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### Messages - yuxuan li

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##### 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*}

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##### 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*}

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##### 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*}

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##### 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}

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##### 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 $x = \frac{7}{2}$

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##### 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}

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##### 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

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

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

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