+i\sin{(2x^{2})}\notag\\
&=x'+iy'\text{, set } x'=\cos{(2x^{2})}\text{ and }y'=\sin{(2x^{2})}\text{,}\notag\\
&\text{here r = 1, and }x'^{2}+y'^{2}=1\notag\\
\text{     when x = -y: } \notag\\
\text{    }g(z)&=g(x-ix)\notag\\
&=e^{x^{2}-2ix^{2}-x^{2}}\notag\\
&=e^{-2ix^{2}}\notag\\
&=\cos{(-2x^{2})}+i\sin{(-2x^{2})}\notag\\
&=x'+iy'\text{, set } x'=\cos{(-2x^{2})}\text{ and }y'=\sin{(-2x^{2})}\text{,}\notag\\
&\text{here r = 1, and }x'^{2}+y'^{2}=1\notag\\
&\text{Therefore, g(z) maps x = y and x = -y onto {w: |w| = 1}}\notag\\
\end{align}
$
$
\begin{align}

\text{   when }x^{2}>y^{2}\text{:}\notag\\
\text{    }g(z)&=g(x+iy)\notag\\
&=e^{{(x+iy)}^{2}}\notag\\
&=e^{x^{2}-y^{2}+2ixy}\notag\\
&=e^{x^{2}-y^{2}}e^{2ixy}\text{,}\notag\\
&\text{here }r=e^{x^{2}-y^{2}}\text{, which is greater than 1 when }x^{2}>y^{2}\notag\\
&\text{Therefore, g(z) maps it onto {w: |w| > 1}}\notag\\

\text{   when }x^{2}<y^{2}\text{:}\notag\\
\text{    }g(z)&=g(x+iy)\notag\\
&=e^{{(x+iy)}^{2}}\notag\\
&=e^{x^{2}-y^{2}+2ixy}\notag\\
&=e^{x^{2}-y^{2}}e^{2ixy}\text{,}\notag\\
&\text{here }r=e^{x^{2}-y^{2}}\text{, which is smaller than 1 when }x^{2}<y^{2}\notag\\
&\text{Therefore, g(z) maps it onto {w: |w| < 1}}\notag\\


\end{align}
$

5
Quiz 1 / Quiz 1 Lec0101 - A
« on: October 03, 2020, 11:48:32 AM »
Describe the locus of points z satisfying the given equation: Im(2iz) = 7.

Let z = x + iy

$
\begin{align}
2iz &=2i(x+iy) \notag\\
&=2ix-2y \notag\\
&=-2y+2xi \notag\\
\end{align}
$

$
\begin{align}
Im(2iz) &=Im(-2y+2xi) \notag\\
&=2x \notag\\
\end{align}

$
=> 2x = 7
=> Vertical line $x = \frac{7}{2}$

6
Quiz 2 / Quiz 2 LEC0101-A
« on: October 02, 2020, 06:27:18 PM »

Question: Find all the value(s) of the given expression.

$
\begin{align}
\log({\sqrt{3}-i}) &=\ln{|\sqrt{3}-i|}+i(arg({\sqrt{3}-i}))\notag\\
&=\ln({\sqrt{{(\sqrt{3})^{2}+(-1)^{2}}}})+i(-\frac{\pi}{6}+2k\pi)\notag\\
&=\ln{2}+i(-\frac{\pi}{6}+2k\pi)\text{, where k} \in \mathbb{Z}\notag\\
\end{align}
$

7
Chapter 1 / Question 3 Chapter 1.4
« on: October 02, 2020, 12:18:11 AM »
Can anyone solve this question?

Find the limit of each sequence that converges; if the sequence diverges, explain why.
3. z_n = n*(i/2)^n

8
Quiz-3 / Quiz 3 TUT0801
« on: October 11, 2019, 04:39:03 PM »
Solution of the quiz.

9
Quiz-2 / TUT0801 - Quiz 2
« on: October 04, 2019, 09:44:32 PM »
Quiz 2 question.

Pages: [1]