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### Messages - Meng Wu

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1
##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 5A
« on: November 27, 2018, 11:19:47 AM »
And how do you prove that two last equations are incompatible?
I don't know if I fully understood what you mean but here's my attempt:

Suppose $$f'(z_0)=10z^4+4=0 \\ \Rightarrow z_0^4=-\frac{2}{5} \Rightarrow z_0=\frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}$$
\begin{align}f(z_0)&=2z_0\cdot z_0^4+4z_0+1=0\\&=2z_0\cdot(-\frac{2}{5})+4z_0+1=0\\&=\frac{16}{5}z_0+1=0\\ \Rightarrow z_0=-\frac{16}{5}\neq \frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}\end{align}

2
##### End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 5A
« on: November 27, 2018, 10:36:20 AM »
$(a)$ $\\$
At $|z|=1$, \begin{align}|2z^5+4z+1+(-4z)|&=|2z^5+1|\\&\leq |2z^5|+1\\&=3\\&<4=|-4z|\end{align}
By Rouche's Theorem,
$2z^5+4z+1$ and $-4z$ has the same number of zeros.$\\$ Since $-4z$ has $1$ zero, therefore $2z^5+4z+1$ has $1$ zero in the disk $\{z\colon |z|<1\}$.
$\\$
$\\$

$(b)$ $\\$
At $|z|=2$, \begin{align}|2z^5+4z+1+(-2z^5)|&=|4z+1|\\&\leq |4z|+1\\&=5\\&<64=|-2z^5|\end{align}
By Rouche's Theorem, $2z^5+4z+1$ and $-2z^5$ has the same number of zeros.$\\$
Since $-2z^5$ has $5$ zeros inside $|z|=2$, therefore $2z^5+4z+1$ has $4$ zeros $(5-1=4)$  in the annulus $\{z\colon 1 <|z| < 2\}$.
$\\$
$\\$

$(c)$ $\\$
Notice that the degree of $f(z)=2z^5+4z+1$ is $5$. Which means it has at most $5$ roots. Now from part(b), all the roots are inside $|z|=2$, therefore there are no roots/zeros in the domain $\{z\colon |z|>2\}$.
$\\$
$\\$
$\\$

Show distinct:
$$f(z_0)=2z_{0}^5+4z_0+1=0 \\ f'(z_0)=10z_0^4+4\neq0$$
Thus the multiplicity is $1$, therefore they are all distinct.

3
##### Quiz-6 / Re: Q6 TUT 0301
« on: November 17, 2018, 04:12:24 PM »
$$\because\sin(z)=\sin(z-\pi+\pi)=-\sin(z-\pi), \space\space\space\space\space\space \space\space\space\space\text{since } \sin(\pi+\theta)=-\sin(\theta).$$
$$\therefore -\sin(z-\pi)=-\sum_{n=0}^\infty \frac{(-1)^n(z-\pi)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}$$
\therefore \begin{align} \frac{\sin(z)}{(z-\pi)^2}= \frac{\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}}{(z-\pi)^2}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}\end{align}
$$\\$$
$$\\$$
$$\frac{\sin(z)}{(z-\pi)^2}=\frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}=-\frac{1}{z-\pi}+\cdots$$
Therefore, the residue of $\frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$ is $-1$, which is the coefficient of $(z-z_0)^{-1}$.

4
##### Quiz-6 / Re: Q6 TUT 0202
« on: November 17, 2018, 04:09:29 PM »
$$\because e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots$$
\begin{align}\therefore e^z-1&=(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)-1\\&=z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots \end{align}
Added omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)
$$g(z)=\frac{1}{f(z)}=\frac{1}{\frac{1}{e^z-1}}=e^z-1\\g(z_0=0)=e^0-1=0\\g'(z)=e^z \Rightarrow g(z_0=0)=e^0=1\neq 0 \\$$
Thus the order of the pole of $f(z)$ at $z_0=0$ is $1$.
Hence we let $$\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots$$
$$\therefore(e^z-1)(\frac{1}{e^z-1})=1\\\therefore (z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots)=1$$
$$\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\frac{1}{2}z+\frac{a_0}{2}z^2+\frac{a_1}{2}z^3+\frac{a_{-1}}{6}z^2+\frac{a_0}{6}z^3+\frac{a_{-1}}{24}z^3+\cdots=1$$
$$\therefore \begin{cases}a_{-1}=1\\a_0z+\frac{a_{-1}}{2}z=0 \Rightarrow a_0+\frac{a_{-1}}{2}=0 \Rightarrow a_0=-\frac{1}{2}\\\frac{a_{-1}}{6}z^2+\frac{a_0}{2}z^2+a_1z^2=0 \Rightarrow \frac{a_{-1}}{6}+\frac{a_0}{2}+a_1=0 \Rightarrow a_1=\frac{1}{12}\\ \frac{a_{-1}}{24}z^3+\frac{a_0}{6}z^3+\frac{a_1}{2}z^3+a_2z^3 \Rightarrow \frac{a_{-1}}{24}+\frac{a_0}{6}+\frac{a_1}{2}+a_2=0 \Rightarrow a_2=0\end{cases}$$
Therefore, the first four terms of the Laurent series:
$$\require{cancel} \cancel{1+\frac{1}{z}+(-\frac{1}{2})z^2+(0)z^3} \\ \text{Typo correction: } \frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2$$
$$\\$$
$$\\$$
The residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.
$$\\$$
$$\\$$
If $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\frac{a_2}{2}z^4+\frac{a_1}{6}z^4+\frac{a_0}{24}z^4+\frac{a_{-1}}{120}z^4 \Rightarrow a_3=-\frac{1}{720}$
$\\$
Hence we have $\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2-\frac{1}{720}z^3$.

5
##### Quiz-5 / Re: Q5 TUT 0203
« on: November 02, 2018, 03:59:37 PM »
We know \begin{align}Log(1-z)&=-\sum_{n=1}^{\infty}\frac{z^n}{n}=-(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)\\ \Rightarrow [Log(1-z)]^2&=[-(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)]^2 \\&=(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)^2 \\ &=(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)\\ &=0+(0)z+(1)z^2+(1)z^3+\frac{11}{12}z^4+\frac{5}{6}z^5+\cdots\end{align}
Series is valid at $|z|<1$.

6
##### Quiz-4 / Re: Q4 TUT 0301
« on: October 26, 2018, 05:54:48 PM »
$|z+1|=2 \text{ is the circle at point -1 with radius }2.$
$$\int_{|z+1|=2}\frac{z^2}{4-z^2}dz=\int_{|z+1|=2}\frac{z^2}{(2-z)(2+z)}dz$$
Let $p=2,q=-2 \text{ where } p \text{ lies outside of the circle |z+1|=2}.$
\begin{align}f(z)=\frac{1}{2\pi i}\int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi \\ \Rightarrow \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi&=2\pi i f(z)\end{align}
$\text{Where } f(\xi)=\frac{\xi^2}{2-\xi}, \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=\int_{|z+1|=2}\frac{\frac{\xi^2}{2-\xi}}{\xi-(-2)}d\xi.$ $\\$ $f(z=-2)=\frac{(-2)^2}{2-(-2)}=1.$ $\\$
Therefore, $$\int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=2\pi i f(z)=2\pi i\cdot 1=2\pi i$$

7
##### Term Test 1 / Re: TT1 Problem 4 (morning)
« on: October 21, 2018, 04:52:46 PM »
For $$\gamma_{1}(t)=3i+3e^{it}, \gamma_{1}(t) \text{ is negatively ortiented.}$$
Thus \begin{align}\int_{\gamma_{1}}(z+\bar z)dz&=\int_\frac{\pi}{2}^0(\gamma_{1}(t)+\overline {\gamma_{1}(t)})\gamma_{1}'(t)dt \\ &=\frac{9}{2}e^{2it}+9it \Bigg|_{\frac{\pi}{2}}^0 \\&=9-\frac{9}{2}\pi i\end{align}
For $$\gamma_2(t)=3+3i-3it=3+i(3-3t), 0\leq t\leq 1.$$
\begin{align}\int_{\gamma_{2}}(z+\bar z)dz&=\int_0^1 (\gamma_{2}(t)+\overline {\gamma_{2}(t)})\gamma_{2}'(t)dt \\&=\int_0^1 ((3+i(3-3it))+(3-i(3-3t)))(-3i)dt \\&=\int_0^1 -18idt \\&=-18it \Big|_0^1 \\&=-18i\end{align}
Therefore, $$\int_L(z+\bar z)dz=\int_{\gamma_{1}}(z+\bar z)dz+\int_{\gamma_{2}}(z+\bar z)dz=9-\frac{9}{2}\pi i -18i$$

8
##### Term Test 1 / Re: TT1 Problem 3 (noon)
« on: October 20, 2018, 10:45:12 PM »
Since $(c)$ gives $u(x,y)+iv(x,y)$, the CR equation is definitely $\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$.
$$\\$$
$(b).$
$$-\frac{\partial v}{\partial x}=-(e^x\sin(y)+xe^x\sin(y)+ye^x\cos(y))=\frac{\partial u}{\partial y}$$
\begin{align}\Rightarrow u(x,y)&=\int (-e^x\sin(y)-xe^x\sin(y)-ye^x\cos(y))dy \\&= xe^x\cos(y)-ye^x\sin(y)+h(x)\end{align}
Hence,
\begin{align}\frac{\partial u}{\partial x}&=e^x\cos(y)+xe^x\cos(y)-ye^x\sin(y)+h'(x)\\&=\frac{\partial v}{\partial y}=xe^x\cos(y)+e^x\cos(y)-ye^x\sin(y)\end{align}
$$\Rightarrow h'(x)=0\\\Rightarrow h(x)=C$$
where $C$ is an arbitrary real constant.$\\$
Therefore, $$u(x,y)=xe^x\cos(y)-ye^x\sin(y)+C$$
$$\\$$
$(c).$
$$\\$$
\begin{align}u(x,y)+iv(x,y)&=xe^x\cos(y)-ye^x\sin(y)+C+i(xe^x \sin (y) +y e^x\cos(y))\\&=xe^x\cos(y)+ixe^x\sin(y)+iye^x\cos(y)-ye^x\sin(y)+C\\&=xe^{x+iy}+iye^{x+iy}+C\\&=e^{x+iy}(x+iy)+C\end{align}
Therefore, \begin{align}f(z)&=ze^z+C\end{align}

9
##### Term Test 1 / Re: TT1 Problem 3 (noon)
« on: October 19, 2018, 11:26:20 PM »
.

10
##### Term Test 1 / Re: TT1 Problem 3 (night)
« on: October 19, 2018, 10:07:07 AM »
I think $h'(x)=-2 \Rightarrow h(x)=-2x+C$, where $C$ is an arbitrary real constant.
Thus, $v(x,y)=3x^2y-y^3+3y-2x+C$.
Therefore, $f(z)=f(z)=z^3+3z-2iz+iC$

11
##### Term Test 1 / Re: TT1 Problem 3 (noon)
« on: October 19, 2018, 09:59:30 AM »
For part$(b)$:
CR-equation is:
$$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

12
##### Term Test 1 / Re: TT1 Problem 1 (morning)
« on: October 19, 2018, 07:50:22 AM »
$\textbf{Alternative Method}:$
$$\\$$
Since,
$$z=x+iy, \text{ where } x,y \in \mathbb{R}.$$
$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$$
Then,
\begin{align}\cos(z)=\cos(x+iy)&=\frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2}\\&= \frac{e^{-y+ix}+e^{y-ix}}{2}\\&=\frac{e^{-y}[\cos(x)+i\sin(x)]+e^{y}[\cos(-x)+i\sin(-x)]}{2}\\&=\frac{e^{-y}\cos(x)+ie^{-y}\sin(x)+e^{y}\cos(x)-e^{y}\sin(x)}{2}, \\\text{since } \cos(-x)=\cos(x) \text{ and }\sin(-x)=-\sin(x).\\&=\frac{\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)}{2}\end{align}
Hence,
\begin{align}|\sin(z)|=|\sin(x+iy)|&=\Bigg|\frac{\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)}{2} \Bigg|\\&=\frac{\Bigg|\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)\Bigg|}{\Big|2\Big|}\\&=\frac{\sqrt{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}}{2}\end{align}
Thus,
\begin{align}|\sin(z)|^2=|\sin(x+iy)|^2&=\Bigg(\frac{\sqrt{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}}{2}\Bigg)^2\\&=\frac{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}{4}\\&=\frac{[\cos^2(x)(e^{-2y}+e^{2y}+2)]+[\sin^2(x)(e^{-2y}+e^2y-2)]}{4}\\&=\frac{e^{-2y}(\cos^2(x)+\sin^2(x))+e^{2y}(\cos^2(x)+\sin^2(x))+2(\cos^2(x)-\sin^2(x))}{4}\\&=\frac{e^{-2y}+e^{2y}+2(2\cos^2(x)-1)}{4}\\&=\frac{e^{-2y}+e^{2y}-2+4\cos^2(x)}{4}\\&=\frac{e^{-2y}+e^{2y}-2}{4}+\frac{4\cos^2(x)}{4}\\&=\sinh^2(y)+\cos^2(x)\end{align}
Note:
$$\sinh(y)=\frac{e^{y}-e^{-y}}{2}, \text{ where } y\in \mathbb{R}.$$
\begin{align}\sinh^2(y)&=\Bigg(\frac{e^{y}-e^{-y}}{2}\Bigg)^2\\&=\frac{e^{2y}+e^{-2y}-2}{4}\end{align}
Therefore, we have proven $|\cos(z)|^2=\cos^2(x)+\sinh^2(y)$ for all $z=x+iy$.

13
##### Term Test 1 / Re: TT1 Problem 1 (morning)
« on: October 19, 2018, 07:26:54 AM »
\begin{align}\cos(z)=\cos(x+iy)&=\cos(x)\cos(iy)-\sin(x)\sin(iy) \\&= \cos(x)\cosh(y)-i\sin(x)\sinh(y)\end{align}
\require{cancel}\begin{align}|\cos(z)|^2=|\cos(x+iy)|^2&=\Bigg(\sqrt{\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)}\Bigg)^2 \\ &=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y) \\&= \cos^2(x)(1+\sinh^2(x))+(1-\cos^2(x))\sinh^2(y)\\&=\cos^2(x)+\cancel{\cos^2(x)\sinh^2(x)}+\sinh^2(y)-\cancel{\cos^2(x)\sinh^2(y)}\\&=\cos^2(x)+\sinh^2(y)\end{align}
$$\space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \Box$$

14
##### Term Test 1 / Re: TT1 Problem 3 (night)
« on: October 19, 2018, 06:35:55 AM »
$(a).$
\begin{align}\frac{\partial u(x,y)}{\partial x}&=3x^2-3y^2+3\\ \frac{\partial^2u(x,y)}{\partial x^2}&=6x\ \\ \frac{\partial u(x,y)}{\partial y}&=-6xy+2\\ \frac{\partial^2u(x,y)}{\partial y^2}&=-6x\end{align}
Where first and second partial derivatives are continuous with respect to both $x$ and $y$.
$$\Delta u= \frac{\partial^2u(x,y)}{\partial x^2}+ \frac{\partial^2u(x,y)}{\partial y^2}= 6x+(-6x)=0$$
Therefore, $u(x,y)=x^3-3xy^2+2y+3x$ is a harmonic function.
$\\$
$\\$
$(b).$
$\\$
Use CR-equation to find the harmonic conjugate.
$$\frac{\partial u}{\partial x }=3x^2-3y^2+3= \frac{\partial v}{\partial y }$$
\begin{align}\Rightarrow v(x,y)&= \int(3x^2-3y^2+3)dy \\&=3x^2y-y^3+3y+h(x) \\ \Rightarrow \frac{\partial v}{\partial x }=6x+h'(x)\end{align}
Hence,$$\frac{\partial u}{\partial y } =-6xy+2=-\frac{\partial v}{\partial x }=-(6x+h'(x)) \\ \Rightarrow h'(x)=-2 \\ \Rightarrow h(x)= -2x$$
Therefore the harmonic conjugate $v(x,y)=3x^2y-y^3+3y-2x.$
$\\$
$\\$
$(c).$
$\\$
\begin{align}u(x,y)+iv(x,y)&=x^3-3xy^2+2y+3x+i(3x^2y-y^3+3y-2x)\\ &=x^3-3xy^2+i3x^2y-iy^3+3x+i3y+2y-i2x\end{align}
Consider $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
Thus $$(x+iy)^3=x^3+i3x^2y-3xy^2-iy^3=z^3$$
$$3x+i3y=3(x+iy)=3z$$
$$2y-i2x=-i2(x+iy)=-2iz$$
Therefore, $f(z)=z^3+3z-2iz$.

15
##### Quiz-3 / Re: Q3 TUT 0101
« on: October 12, 2018, 06:23:13 PM »
Let $$\gamma(t)=p+Re^{it}=-4+e^{it}, \text{ where } 0\leq t \leq 2\pi.$$
$$f(z)=\frac{1}{z+4}$$
Thus $$\gamma'(t)=ie^{it}$$
\begin{align}\int_\gamma f(z)dz&=\int_{0}^{2\pi}f(\gamma(t))\gamma'(t)dt\\&=\int_{0}^{2\pi}\frac{1}{-4+e^{it}+4}(ie^{it})dt\\&=\int_{0}^{2\pi}{e^{-it}}(ie^{it})dt\\&=\int_{0}^{2\pi}idt\\&=it\Big|_0^{2\pi}\\&=2\pi i\end{align}

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