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### Messages - Jiaqi Bi

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##### Quiz 1 / Quiz 1-0201 C Problem 2
« on: January 28, 2021, 12:28:16 PM »
Problem 2: Find the general solutions to the following equation:
$$u_{xyz}=x+y+z$$
Solution:

\begin{align*}
u_{xyz}&=x+y+z\\
u_{xy}&=xz+yz+\frac{1}{2}z^2+f(x,y)\\
u_{x}&=xyz+\frac{1}{2}y^2z+\frac{1}{2}yz^2+f'(x,y)+g(x,z)\\
u&=\frac{1}{2}x^2yz+\frac{1}{2}xy^2z+\frac{1}{2}xyz^2+f''(x,y)+g'(x,z)+h(y,z)\\
&=\frac{1}{2}x^2yz+\frac{1}{2}xy^2z+\frac{1}{2}xyz^2+\phi (x,y)+G(x,z)+h(y,z)\ \ \ \ \ \ \ \  (\text{Say}\ \phi (x,y)=f''(x,y), G(x,z)=g'(x,z))
\end{align*}

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##### Quiz 1 / Quiz 1-0201 C Problem 1
« on: January 28, 2021, 12:26:19 PM »
Problem 1: Determine if they are linear homogeneous, linear inhomogeneous, or nonlinear ($u$ is an unknown function); for nonlinear equations, indicate if they are also semilinear, or quasilinear:
$$u_t+xu_x=0$$
Solution:
This is linear homogeneous because it has $f(x)=0$ on the RHS, and $Lu$ is a partial differential expression linear, such that it contains only coefficient $1$ and $x$ for $u_t$ and $u_x$.

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##### Chapter 1 / Math Notation Question
« on: January 23, 2021, 10:53:54 AM »
For example, I am solving an equation that leads me to this step: $u_x=e^u\cdot g(x)$,
Should I use $\frac{\partial u}{\partial x}$ as my left-hand side or use $\frac{du}{dx}$?

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##### Quiz 1 / Re: 0101 quiz1
« on: December 20, 2020, 01:15:56 PM »
$\text{Im}(2iz)=2x=7\Rightarrow x=\frac{7}{2}$ (Typos at this line.)

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##### Quiz 1 / Re: Quiz1-6101 A
« on: December 20, 2020, 01:08:17 PM »
The second line of the answer, left part of your equation should be $|x+iy-i|$ instead of $|x+iy-1|$

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##### Quiz 7 / Re: Lec 0101- Quiz 7 C
« on: December 17, 2020, 07:43:36 AM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:
$z+a=e^z, (a>0)$

(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)
(The graph will be attached.)
Solution:
\begin{align} h(iy)&=iy+a-e^{iy}\\ &=iy+a-cos(y)-isin(y)\\ &=(a-cos(y))+i(y-sin(y)) \end{align}
If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.
$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.
\begin{align} h(\text{Re}^{it})=\text{Re}^{it}+a=e^{\text{Re}^{it}} \end{align}
Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.
Therefore, the argument for $h(z)$ should be $2\pi$.
By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.
Hence, $h(z)$ has a total of one zero in this plane.

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##### Quiz 7 / Lec 0101- Quiz 7 C
« on: December 17, 2020, 06:51:38 AM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:
$z+a=e^z, (a>0)$

(The graph will be attached.)
(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)
Solution:
\begin{align} h(iy)&=iy+a-e^{iy}\\ \nonumber &=iy+a-cos(y)-isin(y)\\ \nonumber &=(a-cos(y))+i(y-sin(y)) \end{align}
If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.
$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.
\begin{align} h(\text{Re}^{it})=\text{Re}^{it}+a=e^{\text{Re}^{it}} \end{align}
Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.
Therefore, the argument for $h(z)$ should be $2\pi$.
By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.
Hence, $h(z)$ has a total of one zero in this plane.

8
##### Test 4 / Re: Spring 2020 Test 2 Monday Sitting Problem 3
« on: December 02, 2020, 03:34:35 PM »
What I got on these poles were the same as you did $n \neq 0\ \& -1$. I believe that was a typo.

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##### Quiz 5 / Quiz-5-6101 A
« on: November 05, 2020, 08:24:07 AM »
$\textbf{Problem}$:
Give the order of each of the zeros of the given function:
$$e^{2z}-3e^z-4$$

$\textbf{Solution}$:
$f(z)=e^{2z}-3e^z-4=0$
Let $w=e^z$, we get $w^2-3w-4=(w-4)(w+1)$
Then resubstitute $e^z=w$
We get $(e^z-4)(e^z+1)=0 \Rightarrow e^z=4\ \text{and}\ e^z=-1$
Solve for $z$, we get $z=log(4)=ln(4)+i(2k\pi)\ \text{and}\ z=log(-1) =ln(-1)+i(2k\pi) =\pi i+i(2k\pi) =i(\pi+2k\pi)$
To find orders: $f'(z)=2e^{2z}-3e^z$
Substitute $e^z=4\ \text{and}\ -1$ into $f'(z)$
$2\times 16-3\times 4 = 20 \neq 0, 2+3=5\neq 0$
Since first derivative does not equal to $0$ for both $z$, we conclude order $= 1$
Hence, the order is $1$ for $z=ln(4)+i(2k\pi)\ \text{and}\ z=i(\pi+2k\pi)$

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##### Quiz 4 / Quiz-4-6101 C
« on: October 23, 2020, 08:37:43 AM »
Question: $$\int_\gamma sin(z)dz,$$
where $\gamma$ is any curve joining $i$ to $\pi$
Solution:\begin{align*}
&\int_\gamma sin(z)dz\\
&=\int_{i}^{\pi} sin(z)dz\\
&=-cos(z) \Big|_{i}^{\pi}\\
&=-(cos(\pi)-cos(i))\\
&=1+cos(i)
\end{align*}

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##### Quiz 2 / Quiz 2 Session 6101
« on: October 01, 2020, 04:46:51 PM »
Question: $$\sum_{n=1}^{\infty} \frac{1}{n} (\frac{1+i}{\sqrt{2}})^{n}$$
$$\lim_{n \to \infty} \frac{1}{n}(\frac{1+i}{\sqrt{2}})^{n} \\ =\lim_{n \to \infty} \frac{1}{n}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\ =\lim_{n \to \infty} \frac{1}{n}\cdot \lim_{n \to \infty}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\$$
Since $$\lim_{n \to \infty}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^{n} \\ =\lim_{n \to \infty}(\frac{\sqrt{2}\ (cos\frac{\pi}{4}+isin\frac{\pi}{4})}{\sqrt{2}})^{n}\\ =\lim_{n \to \infty}(cos \frac{n\pi}{4}+isin \frac{n\pi}{4}) \ \ (\text{By De Moivres Theorem})\\ =\text{DNE}\ \text{and it diverges}$$